Partiton N into M parts such that difference between Max and Min part is smallest

Given two integers N and M, partiton 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 minimised.

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 atmost 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++

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// C++ program to partiton N into M parts
// such that difference Max and Min
// part is smallest
  
#include <bits/stdc++.h>
using namespace std;
  
// Funtion to partiton N into M parts such
// that difference Max and Min part
// is smallest
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++)
        cout << k + 1 << " ";
  
    for (; i <= m; i++)
        cout << k << " ";
}
  
// Driver Code
int main()
{
    int n = 5, m = 2;
  
    printPartition(n, m);
  
    return 0;
}

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Java

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// Java program to partiton N into M parts
// such that difference Max and Min
// part is smallest
  
import java.io.*;
  
class GFG {
  
  
// Funtion to partiton N into M parts such
// that difference Max and Min part
// is smallest
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++)
        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..

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Python3

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# Python 3 program to partiton N into M parts
# such that difference Max and Min
# part is smallest
  
# Funtion to partiton N into M parts such
# that difference Max and Min part
# is smallest
def 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 Code
if __name__ == '__main__':
    n = 5
    m = 2
    printPartition(n, m)
  
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# program to partiton N into M parts
// such that difference Max and Min
// part is smallest
using 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 Code
static public void Main () 
{
    int n = 5, m = 2;
  
    printPartition(n, m);
}
}
  
// This code is contributed by Sachin

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PHP

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<?php
// PHP program to partiton N into 
// M parts such that difference
// Max and Min part is smallest
  
// Function to partiton N into M 
// parts such that difference Max 
// and Min part is smallest
function 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 conributed 
// by Akanksha Rai
?>

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Output:

3 2

Time Complexity: O(M)



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