Given two integers N and K, the task is to find the smallest value for maximum element of an array of size N consisting of positive integers whose sum of elements is divisible by K.
Examples:
Input: N = 4, K = 3
Output: 2
Explanation:
Let the array be [2, 2, 1, 1]. Here, sum of elements of this array is divisible by K=3, and maximum element is 2.Input: N = 3, K = 5
Output: 2
Approach: To find the smallest maximum of an array of size N and having sum divisible by K, try to create an array with the minimum sum possible.
- The minimum sum of N elements (each having a value greater than 0) that is divisible by K is:
sum = K * ceil(N/K)
- Now, if the sum is divisible by N then the maximum element will be sum/N otherwise it is (sum/N + 1).
Below is the implementation of above approach.
C++
// C++ program for the above approach.#include <iostream>using namespace std;// Function to find smallest maximum number// in an array whose sum is divisible by K.int smallestMaximum(int N, int K){ // Minimum possible sum possible // for an array of size N such that its // sum is divisible by K int sum = ((N + K - 1) / K) * K; // If sum is not divisible by N if (sum % N != 0) return (sum / N) + 1; // If sum is divisible by N else return sum / N;}// Driver code.int main(){ int N = 4; int K = 3; cout << smallestMaximum(N, K) << endl; return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.util.*;class GFG{ // Function to find smallest maximum number// in an array whose sum is divisible by K.static int smallestMaximum(int N, int K){ // Minimum possible sum possible // for an array of size N such that its // sum is divisible by K int sum = ((N + K - 1) / K) * K; // If sum is not divisible by N if (sum % N != 0) return (sum / N) + 1; // If sum is divisible by N else return sum / N;}// Driver codepublic static void main(String args[]){ int N = 4; int K = 3; System.out.println(smallestMaximum(N, K));}}// This code is contributed by code_hunt. |
Python3
# python program for the above approach.# Function to find smallest maximum number# in an array whose sum is divisible by K.def smallestMaximum(N,K): # Minimum possible sum possible # for an array of size N such that its # sum is divisible by K sum = ((N + K - 1) // K) * K # If sum is not divisible by N if (sum % N != 0): return (sum // N) + 1 # If sum is divisible by N else: return sum // N# Driver code.if __name__ == "__main__": N = 4 K = 3 print(smallestMaximum(N, K)) # This code is contributed by anudeep23042002. |
C#
// C# program for the above approach.using System;using System.Collections.Generic;class GFG{// Function to find smallest maximum number// in an array whose sum is divisible by K.static int smallestMaximum(int N, int K){ // Minimum possible sum possible // for an array of size N such that its // sum is divisible by K int sum = ((N + K - 1) / K) * K; // If sum is not divisible by N if (sum % N != 0) return (sum / N) + 1; // If sum is divisible by N else return sum / N;}// Driver code.public static void Main(){ int N = 4; int K = 3; Console.Write(smallestMaximum(N, K));}}// This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to find smallest maximum number // in an array whose sum is divisible by K. function smallestMaximum(N, K) { // Minimum possible sum possible // for an array of size N such that its // sum is divisible by K let sum = Math.floor((N + K - 1) / K) * K; // If sum is not divisible by N if (sum % N != 0) return Math.floor(sum / N) + 1; // If sum is divisible by N else return Math.floor(sum / N); } // Driver code. let N = 4; let K = 3; document.write(smallestMaximum(N, K));// This code is contributed by Potta Lokesh </script> |
Output:
2
Time Complexity: O(1)
Auxiliary Space: O(1)
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