Given two integers N and M, partition N into M integers such that the difference between the maximum and minimum integer obtained by the partition is as small as possible.
Print the M numbers A1, A2….Am, such that:
- sum(A) = N.
- max(A)-min(A) is minimized.
Examples:
Input : N = 11, M = 3
Output : A[] = {4, 4, 3}
Input : N = 8, M = 4
Output : A[] = {2, 2, 2, 2}
To minimize the difference between the terms, we should have all of them as close to each other as possible. Let’s say, we could print any floating values instead of integers, the answer in that case would be 0 (print N/M M times). But since we need to print integers, we can divide it into 2 parts, floor(N/M) and floor(N/M)+1 which would give us the answer at most 1.
How many terms do we need to print of each type?
Let’s say we print floor(N/M) M times, the sum would be equal to N – (N%M). So we need to choose N%M terms and increase it by 1.
Below is the implementation of the above approach:
C++
// C++ program to partition N into M parts// such that difference Max and Min// part is smallest#include <bits/stdc++.h>using namespace std;// Function to partition N into M parts such// that difference Max and Min part// is smallestvoid printPartition(int n, int m){ int k = n / m; // Minimum value int ct = n % m; // Number of (K+1) terms int i; for (i = 1; i <= ct; i++) cout << k + 1 << " "; for (; i <= m; i++) cout << k << " ";}// Driver Codeint main(){ int n = 5, m = 2; printPartition(n, m); return 0;} |
Java
// Java program to partition N into M parts// such that difference Max and Min// part is smallestimport java.io.*;class GFG {// Function to partition N into M parts such// that difference Max and Min part// is smalleststatic void printPartition(int n, int m){ int k = n / m; // Minimum value int ct = n % m; // Number of (K+1) terms int i; for (i = 1; i <= ct; i++) System.out.print( k + 1 + " "); for (; i <= m; i++) System.out.print( k + " ");}// Driver Code public static void main (String[] args) { int n = 5, m = 2; printPartition(n, m); }}// This code is contributed by anuj_67.. |
Python3
# Python 3 program to partition N into M parts# such that difference Max and Min# part is the smallest# Function to partition N into M parts such# that difference Max and Min part# is smallestdef printPartition(n, m): k = int(n / m) # Minimum value ct = n % m # Number of (K+1) terms for i in range(1,ct+1,1): print(k + 1,end= " ") count = i for i in range(count,m,1): print(k,end=" ")# Driver Codeif __name__ == '__main__': n = 5 m = 2 printPartition(n, m)# This code is contributed by# Surendra_Gangwar |
C#
// C# program to partition N into M parts// such that difference Max and Min// part is smallestusing System;class GFG{static void printPartition(int n, int m){ int k = n / m; // Minimum value int ct = n % m; // Number of (K+1) terms int i; for (i = 1; i <= ct; i++) Console.Write( k + 1 + " "); for (; i <= m; i++) Console.Write( k + " ");}// Driver Codestatic public void Main () { int n = 5, m = 2; printPartition(n, m);}}// This code is contributed by Sachin |
PHP
<?php// PHP program to partition N into // M parts such that difference// Max and Min part is smallest// Function to partition N into M // parts such that difference Max // and Min part is smallestfunction printPartition($n, $m){ $k = (int)($n / $m); // Minimum value $ct = $n % $m; // Number of (K+1) terms for ($i = 1; $i <= $ct; $i++) echo $k + 1 . " "; for (; $i <= $m; $i++) echo $k . " ";}// Driver Code$n = 5; $m = 2;printPartition($n, $m);// This code is contributed // by Akanksha Rai?> |
Javascript
<script>// Javascript program to partition N into M parts// such that difference Max and Min// part is smallest// Function to partition N into M parts such// that difference Max and Min part// is smallestfunction prletPartition(n, m){ let k = Math.floor(n / m); // Minimum value let ct = n % m; // Number of (K+1) terms let i; for (i = 1; i <= ct; i++) document.write( k + 1 + " "); for (; i <= m; i++) document.write( k + " ");}// driver program let n = 5, m = 2; prletPartition(n, m); </script> |
3 2
Time Complexity: O(n + m), where n and m are the values of the given integers.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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