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

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 = 3
Output : A[] = {4, 4, 3}

Input : N = 8, M = 4
Output : A[] = {2, 2, 2, 2}

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 using namespace std;    // Function 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; }

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 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

PHP



Output:

3 2

Time Complexity: O(M)

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.