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 minimised.

**Examples**:

Input: N = 11, M = 3Output: A[] = {4, 4, 3}Input: N = 8, M = 4Output: 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 conributed` `// 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 leto 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(M)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend live classes with industry experts, please refer **Geeks Classes Live** and **Geeks Classes Live USA**