Split the number into N parts such that difference between the smallest and the largest part is minimum

• Difficulty Level : Medium
• Last Updated : 07 Sep, 2021

Given two integers ‘X’ and ‘N’, the task is to split the integer ‘X’ into exactly ‘N’ parts such that:
X1 + X2 + X3 + … + Xn = X and the difference between the maximum and the minimum number from the sequence is minimized.
Print the sequence in the end, if the number cannot be divided into exactly ‘N’ parts then print ‘-1’ instead.

Examples:

Input: X = 5, N = 3
Output: 1 2 2
Divide 5 into 3 parts such that the difference between the largest and smallest integer among
them is as minimal as possible. So we divide 5 as 1 + 2 + 2.

Input: X = 25, N = 5
Output: 5 5 5 5 5

Approach: There is always a way of splitting the number if X >= N.

• If the number is being split into exactly ‘N’ parts then every part will have the value X/N and the remaining X%N part can be distributed among any X%N numbers.
• Thus, if X % N == 0 then the minimum difference will always be ‘0’ and the sequence will contain all equal numbers i.e. x/n.
• Else, the difference will be ‘1’ and the sequence will be X/N, X/N, …, (X/N)+1, (X/N)+1..

Below is the implementation of the above approach:

C++

 // CPP implementation of the approach#includeusing namespace std;; // Function that prints// the required sequencevoid split(int x, int n){ // If we cannot split the// number into exactly 'N' partsif(x < n)cout<<"-1"<<" ";               // If x % n == 0 then the minimum    // difference is 0 and all    // numbers are x / n    else if (x % n == 0)    {        for(int i=0;i= zp)            cout<<(pp + 1)<<" ";            else            cout<

Java

 // Java implementation of the approach  class GFG{// Function that prints// the required sequencestatic void split(int x, int n){  // If we cannot split the// number into exactly 'N' partsif(x < n)System.out.print("-1 ");                  // If x % n == 0 then the minimum    // difference is 0 and all    // numbers are x / n    else if (x % n == 0)    {        for(int i=0;i= zp)            System.out.print((pp + 1)+" ");            else            System.out.print(pp+" ");        }    }}      // Driver codepublic static void main(String[] args){          int x = 5;int n = 3;split(x, n);  }}//This code is contributed by mits

Python3

 # Python3 implementation of the approach # Function that prints# the required sequencedef split(x, n):     # If we cannot split the    # number into exactly 'N' parts    if(x < n):        print(-1)     # If x % n == 0 then the minimum    # difference is 0 and all    # numbers are x / n    elif (x % n == 0):        for i in range(n):            print(x//n, end =" ")    else:        # upto n-(x % n) the values        # will be x / n        # after that the values        # will be x / n + 1        zp = n - (x % n)        pp = x//n        for i in range(n):            if(i>= zp):                print(pp + 1, end =" ")            else:                print(pp, end =" ")       # Driver code         x = 5n = 3split(x, n)

C#

 // C# implementation of the approachusing System; public class GFG{    // Function that prints// the required sequencestatic void split(int x, int n){ // If we cannot split the// number into exactly 'N' partsif(x < n)Console.WriteLine("-1 ");               // If x % n == 0 then the minimum    // difference is 0 and all    // numbers are x / n    else if (x % n == 0)    {        for(int i=0;i= zp)            Console.Write((pp + 1)+" ");            else            Console.Write(pp+" ");        }    }}     // Driver codestatic public void Main (){ int x = 5;int n = 3;split(x, n); }}//This code is contributed by Sachin.

PHP

 = \$zp)            {                echo (int)\$pp + 1;                echo (" ");            }            else            {                echo (int)\$pp;                echo (" ");            }        }    }} // Driver code    \$x = 5;\$n = 3;split( \$x, \$n); // This code is contributed// by Shivi_Aggarwal?>

Javascript


Output:
1 2 2

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