Given two arrays A[] and B[] consisting of N positive integers and an integer K, the task is to find the Kth smallest element in the array formed by taking the ith element from the array A[] exactly B[i] times. If there exists no such element, then print -1.
Examples:
Input: K = 4, A[] = {1, 2, 3}, B[] = {1, 2, 3}
Output: 3
Explanation:
The array obtained by taking A[0], B[0] (= 1) time, A[1], B[1] (= 2) times, A[2], B[2]( = 3) times is {1, 2, 2, 3, 3, 3}. Therefore, the 4th element of the array is 3.Input: K = 4, A[] = {3, 4, 5}, B[] = {2, 1, 3}
Output: 3
Explanation:The array formed is {3, 3, 4, 5, 5, 5}. Therefore, the 4th element of the array i.e 5.
Naive Approach: The simplest approach is to iterate over the range [0, N – 1] and push the element at the ith index of the array, B[i] times into the new array. Finally, print the Kth element of the obtained array after sorting the array in ascending order.
Time Complexity: O(N*log(N)), where N is the number of elements in the new array.
Auxiliary Space: O(N)
Efficient Approach: The above approach can be optimized by using a frequency array to keep the count of every element. Follow the steps below to solve the problem:
- Find the maximum element of the array A[] and store it in a variable, say M.
- Initialize an array, say freq[] of size M + 1 with {0}, to store the frequency of every element.
- Iterate in the range [0, N-1] using the variable i:
- Add B[i] in freq[A[i]].
- Initialize a variable, say sum as 0, to store the prefix sum up to a particular index.
- Iterate over the range [0, N – 1] using a variable, say i:
- Add freq[i] in sum.
- If sum is greater than or equal to K, then return i.
- Finally, return -1.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the Kth smallest element// that contains A[i] exactly B[i] timesint KthSmallest(int A[], int B[], int N, int K){ int M = 0; // Traverse the given array for (int i = 0; i < N; i++) { M = max(A[i], M); } // Stores the frequency // of every elements int freq[M + 1] = { 0 }; // Traverse the given array for (int i = 0; i < N; i++) { freq[A[i]] += B[i]; } // Initialize a variable to // store the prefix sums int sum = 0; // Iterate over the range [0, M] for (int i = 0; i <= M; i++) { // Increment sum by freq[i] sum += freq[i]; // If sum is greater // than or equal to K if (sum >= K) { // Return the current // element as answer return i; } } // Return -1 return -1;}// Driver Codeint main(){ // Given Input int A[] = { 3, 4, 5 }; int B[] = { 2, 1, 3 }; int N = sizeof(A) / sizeof(A[0]); int K = 4; // Function call cout << KthSmallest(A, B, N, K); return 0;} |
Java
// Java program for the above approachpublic class GFG_JAVA { // Function to find the Kth smallest element // that contains A[i] exactly B[i] times static int KthSmallest(int A[], int B[], int N, int K) { int M = 0; // Traverse the given array for (int i = 0; i < N; i++) { M = Math.max(A[i], M); } // Stores the frequency // of every elements int freq[] = new int[M + 1]; // Traverse the given array for (int i = 0; i < N; i++) { freq[A[i]] += B[i]; } // Initialize a variable to // store the prefix sums int sum = 0; // Iterate over the range [0, M] for (int i = 0; i <= M; i++) { // Increment sum by freq[i] sum += freq[i]; // If sum is greater // than or equal to K if (sum >= K) { // Return the current // element as answer return i; } } // Return -1 return -1; } // Driver code public static void main(String[] args) { // Given Input int A[] = { 3, 4, 5 }; int B[] = { 2, 1, 3 }; int N = A.length; int K = 4; // Function call System.out.println(KthSmallest(A, B, N, K)); }}// This code is contributed by abhinavjain194 |
Python3
# Python3 program for the above approach# Function to find the Kth smallest element# that contains A[i] exactly B[i] timesdef KthSmallest(A, B, N, K): M = 0 # Traverse the given array for i in range(N): M = max(A[i], M) # Stores the frequency # of every elements freq = [0] * (M + 1) # Traverse the given array for i in range(N): freq[A[i]] += B[i] # Initialize a variable to # store the prefix sums sum = 0 # Iterate over the range [0, M] for i in range(M + 1): # Increment sum by freq[i] sum += freq[i] # If sum is greater # than or equal to K if (sum >= K): # Return the current # element as answer return i # Return -1 return -1# Driver Codeif __name__ == "__main__" : # Given Input A = [ 3, 4, 5 ] B = [ 2, 1, 3 ] N = len(A) K = 4 # Function call print(KthSmallest(A, B, N, K)) # This code is contributed by AnkThon |
C#
// C# program for the above approachusing System;class GFG{// Function to find the Kth smallest element// that contains A[i] exactly B[i] timesstatic int KthSmallest(int []A, int []B, int N, int K){ int M = 0; // Traverse the given array for(int i = 0; i < N; i++) { M = Math.Max(A[i], M); } // Stores the frequency // of every elements int []freq = new int[M + 1]; // Traverse the given array for(int i = 0; i < N; i++) { freq[A[i]] += B[i]; } // Initialize a variable to // store the prefix sums int sum = 0; // Iterate over the range [0, M] for(int i = 0; i <= M; i++) { // Increment sum by freq[i] sum += freq[i]; // If sum is greater // than or equal to K if (sum >= K) { // Return the current // element as answer return i; } } // Return -1 return -1;}// Driver codepublic static void Main(String[] args){ // Given Input int []A = { 3, 4, 5 }; int []B = { 2, 1, 3 }; int N = A.Length; int K = 4; // Function call Console.Write(KthSmallest(A, B, N, K));}}// This code is contributed by shivanisinghss2110 |
Javascript
<script> // JavaScript program for the above approach // Function to find the Kth smallest element // that contains A[i] exactly B[i] times function KthSmallest(A, B, N, K) { let M = 0; // Traverse the given array for (let i = 0; i < N; i++) { M = Math.max(A[i], M); } // Stores the frequency // of every elements let freq = Array.from({length: M + 1}, (_, i) => 0); // Traverse the given array for (let i = 0; i < N; i++) { freq[A[i]] += B[i]; } // Initialize a variable to // store the prefix sums let sum = 0; // Iterate over the range [0, M] for (let i = 0; i <= M; i++) { // Increment sum by freq[i] sum += freq[i]; // If sum is greater // than or equal to K if (sum >= K) { // Return the current // element as answer return i; } } // Return -1 return -1; }// Driver Code // Given Input let A = [ 3, 4, 5 ]; let B = [ 2, 1, 3 ]; let N = A.length; let K = 4; // Function call document.write(KthSmallest(A, B, N, K)); </script> |
Output:
5
Time Complexity: O(N), where N is the size of arrays A[] and B[].
Auxiliary Space: O(M), where M is the maximum element of the array A[].
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