Data Modelling & AI Data Structure & Algorithm

Kth smallest element in an array that contains A[i] exactly B[i] times

23 June 2025

0

Given two arrays A[] and B[] consisting of N positive integers and an integer K, the task is to find the K^thsmallest element in the array formed by taking the i^th 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 4^th 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 4^th 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 i^th index of the array, B[i] times into the new array. Finally, print the K^th 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] times
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 = 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 Code
int 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 approach
public 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] times
def 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 Code
if __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 approach
using System;
 
class GFG{
 
// 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
    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:

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[].

Recommended

Solve DSA problems on GfG Practice.

Solve Problems

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

Kth smallest element in an array that contains A[i] exactly B[i] times

C++

Java

Python3

C#

Javascript

LangExtract + Milvus: A Practical Guide to Building a Hybrid Document Processing and Search System

Stop Your AI Assistant from Writing Outdated Code with Milvus SDK Code Helper

A Practical Guide for Choosing the Right Vector Database for Your AI Applications

LEAVE A REPLY Cancel reply

Most Popular

TCL debuts the first smart TVs with Google Gemini

Android 16 QPR2 Beta 1 introduces ‘expanded’ dark theme settings

How to record calls on your Samsung Galaxy phone

How to measure (and understand) your smartphone’s charging speed

EDITOR PICKS

TCL debuts the first smart TVs with Google Gemini

Android 16 QPR2 Beta 1 introduces ‘expanded’ dark theme settings

How to record calls on your Samsung Galaxy phone

POPULAR POSTS

TCL debuts the first smart TVs with Google Gemini

Android 16 QPR2 Beta 1 introduces ‘expanded’ dark theme settings

How to record calls on your Samsung Galaxy phone

POPULAR CATEGORY

ABOUT US

FOLLOW US