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Max sum of M non-overlapping subarrays of size K
  • Difficulty Level : Hard
  • Last Updated : 09 May, 2020

Given an array and two numbers M and K. We need to find sum of max M subarrays of size K (non-overlapping) in the array. (Order of array remains unchanged). K is the size of subarrays and M is the count of subarray. It may be assumed that size of array is more than m*k. If total array size is not multiple of k, then we can take partial last array.

Examples :

Input: N = 7, M = 3, K = 1 
       arr[] = {2, 10, 7, 18, 5, 33, 0}; 
Output: 61 
Explanation: subsets are: 33, 18, 10 (3 subsets of size 1) 

Input: N = 4, M = 2, K = 2 
       arr[] = {3, 2, 100, 1}; 
Output:  106 
Explanation: subsets are: (3, 2), (100, 1) 2 subsets of size 2

Here we can see that we need to find M subarrays each of size K so,
1. We create a presum array, which contains in each index sum of all elements from ‘index‘ to ‘index + K’ in the given array. And size of the sum array will be n+1-k.
2. Now if we include the subarray of size k, then we can not include any of the elements of that subarray again in any other subarray as it will create overlapping subarrays. So we make recursive call by excluding the k elements of included subarray.
3. if we exclude a subarray then we can use the next k-1 elements of that subarray in other subarrays so we will make recursive call by just excluding the first element of that subarray.
4. At last return the max(included sum, excluded sum).

C++




// C++ program to find Max sum of M non-overlapping
// subarray of size K in an Array
#include <bits/stdc++.h>
using namespace std;
  
// calculating presum of array. presum[i]
// is going to store prefix sum of subarray
// of size k beginning with arr[i]
void calculatePresumArray(int presum[],
                int arr[], int n, int k)
{    
    for (int i=0; i<k; i++)
       presum[0] += arr[i];
  
    // store sum of array index i to i+k 
    // in presum array at index i of it.
    for (int i = 1; i <= n - k; i++) 
       presum[i] += presum[i-1] + arr[i+k-1] - 
                                  arr[i-1];
}
  
// calculating maximum sum of m non overlapping array
int maxSumMnonOverlappingSubarray(int presum[],
              int m, int size, int k, int start)
{
    // if m is zero then no need any array
    // of any size so return 0.
    if (m == 0)
        return 0;
  
    // if start is greater then the size
    // of presum array return 0.
    if (start > size - 1)
        return 0;
  
    int mx = 0;
  
    // if including subarray of size k
    int includeMax = presum[start] + 
            maxSumMnonOverlappingSubarray(presum,
                        m - 1, size, k, start + k);
  
    // if excluding element and searching 
    // in all next possible subarrays
    int excludeMax = 
            maxSumMnonOverlappingSubarray(presum,
                            m, size, k, start + 1);
  
    // return max
    return max(includeMax, excludeMax);
}
  
// Driver code
int main()
{
    int arr[] = { 2, 10, 7, 18, 5, 33, 0 };
    int n = sizeof(arr)/sizeof(arr[0]);
      
     int m = 3, k = 1;
  
    int presum[n + 1 - k] = { 0 };
    calculatePresumArray(presum, arr, n, k);
  
    // resulting presum array will have a size = n+1-k
    cout << maxSumMnonOverlappingSubarray(presum,
                               m, n + 1 - k, k, 0);
  
    return 0;
}

Java




// Java program to find Max sum 
// of M non-overlapping subarray
// of size K in an Array
  
import java.io.*;
  
class GFG 
{
// calculating presum of array. 
// presum[i] is going to store 
// prefix sum of subarray of 
// size k beginning with arr[i]
static void calculatePresumArray(int presum[],
                                 int arr[], 
                                 int n, int k)
    for (int i = 0; i < k; i++)
    presum[0] += arr[i];
  
    // store sum of array index i to i+k 
    // in presum array at index i of it.
    for (int i = 1; i <= n - k; i++) 
    presum[i] += presum[i - 1] + arr[i + k - 1] - 
                                 arr[i - 1];
}
  
// calculating maximum sum of
// m non overlapping array
static int maxSumMnonOverlappingSubarray(int presum[],
                                         int m, int size, 
                                         int k, int start)
{
    // if m is zero then no need 
    // any array of any size so
    // return 0.
    if (m == 0)
        return 0;
  
    // if start is greater then the 
    // size of presum array return 0.
    if (start > size - 1)
        return 0;
  
    int mx = 0;
  
    // if including subarray of size k
    int includeMax = presum[start] + 
            maxSumMnonOverlappingSubarray(presum,
                      m - 1, size, k, start + k);
  
    // if excluding element and searching 
    // in all next possible subarrays
    int excludeMax = 
            maxSumMnonOverlappingSubarray(presum,
                          m, size, k, start + 1);
  
    // return max
    return Math.max(includeMax, excludeMax);
}
  
// Driver code
public static void main (String[] args) 
{
    int arr[] = { 2, 10, 7, 18, 5, 33, 0 };
    int n = arr.length;
    int m = 3, k = 1;
    int presum[] = new int[n + 1 - k] ;
    calculatePresumArray(presum, arr, n, k);
      
    // resulting presum array
    // will have a size = n+1-k
    System.out.println(maxSumMnonOverlappingSubarray(presum,
                                        m, n + 1 - k, k, 0));
}
}
  
// This code is contributed by anuj_67.

Python




# Python3 program to find Max sum of M non-overlapping
# subarray of size K in an Array
  
# calculating presum of array. presum[i]
# is going to store prefix sum of subarray
# of size k beginning with arr[i]
def calculatePresumArray(presum,arr, n, k):
  
    for i in range(k):
        presum[0] += arr[i]
  
    # store sum of array index i to i+k 
    # in presum array at index i of it.
    for i in range(1,n - k + 1): 
        presum[i] += presum[i - 1] + arr[i + k - 1] - arr[i - 1]
  
# calculating maximum sum of m non overlapping array
def maxSumMnonOverlappingSubarray(presum, m, size, k, start):
      
    # if m is zero then no need any array
    # of any size so return 0.
    if (m == 0):
        return 0
  
    # if start is greater then the size
    # of presum array return 0.
    if (start > size - 1):
        return 0
  
    mx = 0
  
    # if including subarray of size k
    includeMax = presum[start] + maxSumMnonOverlappingSubarray(presum,m - 1, size, k, start + k)
  
    # if excluding element and searching 
    # in all next possible subarrays
    excludeMax = maxSumMnonOverlappingSubarray(presum,m, size, k, start + 1)
  
    # return max
    return max(includeMax, excludeMax)
  
# Driver code
arr = [2, 10, 7, 18, 5, 33, 0]
n = len(arr)
  
m, k = 3, 1
  
presum = [0 for i in range(n + 1 - k)]
calculatePresumArray(presum, arr, n, k)
  
# resulting presum array will have a size = n+1-k
print(maxSumMnonOverlappingSubarray(presum, m, n + 1 - k, k, 0))
  
# This code is contributed by mohit kumar

C#




// C# program to find Max sum of M
// non-overlapping subarray of size 
// K in an Array
using System;
  
class GFG {
      
    // calculating presum of array. 
    // presum[i] is going to store 
    // prefix sum of subarray of 
    // size k beginning with arr[i]
    static void calculatePresumArray(int []presum,
                          int []arr, int n, int k)
    
        for (int i = 0; i < k; i++)
        presum[0] += arr[i];
      
        // store sum of array index i to i+k 
        // in presum array at index i of it.
        for (int i = 1; i <= n - k; i++) 
        presum[i] += presum[i - 1] + arr[i + k - 1]
                                      - arr[i - 1];
    }
      
    // calculating maximum sum of
    // m non overlapping array
    static int maxSumMnonOverlappingSubarray(
                     int []presum, int m, int size, 
                                  int k, int start)
    {
          
        // if m is zero then no need 
        // any array of any size so
        // return 0.
        if (m == 0)
            return 0;
      
        // if start is greater then the 
        // size of presum array return 0.
        if (start > size - 1)
            return 0;
      
        //int mx = 0;
      
        // if including subarray of size k
        int includeMax = presum[start] + 
                maxSumMnonOverlappingSubarray(presum,
                          m - 1, size, k, start + k);
      
        // if excluding element and searching 
        // in all next possible subarrays
        int excludeMax = 
                maxSumMnonOverlappingSubarray(presum,
                              m, size, k, start + 1);
      
        // return max
        return Math.Max(includeMax, excludeMax);
    }
      
    // Driver code
    public static void Main () 
    {
        int []arr = { 2, 10, 7, 18, 5, 33, 0 };
        int n = arr.Length;
        int m = 3, k = 1;
        int []presum = new int[n + 1 - k] ;
        calculatePresumArray(presum, arr, n, k);
          
        // resulting presum array
        // will have a size = n+1-k
        Console.WriteLine(
              maxSumMnonOverlappingSubarray(presum,
                              m, n + 1 - k, k, 0));
    }
}
  
// This code is contributed by anuj_67.
Output :
61



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