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

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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 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;
}

Java

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

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 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

C#

// C# program to partition 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

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 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 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 smallest
function 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>

Output:

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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