Given two sorted arrays of size m and n respectively, you are tasked with finding the element that would be at the k’th position of the final sorted array.
Examples:
Input : Array 1 - 2 3 6 7 9
Array 2 - 1 4 8 10
k = 5
Output : 6
Explanation: The final sorted array would be -
1, 2, 3, 4, 6, 7, 8, 9, 10
The 5th element of this array is 6.
Input : Array 1 - 100 112 256 349 770
Array 2 - 72 86 113 119 265 445 892
k = 7
Output : 256
Explanation: Final sorted array is -
72, 86, 100, 112, 113, 119, 256, 265, 349, 445, 770, 892
7th element of this array is 256.
Basic Approach
Since we are given two sorted arrays, we can use the merging technique to get the final merged array. From this, we simply go to the k’th index.
C++
// Program to find kth element from two sorted arrays#include <iostream>using namespace std;int kth(int arr1[], int arr2[], int m, int n, int k){ int sorted1[m + n]; int i = 0, j = 0, d = 0; while (i < m && j < n) { if (arr1[i] < arr2[j]) sorted1[d++] = arr1[i++]; else sorted1[d++] = arr2[j++]; } while (i < m) sorted1[d++] = arr1[i++]; while (j < n) sorted1[d++] = arr2[j++]; return sorted1[k - 1];}// Driver Codeint main(){ int arr1[5] = {2, 3, 6, 7, 9}; int arr2[4] = {1, 4, 8, 10}; int k = 5; cout << kth(arr1, arr2, 5, 4, k); return 0;} |
Java
// Java Program to find kth element // from two sorted arraysclass Main{ static int kth(int arr1[], int arr2[], int m, int n, int k) { int[] sorted1 = new int[m + n]; int i = 0, j = 0, d = 0; while (i < m && j < n) { if (arr1[i] < arr2[j]) sorted1[d++] = arr1[i++]; else sorted1[d++] = arr2[j++]; } while (i < m) sorted1[d++] = arr1[i++]; while (j < n) sorted1[d++] = arr2[j++]; return sorted1[k - 1]; } // Driver Code public static void main (String[] args) { int arr1[] = {2, 3, 6, 7, 9}; int arr2[] = {1, 4, 8, 10}; int k = 5; System.out.print(kth(arr1, arr2, 5, 4, k)); }}/* This code is contributed by Harsh Agarwal */ |
Python3
# Program to find kth element# from two sorted arraysdef kth(arr1, arr2, m, n, k): sorted1 = [0] * (m + n) i = 0 j = 0 d = 0 while (i < m and j < n): if (arr1[i] < arr2[j]): sorted1[d] = arr1[i] i += 1 else: sorted1[d] = arr2[j] j += 1 d += 1 while (i < m): sorted1[d] = arr1[i] d += 1 i += 1 while (j < n): sorted1[d] = arr2[j] d += 1 j += 1 return sorted1[k - 1]# Driver codearr1 = [2, 3, 6, 7, 9]arr2 = [1, 4, 8, 10]k = 5print(kth(arr1, arr2, 5, 4, k))# This code is contributed by Smitha Dinesh Semwal |
C#
// C# Program to find kth element// from two sorted arraysclass GFG { static int kth(int[] arr1, int[] arr2, int m, int n, int k) { int[] sorted1 = new int[m + n]; int i = 0, j = 0, d = 0; while (i < m && j < n) { if (arr1[i] < arr2[j]) sorted1[d++] = arr1[i++]; else sorted1[d++] = arr2[j++]; } while (i < m) sorted1[d++] = arr1[i++]; while (j < n) sorted1[d++] = arr2[j++]; return sorted1[k - 1]; } // Driver Code static void Main() { int[] arr1 = { 2, 3, 6, 7, 9 }; int[] arr2 = { 1, 4, 8, 10 }; int k = 5; System.Console.WriteLine(kth(arr1, arr2, 5, 4, k)); }}// This code is contributed by mits |
PHP
<?php // Program to find kth// element from two // sorted arraysfunction kth($arr1, $arr2, $m, $n, $k){ $sorted1[$m + $n] = 0; $i = 0; $j = 0; $d = 0; while ($i < $m && $j < $n) { if ($arr1[$i] < $arr2[$j]) $sorted1[$d++] = $arr1[$i++]; else $sorted1[$d++] = $arr2[$j++]; } while ($i < $m) $sorted1[$d++] = $arr1[$i++]; while ($j < $n) $sorted1[$d++] = $arr2[$j++]; return $sorted1[$k - 1];}// Driver Code$arr1 = array(2, 3, 6, 7, 9);$arr2 = array(1, 4, 8, 10);$k = 5;echo kth($arr1, $arr2, 5, 4, $k);// This code is contributed // by ChitraNayal?> |
Javascript
<script>// JavaScript Program to find kth element // from two sorted arrays function kth(arr1 , arr2 , m , n , k) { var sorted1 = Array(m + n).fill(0); var i = 0, j = 0, d = 0; while (i < m && j < n) { if (arr1[i] < arr2[j]) sorted1[d++] = arr1[i++]; else sorted1[d++] = arr2[j++]; } while (i < m) sorted1[d++] = arr1[i++]; while (j < n) sorted1[d++] = arr2[j++]; return sorted1[k - 1]; } // Driver Code var arr1 = [ 2, 3, 6, 7, 9 ]; var arr2 = [ 1, 4, 8, 10 ]; var k = 5; document.write(kth(arr1, arr2, 5, 4, k));// This code contributed by umadevi9616 </script> |
Output
6
Time Complexity: O(n)
Auxiliary Space : O(m + n)
Space Optimized Version of above approach: We can avoid the use of extra array.
C++
// C++ program to find kth element // from two sorted arrays#include <bits/stdc++.h>using namespace std;int find(int A[], int B[], int m, int n, int k_req) { int k = 0, i = 0, j = 0; // Keep taking smaller of the current // elements of two sorted arrays and // keep incrementing k while(i < m && j < n) { if(A[i] < B[j]) { k++; if(k == k_req) return A[i]; i++; } else { k++; if(k == k_req) return B[j]; j++; } } // If array B[] is completely traversed while(i < m) { k++; if(k == k_req) return A[i]; i++; } // If array A[] is completely traversed while(j < n) { k++; if(k == k_req) return B[j]; j++; }} // Driver Codeint main(){ int A[5] = { 2, 3, 6, 7, 9 }; int B[4] = { 1, 4, 8, 10 }; int k = 5; cout << find(A, B, 5, 4, k); return 0;} // This code is contributed by Sreejith S |
Java
import java.io.*;class GFG { public static int find(int A[], int B[], int m, int n, int k_req) { int k = 0, i = 0, j = 0; // Keep taking smaller of the current // elements of two sorted arrays and // keep incrementing k while (i < m && j < n) { if (A[i] < B[j]) { k++; if (k == k_req) return A[i]; i++; } else { k++; if (k == k_req) return B[j]; j++; } } // If array B[] is completely traversed while (i < m) { k++; if (k == k_req) return A[i]; i++; } // If array A[] is completely traversed while (j < n) { k++; if (k == k_req) return B[j]; j++; } return -1; } // Driver Code public static void main(String[] args) { int[] A = { 2, 3, 6, 7, 9 }; int[] B = { 1, 4, 8, 10 }; int k = 5; System.out.println(find(A, B, 5, 4, k)); }} |
Python3
# Python3 Program to find kth element # from two sorted arraysdef find(A, B, m, n, k_req): i, j, k = 0, 0, 0 # Keep taking smaller of the current # elements of two sorted arrays and # keep incrementing k while i < len(A) and j < len(B): if A[i] < B[j]: k += 1 if k == k_req: return A[i] i += 1 else: k += 1 if k == k_req: return B[j] j += 1 # If array B[] is completely traversed while i < len(A): k += 1 if k == k_req: return A[i] i += 1 # If array A[] is completely traversed while j < len(B): k += 1 if k == k_req: return B[j] j += 1# driver codeA = [2, 3, 6, 7, 9]B = [1, 4, 8, 10]k = 5;print(find(A, B, 5, 4, k))# time complexity of O(k) |
C#
// C# program to find kth element // from two sorted arraysusing System;public class GFG{ public static int find(int[] A, int[] B, int m, int n,int k_req) { int k = 0, i = 0, j = 0; // Keep taking smaller of the current // elements of two sorted arrays and // keep incrementing k while (i < m && j < n) { if (A[i] < B[j]) { k++; if (k == k_req) return A[i]; i++; } else { k++; if (k == k_req) return B[j]; j++; } } // If array B[] is completely traversed while (i < m) { k++; if (k == k_req) return A[i]; i++; } // If array A[] is completely traversed while (j < n) { k++; if (k == k_req) return B[j]; j++; } return -1; } // Driver Code static public void Main (){ int[] A = { 2, 3, 6, 7, 9 }; int[] B = { 1, 4, 8, 10 }; int k = 5; Console.WriteLine(find(A, B, 5, 4, k)); }}// This code is contributed by rag2127 |
Javascript
<script> function find(A, B, m, n, k_req) { let k = 0, i = 0, j = 0; // Keep taking smaller of the current // elements of two sorted arrays and // keep incrementing k while (i < m && j < n) { if (A[i] < B[j]) { k++; if (k == k_req) return A[i]; i++; } else { k++; if (k == k_req) return B[j]; j++; } } // If array B[] is completely traversed while (i < m) { k++; if (k == k_req) return A[i]; i++; } // If array A[] is completely traversed while (j < n) { k++; if (k == k_req) return B[j]; j++; } return -1; } // Driver Code let A = [ 2, 3, 6, 7, 9 ]; let B = [ 1, 4, 8, 10 ]; let k = 5; document.write(find(A, B, 5, 4, k)); // This code is contributed by Dharanendra L V. </script> |
Output
6
Time Complexity: O(k)
Auxiliary Space: O(1)
Divide And Conquer Approach 1:
While the previous method works, can we make our algorithm more efficient? The answer is yes. By using a divide and conquer approach, similar to the one used in binary search, we can attempt to find the k’th element in a more efficient way.
- Compare the middle elements of arrays arr1 and arr2, let us call these indices mid1 and mid2 respectively. Let us assume arr1[mid1] > arr2[mid2], then clearly the elements after mid2 cannot be the required element. Set the last element of arr2 to be arr2[mid2].
In this way, define a new subproblem with half the size of one of the arrays.
C++
// Program to find k-th element from two sorted arrays#include <iostream>using namespace std;int kth(int *arr1, int *arr2, int *end1, int *end2, int k){ if (arr1 == end1) return arr2[k]; if (arr2 == end2) return arr1[k]; int mid1 = (end1 - arr1) / 2; int mid2 = (end2 - arr2) / 2; if (mid1 + mid2 < k) { if (arr1[mid1] > arr2[mid2]) return kth(arr1, arr2 + mid2 + 1, end1, end2, k - mid2 - 1); else return kth(arr1 + mid1 + 1, arr2, end1, end2, k - mid1 - 1); } else { if (arr1[mid1] > arr2[mid2]) return kth(arr1, arr2, arr1 + mid1, end2, k); else return kth(arr1, arr2, end1, arr2 + mid2, k); }}int main(){ int arr1[5] = {2, 3, 6, 7, 9}; int arr2[4] = {1, 4, 8, 10}; int k = 5; cout << kth(arr1, arr2, arr1 + 5, arr2 + 4, k - 1); return 0;} |
Java
/*package whatever //do not write package name here */import java.io.*;import java.util.Arrays;public class MainClass { // kth function to find kth element in the merged array static int kth(int[] arr1, int[] arr2, int n, int m, int k) { // base cases if (n == 1 || m == 1) { // if one array is empty or only has one element, find the kth element in the other array if (m == 1) { int[] temp = arr1; arr1 = arr2; arr2 = temp; m = n; } if (k == 1) { return Math.min(arr1[0], arr2[0]); } else if (k == m + 1) { return Math.max(arr1[0], arr2[0]); } else { if (arr2[k - 1] < arr1[0]) { return arr2[k - 1]; } else { return Math.max(arr1[0], arr2[k - 2]); } } } int mid1 = (n - 1) / 2; int mid2 = (m - 1) / 2; if (mid1 + mid2 + 1 < k) { // if k is greater than the sum of midpoints, discard the smaller half of the arrays and recurse if (arr1[mid1] < arr2[mid2]) { return kth( Arrays.copyOfRange(arr1, mid1 + 1, n), arr2, n - (mid1 + 1), m, k - (mid1 + 1) ); } else { return kth( arr1, Arrays.copyOfRange(arr2, mid2 + 1, m), n, m - (mid2 + 1), k - (mid2 + 1) ); } } else { if (arr1[mid1] < arr2[mid2]) { return kth( arr1, Arrays.copyOfRange(arr2, 0, mid2 + 1), n, mid2 + 1, k ); } else { return kth( Arrays.copyOfRange(arr1, 0, mid1 + 1), arr2, mid1 + 1, m, k ); } } } public static void main(String[] args) { int[] arr1 = {2, 3, 6, 7, 9}; int[] arr2 = {1, 4, 8, 10}; int k = 5; System.out.println(kth(arr1, arr2, 5, 4, k)); }} |
Python3
# Python program to find k-th element from two sorted arraysdef kth(arr1, arr2, n, m, k): if n == 1 or m == 1: if m == 1: arr2, arr1 = arr1, arr2 m = n if k == 1: return min(arr1[0], arr2[0]) elif k == m + 1: return max(arr1[0], arr2[0]) else: if arr2[k - 1] < arr1[0]: return arr2[k - 1] else: return max(arr1[0], arr2[k - 2]) mid1 = (n - 1)//2 mid2 = (m - 1)//2 if mid1+mid2+1 < k: if arr1[mid1] < arr2[mid2]: return kth(arr1[mid1 + 1:], arr2, n - mid1 - 1, m, k - mid1 - 1) else: return kth(arr1, arr2[mid2 + 1:], n, m - mid2 - 1, k - mid2 - 1) else: if arr1[mid1] < arr2[mid2]: return kth(arr1, arr2[:mid2 + 1], n, mid2 + 1, k) else: return kth(arr1[:mid1 + 1], arr2, mid1 + 1, m, k)if __name__ == "__main__": arr1 = [2, 3, 6, 7, 9] arr2 = [1, 4, 8, 10] k = 5 print(kth(arr1, arr2, 5, 4, k))# This code is contributed by harshitkap00r |
C#
using System;public class Program { // kth function to find kth element in the merged array static int kth(int[] arr1, int[] arr2, int n, int m, int k) { // base cases if (n == 1 || m == 1) { // if one array is empty or only has one element, find the kth element in the other array if (m == 1) { int[] temp = arr1; arr1 = arr2; arr2 = temp; m = n; } if (k == 1) { return Math.Min(arr1[0], arr2[0]); } else if (k == m + 1) { return Math.Max(arr1[0], arr2[0]); } else { if (arr2[k - 1] < arr1[0]) { return arr2[k - 1]; } else { return Math.Max(arr1[0], arr2[k - 2]); } } } int mid1 = (n - 1) / 2; int mid2 = (m - 1) / 2; if (mid1 + mid2 + 1 < k) { // if k is greater than the sum of midpoints, discard the smaller half of the arrays and recurse if (arr1[mid1] < arr2[mid2]) { return kth( arr1[(mid1 + 1)..], arr2, n - (mid1 + 1), m, k - (mid1 + 1) ); } else { return kth( arr1, arr2[(mid2 + 1)..], n, m - (mid2 + 1), k - (mid2 + 1) ); } } else { if (arr1[mid1] < arr2[mid2]) { return kth( arr1, arr2[..(mid2 + 1)], n, mid2 + 1, k ); } else { return kth( arr1[..(mid1 + 1)], arr2, mid1 + 1, m, k ); } } } public static void Main() { int[] arr1 = { 2, 3, 6, 7, 9 }; int[] arr2 = { 1, 4, 8, 10 }; int k = 5; Console.WriteLine(kth(arr1, arr2, 5, 4, k)); }} |
Javascript
function kth(arr1, start1, end1, arr2, start2, end2, k) { if (start1 === end1) { return arr2[start2 + k]; } if (start2 === end2) { return arr1[start1 + k]; } let mid1 = Math.floor((start1 + end1) / 2); let mid2 = Math.floor((start2 + end2) / 2); if (mid1 - start1 + mid2 - start2 < k) { if (arr1[mid1] > arr2[mid2]) { return kth(arr1, start1, end1, arr2, mid2 + 1, end2, k - (mid2 - start2) - 1); } else { return kth(arr1, mid1 + 1, end1, arr2, start2, end2, k - (mid1 - start1) - 1); } } else { if (arr1[mid1] > arr2[mid2]) { return kth(arr1, start1, mid1, arr2, start2, end2, k); } else { return kth(arr1, start1, end1, arr2, start2, mid2, k); } }}let arr1 = [2, 3, 6, 7, 9];let arr2 = [1, 4, 8, 10];let k = 5;console.log(kth(arr1, 0, arr1.length, arr2, 0, arr2.length, k - 1)); |
Output
6
Note that in the above code, k is 0 indexed, which means if we want a k that’s 1 indexed, we have to subtract 1 when passing it to the function.
Time Complexity: O(log n + log m)
Auxiliary Space: O(logn + logm)
Divide And Conquer Approach 2:
While the above implementation is very efficient, we can still get away with making it more efficient. Instead of dividing the array into segments of n / 2 and m / 2 and then recursing, we can divide them both by k / 2 and recurse.
The below implementation displays this.
Explanation: Instead of comparing the middle element of the arrays, we compare the k / 2nd element. Let arr1 and arr2 be the arrays. Now, if arr1[k / 2] arr1[1] New subproblem: Array 1 - 6 7 9 Array 2 - 1 4 8 10 k = 5 - 2 = 3 floor(k / 2) = 1 arr1[1] = 6 arr2[1] = 1 arr1[1] > arr2[1] New subproblem: Array 1 - 6 7 9 Array 2 - 4 8 10 k = 3 - 1 = 2 floor(k / 2) = 1 arr1[1] = 6 arr2[1] = 4 arr1[1] > arr2[1] New subproblem: Array 1 - 6 7 9 Array 2 - 8 10 k = 2 - 1 = 1 Now, we directly compare first elements, since k = 1. arr1[1] < arr2[1] Hence, arr1[1] = 6 is the answer.
C++
// C++ Program to find kth element from two sorted arrays#include <iostream>using namespace std;int kth(int arr1[], int arr2[], int m, int n, int k, int st1 = 0, int st2 = 0){ // In case we have reached end of array 1 if (st1 == m) return arr2[st2 + k - 1]; // In case we have reached end of array 2 if (st2 == n) return arr1[st1 + k - 1]; // k should never reach 0 or exceed sizes // of arrays if (k == 0 || k > (m - st1) + (n - st2)) return -1; // Compare first elements of arrays and return if (k == 1) return (arr1[st1] < arr2[st2]) ? arr1[st1] : arr2[st2]; int curr = k / 2; // Size of array 1 is less than k / 2 if (curr - 1 >= m - st1) { // Last element of array 1 is not kth // We can directly return the (k - m)th // element in array 2 if (arr1[m - 1] < arr2[st2 + curr - 1]) return arr2[st2 + (k - (m - st1) - 1)]; else return kth(arr1, arr2, m, n, k - curr, st1, st2 + curr); } // Size of array 2 is less than k / 2 if (curr-1 >= n-st2) { if (arr2[n - 1] < arr1[st1 + curr - 1]) return arr1[st1 + (k - (n - st2) - 1)]; else return kth(arr1, arr2, m, n, k - curr, st1 + curr, st2); } else { // Normal comparison, move starting index // of one array k / 2 to the right if (arr1[curr + st1 - 1] < arr2[curr + st2 - 1]) return kth(arr1, arr2, m, n, k - curr, st1 + curr, st2); else return kth(arr1, arr2, m, n, k - curr, st1, st2 + curr); }}// Driver codeint main(){ int arr1[5] = {2, 3, 6, 7, 9}; int arr2[4] = {1, 4, 8, 10}; int k = 5; cout << kth(arr1, arr2, 5, 4, k); return 0;} |
Java
// Java Program to find kth element from two sorted arraysclass GFG { static int kth(int arr1[], int arr2[], int m, int n, int k, int st1, int st2) { // In case we have reached end of array 1 if (st1 == m) { return arr2[st2 + k - 1]; } // In case we have reached end of array 2 if (st2 == n) { return arr1[st1 + k - 1]; } // k should never reach 0 or exceed sizes // of arrays if (k == 0 || k > (m - st1) + (n - st2)) { return -1; } // Compare first elements of arrays and return if (k == 1) { return (arr1[st1] < arr2[st2]) ? arr1[st1] : arr2[st2]; } int curr = k / 2; // Size of array 1 is less than k / 2 if (curr - 1 >= m - st1) { // Last element of array 1 is not kth // We can directly return the (k - m)th // element in array 2 if (arr1[m - 1] < arr2[st2 + curr - 1]) { return arr2[st2 + (k - (m - st1) - 1)]; } else { return kth(arr1, arr2, m, n, k - curr, st1, st2 + curr); } } // Size of array 2 is less than k / 2 if (curr - 1 >= n - st2) { if (arr2[n - 1] < arr1[st1 + curr - 1]) { return arr1[st1 + (k - (n - st2) - 1)]; } else { return kth(arr1, arr2, m, n, k - curr, st1 + curr, st2); } } else // Normal comparison, move starting index // of one array k / 2 to the right if (arr1[curr + st1 - 1] < arr2[curr + st2 - 1]) { return kth(arr1, arr2, m, n, k - curr, st1 + curr, st2); } else { return kth(arr1, arr2, m, n, k - curr, st1, st2 + curr); } } // Driver code public static void main(String[] args) { int arr1[] = {2, 3, 6, 7, 9}; int arr2[] = {1, 4, 8, 10}; int k = 5; int st1 = 0, st2 = 0; System.out.println(kth(arr1, arr2, 5, 4, k, st1, st2)); }} // This code is contributed by 29AjayKumar |
Python3
# Python3 program to find kth element from# two sorted arraysdef kth(arr1, arr2, m, n, k, st1 = 0, st2 = 0): # In case we have reached end of array 1 if (st1 == m): return arr2[st2 + k - 1] # In case we have reached end of array 2 if (st2 == n): return arr1[st1 + k - 1] # k should never reach 0 or exceed sizes # of arrays if (k == 0 or k > (m - st1) + (n - st2)): return -1 # Compare first elements of arrays and return if (k == 1): if(arr1[st1] < arr2[st2]): return arr1[st1] else: return arr2[st2] curr = int(k / 2) # Size of array 1 is less than k / 2 if(curr - 1 >= m - st1): # Last element of array 1 is not kth # We can directly return the (k - m)th # element in array 2 if (arr1[m - 1] < arr2[st2 + curr - 1]): return arr2[st2 + (k - (m - st1) - 1)] else: return kth(arr1, arr2, m, n, k - curr, st1, st2 + curr) # Size of array 2 is less than k / 2 if (curr - 1 >= n - st2): if (arr2[n - 1] < arr1[st1 + curr - 1]): return arr1[st1 + (k - (n - st2) - 1)] else: return kth(arr1, arr2, m, n, k - curr,st1 + curr, st2) else: # Normal comparison, move starting index # of one array k / 2 to the right if (arr1[curr + st1 - 1] < arr2[curr + st2 - 1]): return kth(arr1, arr2, m, n, k - curr, st1 + curr, st2) else: return kth(arr1, arr2, m, n, k - curr, st1, st2 + curr)# Driver codearr1 = [ 2, 3, 6, 7, 9 ]arr2 = [ 1, 4, 8, 10 ]k = 5print(kth(arr1, arr2, 5, 4, k))# This code is contributed by avanitrachhadiya2155 |
C#
// C# Program to find kth element from two sorted arraysusing System;class GFG { static int kth(int []arr1, int []arr2, int m, int n, int k, int st1, int st2) { // In case we have reached end of array 1 if (st1 == m) { return arr2[st2 + k - 1]; } // In case we have reached end of array 2 if (st2 == n) { return arr1[st1 + k - 1]; } // k should never reach 0 or exceed sizes // of arrays if (k == 0 || k > (m - st1) + (n - st2)) { return -1; } // Compare first elements of arrays and return if (k == 1) { return (arr1[st1] < arr2[st2]) ? arr1[st1] : arr2[st2]; } int curr = k / 2; // Size of array 1 is less than k / 2 if (curr - 1 >= m - st1) { // Last element of array 1 is not kth // We can directly return the (k - m)th // element in array 2 if (arr1[m - 1] < arr2[st2 + curr - 1]) { return arr2[st2 + (k - (m - st1) - 1)]; } else { return kth(arr1, arr2, m, n, k - curr, st1, st2 + curr); } } // Size of array 2 is less than k / 2 if (curr - 1 >= n - st2) { if (arr2[n - 1] < arr1[st1 + curr - 1]) { return arr1[st1 + (k - (n - st2) - 1)]; } else { return kth(arr1, arr2, m, n, k - curr, st1 + curr, st2); } } else // Normal comparison, move starting index // of one array k / 2 to the right if (arr1[curr + st1 - 1] < arr2[curr + st2 - 1]) { return kth(arr1, arr2, m, n, k - curr, st1 + curr, st2); } else { return kth(arr1, arr2, m, n, k - curr, st1, st2 + curr); } } // Driver code public static void Main(String[] args) { int []arr1 = {2, 3, 6, 7, 9}; int []arr2 = {1, 4, 8, 10}; int k = 5; int st1 = 0, st2 = 0; Console.WriteLine(kth(arr1, arr2, 5, 4, k, st1, st2)); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// javascript Program to find kth element from two sorted arrays function kth(arr1 , arr2 , m , n , k , st1 , st2){ // In case we have reached end of array 1 if (st1 == m) { return arr2[st2 + k - 1]; } // In case we have reached end of array 2 if (st2 == n) { return arr1[st1 + k - 1]; } // k should never reach 0 or exceed sizes // of arrays if (k == 0 || k > (m - st1) + (n - st2)) { return -1; } // Compare first elements of arrays and return if (k == 1) { return (arr1[st1] < arr2[st2]) ? arr1[st1] : arr2[st2]; } var curr = parseInt(k / 2); // Size of array 1 is less than k / 2 if (curr - 1 >= m - st1) { // Last element of array 1 is not kth // We can directly return the (k - m)th // element in array 2 if (arr1[m - 1] < arr2[st2 + curr - 1]) { return arr2[st2 + (k - (m - st1) - 1)]; } else { return kth(arr1, arr2, m, n, k - curr, st1, st2 + curr); } } // Size of array 2 is less than k / 2 if (curr - 1 >= n - st2) { if (arr2[n - 1] < arr1[st1 + curr - 1]) { return arr1[st1 + (k - (n - st2) - 1)]; } else { return kth(arr1, arr2, m, n, k - curr, st1 + curr, st2); } } else // Normal comparison, move starting index // of one array k / 2 to the right if (arr1[curr + st1 - 1] < arr2[curr + st2 - 1]) { return kth(arr1, arr2, m, n, k - curr, st1 + curr, st2); } else { return kth(arr1, arr2, m, n, k - curr, st1, st2 + curr); } } // Driver code var arr1 = [ 2, 3, 6, 7, 9 ]; var arr2 = [ 1, 4, 8, 10 ]; var k = 5; var st1 = 0, st2 = 0; document.write(kth(arr1, arr2, 5, 4, k, st1, st2));// This code is contributed by gauravrajput1</script> |
Output
6
Time Complexity: O(log k)
Auxiliary Space: O(logk)
Now, k can take a maximum value of m + n. This means that log k can be in the worst case, log(m + n). Logm + logn = log(mn) by properties of logarithms, and when m, n > 2, log(m + n) < log(mn). Thus this algorithm slightly outperforms the previous algorithm. Also, see another simple implemented log k approach suggested by Raj Kumar.
C++
// C++ Program to find kth // element from two sorted arrays// Time Complexity: O(log k)#include <iostream>using namespace std;int kth(int arr1[], int m, int arr2[], int n, int k){ if (k > (m + n) || k < 1) return -1; // let m <= n if (m > n) return kth(arr2, n, arr1, m, k); // Check if arr1 is empty returning // k-th element of arr2 if (m == 0) return arr2[k - 1]; // Check if k = 1 return minimum of // first two elements of both // arrays if (k == 1) return min(arr1[0], arr2[0]); // Now the divide and conquer part int i = min(m, k / 2), j = min(n, k / 2); if (arr1[i - 1] > arr2[j - 1]) // Now we need to find only // k-j th element since we // have found out the lowest j return kth(arr1, m, arr2 + j, n - j, k - j); else // Now we need to find only // k-i th element since we // have found out the lowest i return kth(arr1 + i, m - i, arr2, n, k - i);}// Driver codeint main(){ int arr1[5] = { 2, 3, 6, 7, 9 }; int arr2[4] = { 1, 4, 8, 10 }; int m = sizeof(arr1) / sizeof(arr1[0]); int n = sizeof(arr2) / sizeof(arr2[0]); int k = 5; int ans = kth(arr1, m, arr2, n, k); if (ans == -1) cout << "Invalid query"; else cout << ans; return 0;}// This code is contributed by Raj Kumar |
Java
// Java Program to find kth element // from two sorted arrays// Time Complexity: O(log k)import java.util.Arrays;class Gfg { static int kth(int arr1[], int m, int arr2[], int n, int k) { if (k > (m + n) || k < 1) return -1; // let m > n if (m > n) return kth(arr2, n, arr1, m, k); // Check if arr1 is empty returning // k-th element of arr2 if (m == 0) return arr2[k - 1]; // Check if k = 1 return minimum of first // two elements of both arrays if (k == 1) return Math.min(arr1[0], arr2[0]); // Now the divide and conquer part int i = Math.min(m, k / 2); int j = Math.min(n, k / 2); if (arr1[i - 1] > arr2[j - 1]) { // Now we need to find only k-j th element // since we have found out the lowest j int temp[] = Arrays.copyOfRange(arr2, j, n); return kth(arr1, m, temp, n - j, k - j); } // Now we need to find only k-i th element // since we have found out the lowest i int temp[] = Arrays.copyOfRange(arr1, i, m); return kth(temp, m - i, arr2, n, k - i); } // Driver code public static void main(String[] args) { int arr1[] = { 2, 3, 6, 7, 9 }; int arr2[] = { 1, 4, 8, 10 }; int m = arr1.length; int n = arr2.length; int k = 5; int ans = kth(arr1, m, arr2, n, k); if (ans == -1) System.out.println("Invalid query"); else System.out.println(ans); }}// This code is contributed by Vivek Kumar Singh |
Python3
# Python3 Program to find kth element from two# sorted arrays. Time Complexity: O(log k)def kth(arr1, m, arr2, n, k): if (k > (m + n) or k < 1): return -1 # Let m <= n if (m > n): return kth(arr2, n, arr1, m, k) # Check if arr1 is empty returning # k-th element of arr2 if (m == 0): return arr2[k - 1] # Check if k = 1 return minimum of # first two elements of both # arrays if (k == 1): return min(arr1[0], arr2[0]) # Now the divide and conquer part i = min(m, k // 2) j = min(n, k // 2) if (arr1[i - 1] > arr2[j - 1]): # Now we need to find only # k-j th element since we # have found out the lowest j return kth(arr1, m, arr2[j:], n - j, k - j) else: # Now we need to find only # k-i th element since we # have found out the lowest i return kth(arr1[i:], m - i, arr2, n, k - i)# Driver codearr1 = [ 2, 3, 6, 7, 9 ]arr2 = [ 1, 4, 8, 10 ]m = len(arr1)n = len(arr2)k = 5ans = kth(arr1, m, arr2, n, k)if (ans == -1): print("Invalid query")else: print(ans)# This code is contributed by Shubham Singh |
C#
// C# Program to find kth element // from two sorted arrays// Time Complexity: O(log k)using System;using System.Collections.Generic;public class GFG{ static int kth(int[] arr1, int m, int[] arr2, int n, int k) { if (k > (m + n) || k < 1) return -1; // let m > n if (m > n) return kth(arr2, n, arr1, m, k); // Check if arr1 is empty returning // k-th element of arr2 if (m == 0) return arr2[k - 1]; // Check if k = 1 return minimum of first // two elements of both arrays if (k == 1) return Math.Min(arr1[0], arr2[0]); // Now the divide and conquer part int i = Math.Min(m, k / 2); int j = Math.Min(n, k / 2); if (arr1[i - 1] > arr2[j - 1]) { // Now we need to find only k-j th element // since we have found out the lowest j int[] temp = new int[n - j]; Array.Copy(arr2, j, temp, 0, n-j); return kth(arr1, m, temp, n - j, k - j); } // Now we need to find only k-i th element // since we have found out the lowest i int[] temp1 = new int[m-i]; Array.Copy(arr1, i, temp1, 0, m-i); return kth(temp1, m - i, arr2, n, k - i); } // Driver code public static void Main() { int[] arr1 = { 2, 3, 6, 7, 9 }; int[] arr2 = { 1, 4, 8, 10 }; int m = arr1.Length; int n = arr2.Length; int k = 5; int ans = kth(arr1, m, arr2, n, k); if (ans == -1) Console.WriteLine("Invalid query"); else Console.WriteLine(ans); }}// This code is contributed by Shubham Singh |
Javascript
<script>// Javascript program to illustrate time// complexity for Nested loopsfunction kth(arr1, m, arr2, n, k){ if (k > (m + n) || k < 1){ return -1; } // let m <= n if (m > n){ return kth(arr2, n, arr1, m, k); } // Check if arr1 is empty returning // k-th element of arr2 if (m == 0){ return arr2[k - 1]; } // Check if k = 1 return minimum of // first two elements of both // arrays if (k == 1){ return Math.min(arr1[0], arr2[0]); } // Now the divide and conquer part let i = Math.min(m, parseInt(k / 2)); let j = Math.min(n, parseInt(k / 2)); if (arr1[i - 1] > arr2[j - 1]){ // Now we need to find only // k-j th element since we // have found out the lowest j let temp = arr2.slice(j,n) return kth(arr1, m, temp, n - j, k - j); } else{ // Now we need to find only // k-i th element since we // have found out the lowest i let temp = arr1.slice(i,m) return kth( temp, m - i, arr2, n, k - i); }}// Driver codelet arr1 = [ 2, 3, 6, 7, 9 ];let arr2 = [ 1, 4, 8, 10 ];let m = 5;let n = 4;let k = 5;let ans = kth(arr1, m, arr2, n, k);if (ans == -1){ document.write("Invalid query");}else{ document.write(ans);}// This code is contributed by Shubham Singh</script> |
Output
6
Time Complexity:O(log k)
Auxiliary Space: O(log k)
Another Approach: (Using Min Heap)
- Push the elements of both arrays to a priority queue (min-heap).
- Pop-out k-1 elements from the front.
- Element at the front of the priority queue is the required answer.
Below is the implementation of the above approach:
C++
// C++ Program to find kth // element from two sorted arrays#include <bits/stdc++.h>using namespace std;// Function to find K-th minint kth(int* a, int* b, int n, int m, int k){ // Declaring a min heap priority_queue<int, vector<int>, greater<int> > pq; // Pushing elements for // array a to min-heap for (int i = 0; i < n; i++) { pq.push(a[i]); } // Pushing elements for // array b to min-heap for (int i = 0; i < m; i++) { pq.push(b[i]); } // Popping-out K-1 elements while (k-- > 1) { pq.pop(); } return pq.top();}//Driver Codeint main(){ int arr1[5] = {2, 3, 6, 7, 9}; int arr2[4] = {1, 4, 8, 10}; int k = 5; cout << kth(arr1, arr2, 5, 4, k); return 0;}// This code is contributed by yashbeersingh42 |
Java
// Java Program to find kth element // from two sorted arraysimport java.util.*;class GFG { // Function to find K-th min static int kth(int a[], int b[], int n, int m, int k) { // Declaring a min heap PriorityQueue<Integer> pq = new PriorityQueue<>(); // Pushing elements for // array a to min-heap for (int i = 0; i < n; i++) { pq.offer(a[i]); } // Pushing elements for // array b to min-heap for (int i = 0; i < m; i++) { pq.offer(b[i]); } // Popping-out K-1 elements while (k-- > 1) { pq.remove(); } return pq.peek(); } // Driver Code public static void main(String[] args) { int arr1[] = { 2, 3, 6, 7, 9 }; int arr2[] = { 1, 4, 8, 10 }; int k = 5; System.out.print(kth(arr1, arr2, 5, 4, k)); }}// This code is contributed by yashbeersingh42 |
Python3
# Python Program to find kth element # from two sorted arrays# Function to find K-th mindef kth(a , b , n , m , k): # Declaring a min heap pq = []; # Pushing elements for # array a to min-heap for i in range(n): pq.append(a[i]); # Pushing elements for # array b to min-heap for i in range(m): pq.append(b[i]); pq = sorted(pq, reverse = True) # Popping-out K-1 elements while (k > 1): k -= 1; pq.pop(); return pq.pop();# Driver Codearr1 = [ 2, 3, 6, 7, 9 ];arr2 = [ 1, 4, 8, 10 ];k = 5;print(kth(arr1, arr2, 5, 4, k));# This code is contributed by Saurabh Jaiswal |
C#
// C# Program to find kth element // from two sorted arraysusing System;using System.Collections.Generic;public class GFG { // Function to find K-th min static int kth(int []a, int []b, int n, int m, int k) { // Declaring a min heap List<int> pq = new List<int>(); // Pushing elements for // array a to min-heap for (int i = 0; i < n; i++) { pq.Add(a[i]); } // Pushing elements for // array b to min-heap for (int i = 0; i < m; i++) { pq.Add(b[i]); } pq.Sort(); // Popping-out K-1 elements while (k-- > 1) { pq.RemoveAt(0); } return pq[0]; } // Driver Code public static void Main(String[] args) { int []arr1 = { 2, 3, 6, 7, 9 }; int []arr2 = { 1, 4, 8, 10 }; int k = 5; Console.Write(kth(arr1, arr2, 5, 4, k)); }}// This code is contributed by gauravrajput1 |
Javascript
<script>// javascript Program to find kth element // from two sorted arrays // Function to find K-th min function kth(a , b , n , m , k) { // Declaring a min heap var pq = []; // Pushing elements for // array a to min-heap for (i = 0; i < n; i++) { pq.push(a[i]); } // Pushing elements for // array b to min-heap for (i = 0; i < m; i++) { pq.push(b[i]); } pq.sort((a,b)=>b-a); // Popping-out K-1 elements while (k-- > 1) { pq.pop(); } return pq.pop(); } // Driver Code var arr1 = [ 2, 3, 6, 7, 9 ]; var arr2 = [ 1, 4, 8, 10 ]; var k = 5; document.write(kth(arr1, arr2, 5, 4, k));// This code is contributed by gauravrajput1 </script> |
Output
6
Time Complexity: O(N*logN)
Auxiliary Space: O(m+n)
Another Approach : (Using Upper Bound STL)
Given two sorted arrays of size m and n respectively, you are tasked with finding the element that would be at the k’th position of the final sorted array.
Examples :
Input : Array 1 – 2 3 6 7 9
Array 2 – 1 4 8 10
k = 5
Output : 6
Explanation: The final sorted array would be –
1, 2, 3, 4, 6, 7, 8, 9, 10
The 5th element of this array is 6, The 1st element of this array is 1. The thing to notice here is upper_bound(6) gives 5, upper_bound(4) gives 4 that is number of element equal to or less than the number we are giving as input to upper_bound().
Here is another example
Input : Array 1 – 100 112 256 349 770
Array 2 – 72 86 113 119 265 445 892
k = 7
Output : 256
Explanation: Final sorted array is –
72, 86, 100, 112, 113, 119, 256, 265, 349, 445, 770, 892
7th element of this array is 256.
Observation required :
The simplest method to solve this question is using upper_bound to check what is the position of a element in the sorted array. The upper_bound function return the pointer to element which is greater than the element we searched.
So to find the kth element we need to just find the element whose upper_bound() is 4. So again now we know what upper_bound() gives us we need 1 last observation to solve this question. If we have been given 2 arrays, We just need to the sum of upper_bound for the 2 arrays
Input : Array 1 – 2 3 6 7 9
Array 2 – 1 4 8 10
k = 5
Value of upper_bound for value(6) in array1 is 3 and for array 2 is 2. This give us a total of 5. which is the answer.
Algorithm :
- We take a mid between [L,R] using the formula mid = (L+R)/2.
- Check if the middle can be the kth element using upper_bound() function
- Find the sum of upper_bound() for both the arrays and if the sum is >= K, It’s a possible value of kth element.
- If sum is >= K then we assign R = mid – 1.
- else if sum <k then the current mid is too small and we assign L = mid+1.
- Repeat from top
- Return the smallest value found.
Here is the implementation for the optimized method :
C++
// C++ program to find the kth element#include <bits/stdc++.h>using namespace std;long long int maxN = 1e10; // the maximum value in the array possible.long long int kthElement(int arr1[], int arr2[], int n, int m, int k){ long long int left = 1, right = maxN; // The range of where ans can lie. long long int ans = 1e15; // We have to find min of all // the ans so take . // using binary search to check all possible values of // kth element while (left <= right) { long long int mid = (left + right) / 2; long long int up_cnt = upper_bound(arr1, arr1 + n, mid) - arr1; up_cnt += upper_bound(arr2, arr2 + m, mid) - arr2; if (up_cnt >= k) { ans = min(ans, mid); // find the min of all answers. right = mid - 1; // Try to find a smaller answer. } else left = mid + 1; // Current mid is too small so // shift right. } return ans;}// Driver codeint main(){ // Example 1 int n = 5, m = 7, k = 7; int arr1[n] = { 100, 112, 256, 349, 770 }; int arr2[m] = { 72, 86, 113, 119, 265, 445, 892 }; cout << kthElement(arr1, arr2, n, m, k) << endl; return 0;} |
Java
// Java program to find the kth elementimport java.util.*;class GFG{static long maxN = (long)1e10; // the maximum value in the array possible.static int upperBound(int[] a, int low, int high, long element){ while(low < high){ int middle = low + (high - low)/2; if(a[middle] > element) high = middle; else low = middle + 1; } return low;}static long kthElement(int arr1[], int arr2[], int n, int m, int k){ long left = 1, right = maxN; // The range of where ans can lie. long ans = (long)1e15; // We have to find min of all // the ans so take . // using binary search to check all possible values of // kth element while (left <= right) { long mid = (left + right) / 2; long up_cnt = upperBound(arr1,0, n, mid); up_cnt += upperBound(arr2, 0, m, mid); if (up_cnt >= k) { ans = Math.min(ans, mid); // find the min of all answers. right = mid - 1; // Try to find a smaller answer. } else left = mid + 1; // Current mid is too small so // shift right. } return ans;}// Driver codepublic static void main(String[] args){ // Example 1 int n = 5, m = 7, k = 7; int arr1[] = { 100, 112, 256, 349, 770 }; int arr2[] = { 72, 86, 113, 119, 265, 445, 892 }; System.out.print(kthElement(arr1, arr2, n, m, k) +"\n");}}// This code is contributed by gauravrajput1 |
Python3
# Python program to find the kth elementmaxN = 10**10 # the maximum value in the array possible.def upperBound(a, low, high, element): while(low < high): middle = low + (high - low)//2 if(a[middle] > element): high = middle else: low = middle + 1 return lowdef kthElement(arr1, arr2, n, m, k): left = 1 right = maxN # The range of where ans can lie. ans = 10**15 # We have to find min of all # the ans so take . # using binary search to check all possible values of # kth element while (left <= right): mid = (left + right) // 2 up_cnt = upperBound(arr1,0, n, mid) up_cnt += upperBound(arr2, 0, m, mid) if (up_cnt >= k): ans = min(ans, mid) # find the min of all answers. right= mid - 1 # Try to find a smaller answer. else: left = mid + 1 # Current mid is too small so # shift right. return ans# Driver code# Example 1n = 5m = 7k = 7arr1 = [100, 112, 256, 349, 770]arr2 = [72, 86, 113, 119, 265, 445, 892]print(kthElement(arr1, arr2, n, m, k))# This code is contributed by Shubham Singh |
C#
// C# program to find the kth elementusing System;public class GFG{ static long maxN = (long)1e10; // the maximum value in the array possible. static int upperBound(int[] a, int low, int high, long element) { while(low < high){ int middle = low + (high - low)/2; if(a[middle] > element) high = middle; else low = middle + 1; } return low; } static long kthElement(int[] arr1, int[] arr2, int n, int m, int k) { long left = 1, right = maxN; // The range of where ans can lie. long ans = (long)1e15; // We have to find min of all // the ans so take . // using binary search to check all possible values of // kth element while (left <= right) { long mid = (left + right) / 2; long up_cnt = upperBound(arr1,0, n, mid); up_cnt += upperBound(arr2, 0, m, mid); if (up_cnt >= k) { ans = Math.Min(ans, mid); // find the min of all answers. right = mid - 1; // Try to find a smaller answer. } else left = mid + 1; // Current mid is too small so // shift right. } return ans; } // Driver code static public void Main (){ // Example 1 int n = 5, m = 7, k = 7; int[] arr1 = { 100, 112, 256, 349, 770 }; int[] arr2 = { 72, 86, 113, 119, 265, 445, 892 }; Console.Write(kthElement(arr1, arr2, n, m, k) +"\n"); }}// This code is contributed by SHubham Singh |
Javascript
<script>// Javascript program to find the kth elementvar maxN = 10000000000; // the maximum value in the array possible.function upperBound(a, low, high, element){ while(low < high){ var middle = parseInt(low + (high - low)/2); if(a[middle] > element) high = middle; else low = middle + 1; } return low;}function kthElement(arr1, arr2, n, m, k){ var left = 1 var right = maxN; // The range of where ans can lie. var ans = 1000000000000000; // We have to find min of all // the ans so take . // using binary search to check all possible values of // kth element while (left <= right) { var mid = parseInt((left + right) / 2); var up_cnt = upperBound(arr1,0, n, mid); up_cnt += upperBound(arr2, 0, m, mid); if (up_cnt >= k) { ans = Math.min(ans, mid); // find the min of all answers. right = mid - 1; // Try to find a smaller answer. } else left = mid + 1; // Current mid is too small so // shift right. } return ans;}// Driver code// Example 1var n = 5, m = 7, k = 7;var arr1 = [ 100, 112, 256, 349, 770 ];var arr2 = [ 72, 86, 113, 119, 265, 445, 892 ];document.write(kthElement(arr1, arr2, n, m, k));// This code is contributed by Shubham Singh</script> |
Output
256
Time Complexity : O( Log( maxN ).log( N+M ) )
Auxiliary Space : O( 1 )
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