Split Array into K non-overlapping subset such that maximum among all subset sum is minimum

Given an array arr[] consisting of N integers and an integer K, the task is to split the given array into K non-overlapping subsets such that the maximum among the sum of all subsets is minimum.

Examples:

Input: arr[] = {1, 7, 9, 2, 12, 3, 3}, M = 3
Output: 13
Explanation:
One possible way to split the array into 3 non-overlapping subsets is {arr[4], arr[0]}, {arr[2], arr[6]}, and {arr[1], arr[5], arr[3]}.
The sum of each subset is 13, 12 and 12 respectively. Now, the maximum among all the sum of subsets is 13, which is the minimum possible sum.

Input: arr[] = {1, 2, 3, 4, 5}, M = 2
Output: 8

Approach: The given problem can be solved by the Greedy Approach by using the priority queue and sorting the given array. Follow the steps below to solve the problem:

• Initialize a Min-Heap using priority_queue say PQ to store the sum of elements of each group.
• Iterate in the range [1, K] and push 0 into the PQ.
• Sort the array, arr[] in decreasing order.
• Traverse the array arr[] and for each array elements arr[i], add the element arr[i] to the smallest sum group which will be at the top of the priority queue PQ.
• After completing the above steps, print the last element of the priority queue PQ as the resultant minimized maximum sum among all the possible groups.

Below is the implementation of the above approach:

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C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to split the array into M
// groups such that maximum of the sum
// of all elements of all the groups
// is minimized
int findMinimumValue(int arr[], int N,
                     int M)
{
    // Sort the array in decreasing order
    sort(arr, arr + N, greater<int>());

    // Initialize priority queue (Min heap)
    priority_queue<int, vector<int>,
                   greater<int> >
        pq;

    // Push 0 for all the M groups
    for (int i = 1; i <= M; ++i) {
        pq.push(0);
    }

    // Traverse the array, arr[]
    for (int i = 0; i < N; ++i) {

        // Pop the group having the
        // minimum sum
        int val = pq.top();
        pq.pop();

        // Increment val by arr[i]
        val += arr[i];

        // Push the new sum of the
        // group into the pq
        pq.push(val);
    }

    // Iterate while size of the pq
    // is greater than 1
    while (pq.size() > 1) {
        pq.pop();
    }

    // Return result
    return pq.top();
}

// Driver Code
int main()
{
    int arr[] = { 1, 7, 9, 2, 12, 3, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 3;
    cout << findMinimumValue(arr, N, K);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class Main
{
    // Function to split the array into M
    // groups such that maximum of the sum
    // of all elements of all the groups
    // is minimized
    static int findMinimumValue(Vector<Integer> arr, int N, int M)
    {

        // Sort the array in decreasing order
        Collections.sort(arr);
        Collections.reverse(arr);

        // Initialize priority queue (Min heap)
        Vector<Integer> pq = new Vector<Integer>();

        // Push 0 for all the M groups
        for (int i = 1; i <= M; ++i) {
            pq.add(0);
        }

        Collections.sort(pq);

        // Traverse the array, arr[]
        for (int i = 0; i < N; ++i) {

            // Pop the group having the
            // minimum sum
            int val = pq.get(0);
            pq.remove(0);

            // Increment val by arr[i]
            val += arr.get(i);

            // Push the new sum of the
            // group into the pq
            pq.add(val);
            Collections.sort(pq);
        }

        // Iterate while size of the pq
        // is greater than 1
        while (pq.size() > 1) {
            pq.remove(0);
        }

        // Return result
        return pq.get(0);
    }

    public static void main(String[] args) {
        Integer[] arr = { 1, 7, 9, 2, 12, 3, 3 };
        Vector<Integer> Arr = new Vector<Integer>();
        Collections.addAll(Arr, arr);
        int N = Arr.size();
        int K = 3;
        System.out.println(findMinimumValue(Arr, N, K));
    }
}

// This code is contributed by divyesh072019.

Python3

# Python3 program for the above approach

# Function to split the array into M
# groups such that maximum of the sum
# of all elements of all the groups
# is minimized
def findMinimumValue(arr, N, M):

    # Sort the array in decreasing order
    arr.sort()
    arr.reverse()

    # Initialize priority queue (Min heap)
    pq = []

    # Push 0 for all the M groups
    for i in range(1, M + 1):
        pq.append(0)

    pq.sort()

    # Traverse the array, arr[]
    for i in range(N):

        # Pop the group having the
        # minimum sum
        val = pq[0]
        del pq[0]

        # Increment val by arr[i]
        val += arr[i]

        # Push the new sum of the
        # group into the pq
        pq.append(val)
        pq.sort()

    # Iterate while size of the pq
    # is greater than 1
    while (len(pq) > 1) :
        del pq[0]

    # Return result
    return pq[0]

arr = [ 1, 7, 9, 2, 12, 3, 3 ]
N = len(arr)
K = 3
print(findMinimumValue(arr, N, K))

# This code is contributed by suresh07.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {

    // Function to split the array into M
    // groups such that maximum of the sum
    // of all elements of all the groups
    // is minimized
    static int findMinimumValue(int[] arr, int N, int M)
    {

        // Sort the array in decreasing order
        Array.Sort(arr);
        Array.Reverse(arr);

        // Initialize priority queue (Min heap)
        List<int> pq = new List<int>();

        // Push 0 for all the M groups
        for (int i = 1; i <= M; ++i) {
            pq.Add(0);
        }

        pq.Sort();

        // Traverse the array, arr[]
        for (int i = 0; i < N; ++i) {

            // Pop the group having the
            // minimum sum
            int val = pq[0];
            pq.RemoveAt(0);

            // Increment val by arr[i]
            val += arr[i];

            // Push the new sum of the
            // group into the pq
            pq.Add(val);
            pq.Sort();
        }

        // Iterate while size of the pq
        // is greater than 1
        while (pq.Count > 1) {
            pq.RemoveAt(0);
        }

        // Return result
        return pq[0];
    }

  static void Main() {
    int[] arr = { 1, 7, 9, 2, 12, 3, 3 };
    int N = arr.Length;
    int K = 3;
    Console.Write(findMinimumValue(arr, N, K));
  }
}

// This code is contributed by divyeshrabadiya07.

Javascript

<script>
    // Javascript program for the above approach

    // Function to split the array into M
    // groups such that maximum of the sum
    // of all elements of all the groups
    // is minimized
    function findMinimumValue(arr, N, M)
    {

        // Sort the array in decreasing order
        arr.sort(function(a, b){return a - b});
        arr.reverse();

        // Initialize priority queue (Min heap)
        let pq = [];

        // Push 0 for all the M groups
        for (let i = 1; i <= M; ++i) {
            pq.push(0);
        }

        pq.sort(function(a, b){return a - b});

        // Traverse the array, arr[]
        for (let i = 0; i < N; ++i) {

            // Pop the group having the
            // minimum sum
            let val = pq[0];
            pq.shift();

            // Increment val by arr[i]
            val += arr[i];

            // Push the new sum of the
            // group into the pq
            pq.push(val);
            pq.sort(function(a, b){return a - b});
        }

        // Iterate while size of the pq
        // is greater than 1
        while (pq.length > 1) {
            pq.shift();
        }

        // Return result
        return pq[0];
    }

    let arr = [ 1, 7, 9, 2, 12, 3, 3 ];
    let N = arr.length;
    let K = 3;
    document.write(findMinimumValue(arr, N, K));

    // This code is contributed by decode2207.
</script>

Output:

13

Time Complexity: O(N*log K)
Auxiliary Space: O(M)