Given an integer K and a matrix of size N x M, where each matrix element is equal to the product of its indices(i * j), the task is to find the Kth Smallest element in the given Matrix.
Examples:
Input: N = 2, M = 3, K = 5
Output: 4
Explanation:
The matrix possible for given dimensions is {{1, 2, 3}, {2, 4, 6}}
Sorted order of the matrix elements: {1, 2, 2, 3, 4, 6}
Therefore, the 5th smallest element is 4.
Input: N = 1, M = 10, K = 8
Output: 8
Explanation:
The matrix possible for given dimensions is {{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}}
Therefore, the 8th smallest element is 8.
Naive Approach: The simplest approach is to store all the elements of the matrix in an array and then find the Kth smallest element by sorting the array.
Below is the implementation of the approach:
C++
// C++ code for the approach#include <bits/stdc++.h>using namespace std;// Function to find Kth smallest element in matrixint kthSmallest(vector<vector<int>> &mat, int n, int m, int k) { // Create a temporary array int arr[n * m]; // Fill the temporary array int index = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { arr[index++] = mat[i][j]; } } // Sort the temporary array sort(arr, arr + n * m); // Return Kth smallest element return arr[k - 1];}// Driver codeint main() { // Given matrix int n = 2, m = 3, k = 5; vector<vector<int>> mat(n, vector<int>(m,0)); for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { mat[i][j] = (i + 1) * (j + 1); } } // Function call cout << kthSmallest(mat, n, m, k) << endl; return 0;} |
Java
// Java code for the approachimport java.util.*;class GFG { // Function to find Kth smallest element in matrix public static int kthSmallest(int[][] mat, int n, int m, int k) { // Create a temporary array int[] arr = new int[n * m]; // Fill the temporary array int index = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { arr[index++] = mat[i][j]; } } // Sort the temporary array Arrays.sort(arr); // Return Kth smallest element return arr[k - 1]; } // Driver code public static void main(String[] args) { // Given matrix int n = 2, m = 3, k = 5; int[][] mat = new int[n][m]; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { mat[i][j] = (i + 1) * (j + 1); } } // Function call System.out.println(kthSmallest(mat, n, m, k)); }} |
Python
def GFG(mat, n, m, k): arr = [] # Fill temporary array for i in range(n): for j in range(m): arr.append(mat[i][j]) arr.sort() # Return Kth smallest element return arr[k - 1]# Driver codeif __name__ == "__main__": n = 2 m = 3 k = 5 mat = [[0 for j in range(m)] for i in range(n)] for i in range(n): for j in range(m): mat[i][j] = (i + 1) * (j + 1) # Function call print(GFG(mat, n, m, k)) |
C#
using System;using System.Linq;class Program { // Function to find Kth smallest element in matrix static int kthSmallest(int[][] mat, int n, int m, int k) { // Create a temporary array int[] arr = new int[n * m]; // Fill the temporary array int index = 0; for (int i = 0; i < n; i++) { for (int j = 0; j < m; j++) { arr[index++] = mat[i][j]; } } // Sort the temporary array Array.Sort(arr); // Return Kth smallest element return arr[k - 1]; } static void Main(string[] args) { // Given matrix int n = 2, m = 3, k = 5; int[][] mat = new int[n][]; for (int i = 0; i < n; i++) { mat[i] = new int[m]; for (int j = 0; j < m; j++) { mat[i][j] = (i + 1) * (j + 1); } } // Function call Console.WriteLine(kthSmallest(mat, n, m, k)); }}// This code is contributed by user_dtewbxkn77n |
Javascript
// Javascript code addition function kthSmallest(mat, n, m, k) { // Create a temporary array let arr = new Array(n * m); // Fill the temporary array let index = 0; for (let i = 0; i < n; i++) { for (let j = 0; j < m; j++) { arr[index++] = mat[i][j]; } } // Sort the temporary array arr.sort((a, b) => a - b); // Return Kth smallest element return arr[k - 1];}// Driver codelet n = 2, m = 3, k = 5;let mat = [];for (let i = 0; i < n; i++) { mat.push(new Array(m)); for (let j = 0; j < m; j++) { mat[i][j] = (i + 1) * (j + 1); }}// Function callconsole.log(kthSmallest(mat, n, m, k));// The code is contributed by Nidhi goel. |
Output
4
Time Complexity: O(N×M×log(N×M))
Auxiliary Space: O(N×M)
Efficient Approach:
To optimize the naive approach the idea is to use the Binary Search algorithm. Follow the steps below to solve the problem:
- Initialize low as 1 and high as N×M, as the Kth smallest element lies between 1 and N×M.
- Find the mid element between the low and high elements.
- If the number of elements less than mid is greater than or equal to K, then update high to mid-1 as the Kth smallest element lies between low and mid.
- If the number of elements less than mid is less than K, then update low to mid+1 as the Kth smallest element lies between mid and high.
- As the elements in the ith row are the multiple of i, the number of elements less than mid in the ith row can be calculated easily by min(mid/i, M). So, the time complexity to find the number of elements less than mid can be done in only O(N).
- Perform binary search till low is less than or equal to high and return high + 1 as the Kth smallest element of the matrix N×M.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; #define LL long long // Function that returns true if the // count of elements is less than mid bool countLessThanMid(LL mid, LL N, LL M, LL K) { // To store count of elements // less than mid LL count = 0; // Loop through each row for (int i = 1; i <= min((LL)N, mid); ++i) { // Count elements less than // mid in the ith row count = count + min(mid / i, M); } if (count >= K) return false; else return true; } // Function that returns the Kth // smallest element in the NxM // Matrix after sorting in an array LL findKthElement(LL N, LL M, LL K) { // Initialize low and high LL low = 1, high = N * M; // Perform binary search while (low <= high) { // Find the mid LL mid = low + (high - low) / 2; // Check if the count of // elements is less than mid if (countLessThanMid(mid, N, M, K)) low = mid + 1; else high = mid - 1; } // Return Kth smallest element // of the matrix return high + 1; } // Driver Code int main() { LL N = 2, M = 3; LL int K = 5; cout << findKthElement(N, M, K) << endl; return 0; } |
Java
// Java program to implement // the above approach class GFG{ // Function that returns true if the // count of elements is less than mid public static boolean countLessThanMid(int mid, int N, int M, int K) { // To store count of elements // less than mid int count = 0; // Loop through each row for(int i = 1; i <= Math.min(N, mid); ++i) { // Count elements less than // mid in the ith row count = count + Math.min(mid / i, M); } if (count >= K) return false; else return true; } // Function that returns the Kth // smallest element in the NxM // Matrix after sorting in an array public static int findKthElement(int N, int M, int K) { // Initialize low and high int low = 1, high = N * M; // Perform binary search while (low <= high) { // Find the mid int mid = low + (high - low) / 2; // Check if the count of // elements is less than mid if (countLessThanMid(mid, N, M, K)) low = mid + 1; else high = mid - 1; } // Return Kth smallest element // of the matrix return high + 1; } // Driver codepublic static void main(String[] args) { int N = 2, M = 3; int K = 5; System.out.println(findKthElement(N, M, K));}}// This code is contributed by divyeshrabadiya07 |
Python3
# Python3 program for the above approach # Function that returns true if the # count of elements is less than mid def countLessThanMid(mid, N, M, K): # To store count of elements # less than mid count = 0 # Loop through each row for i in range (1, min(N, mid) + 1): # Count elements less than # mid in the ith row count = count + min(mid // i, M) if (count >= K): return False else: return True# Function that returns the Kth # smallest element in the NxM # Matrix after sorting in an array def findKthElement(N, M, K): # Initialize low and high low = 1 high = N * M # Perform binary search while (low <= high): # Find the mid mid = low + (high - low) // 2 # Check if the count of # elements is less than mid if (countLessThanMid(mid, N, M, K)): low = mid + 1 else: high = mid - 1 # Return Kth smallest element # of the matrix return high + 1# Driver Code if __name__ == "__main__": N = 2 M = 3 K = 5 print(findKthElement(N, M, K)) # This code is contributed by Chitranayal |
C#
// C# program to implement // the above approach using System;class GFG{ // Function that returns true if the // count of elements is less than mid public static bool countLessThanMid(int mid, int N, int M, int K) { // To store count of elements // less than mid int count = 0; // Loop through each row for(int i = 1; i <= Math.Min(N, mid); ++i) { // Count elements less than // mid in the ith row count = count + Math.Min(mid / i, M); } if (count >= K) return false; else return true; } // Function that returns the Kth // smallest element in the NxM // Matrix after sorting in an array public static int findKthElement(int N, int M, int K) { // Initialize low and high int low = 1, high = N * M; // Perform binary search while (low <= high) { // Find the mid int mid = low + (high - low) / 2; // Check if the count of // elements is less than mid if (countLessThanMid(mid, N, M, K)) low = mid + 1; else high = mid - 1; } // Return Kth smallest element // of the matrix return high + 1; } // Driver codepublic static void Main(String[] args) { int N = 2, M = 3; int K = 5; Console.WriteLine(findKthElement(N, M, K));}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program for the above approach // Function that returns true if the // count of elements is less than mid function countLessThanMid(mid, N, M, K) { // To store count of elements // less than mid let count = 0; // Loop through each row for (let i = 1; i <= Math.min(N, mid); ++i) { // Count elements less than // mid in the ith row count = count + Math.min(parseInt(mid / i), M); } if (count >= K) return false; else return true; } // Function that returns the Kth // smallest element in the NxM // Matrix after sorting in an array function findKthElement(N, M, K) { // Initialize low and high let low = 1, high = N * M; // Perform binary search while (low <= high) { // Find the mid let mid = low + parseInt((high - low) / 2); // Check if the count of // elements is less than mid if (countLessThanMid(mid, N, M, K)) low = mid + 1; else high = mid - 1; } // Return Kth smallest element // of the matrix return high + 1; } // Driver Code let N = 2, M = 3; let K = 5; document.write(findKthElement(N, M, K)); </script> |
Output:
4
Time Complexity: O(N×log(N×M))
Auxiliary Space: O(1)
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