# Partition into two subarrays of lengths k and (N – k) such that the difference of sums is maximum

Given an array of non-negative integers of length N and an integer k. Partition the given array into two subarrays of length K and N – k so that the difference between the sum of both subarray is maximum.

Examples :

```Input : arr[] = {8, 4, 5, 2, 10}
k = 2
Output : 17
Explanation :
Here, we can make first subarray of length k = {4, 2}
and second subarray of length N - k = {8, 5, 10}. Then,
the max_difference = (8 + 5 + 10) - (4 + 2) = 17.

Input : arr[] = {1, 1, 1, 1, 1, 1, 1, 1}
k = 3
Output : 2
Explanation :
Here, subarrays would be {1, 1, 1, 1, 1} and {1, 1, 1}.
So, max_difference would be 2.
```

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

Choose k numbers with largest possible sum. Then the solution obviously is k largest numbers. So that here greedy algorithm works – at each step we choose the largest possible number until we get all K numbers.

In this problem we should divide the array of N numbers into two subarrays of k and N – k numbers respectively. Consider two cases –

• The subarray with largest sum, among these two subarrays, is subarray of K numbers. Then we want to maximize the sum in it, since the sum in the second subarray will only decrease if the sum in the first subarray will increase. So we are now in sub-problem considered above and should choose k largest numbers.
• The subarray with largest sum, among these two subarray, is subarray of N – k numbers. Similarly to the previous case we then have to choose N – k largest numbers among all numbers.

Now, Let’s think which of the two above cases actually gives the answer. We can easily see that larger difference would be when more numbers are included to the group of largest numbers. Hence we could set M = max(k, N – k), find the sum of M largest numbers (let it be S1) and then the answer is S1 – (S – S1), where S is the sum of all numbers.

Below is the implementation of the above approach :

 `// CPP program to calculate max_difference between ` `// the sum of two subarrays of length k and N - k ` `#include ` `using` `namespace` `std; ` ` `  `// Function to calculate max_difference ` `int` `maxDifference(``int` `arr[], ``int` `N, ``int` `k) ` `{ ` `    ``int` `M, S = 0, S1 = 0, max_difference = 0; ` ` `  `    ``// Sum of the array ` `    ``for` `(``int` `i = 0; i < N; i++) ` `        ``S += arr[i]; ` ` `  `    ``// Sort the array in descending order ` `    ``sort(arr, arr + N, greater<``int``>()); ` `    ``M = max(k, N - k); ` `    ``for` `(``int` `i = 0; i < M; i++) ` `        ``S1 += arr[i]; ` ` `  `    ``// Calculating max_difference ` `    ``max_difference = S1 - (S - S1); ` `    ``return` `max_difference; ` `} ` ` `  `// Driver function ` `int` `main() ` `{ ` `    ``int` `arr[] = { 8, 4, 5, 2, 10 }; ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``int` `k = 2; ` `    ``cout << maxDifference(arr, N, k) << endl; ` `    ``return` `0; ` `} `

 `// Java program to calculate max_difference between ` `// the sum of two subarrays of length k and N - k ` `import` `java.util.*;  ` ` `  `class` `GFG ` `{ ` ` `  `// Function to calculate max_difference ` `static` `int` `maxDifference(``int` `arr[], ``int` `N, ``int` `k) ` `{ ` `    ``int` `M, S = ``0``, S1 = ``0``, max_difference = ``0``; ` ` `  `    ``// Sum of the array ` `    ``for` `(``int` `i = ``0``; i < N; i++) ` `        ``S += arr[i]; ` `    ``int` `temp; ` `     `  `    ``// Sort the array in descending order ` `    ``for` `(``int` `i = ``0``; i < N; i++)  ` `    ``{ ` `        ``for` `(``int` `j = i + ``1``; j < N; j++)  ` `        ``{ ` `            ``if` `(arr[i] < arr[j])  ` `            ``{ ` `                ``temp = arr[i]; ` `                ``arr[i] = arr[j]; ` `                ``arr[j] = temp; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``M = Math.max(k, N - k); ` `    ``for` `(``int` `i = ``0``; i < M; i++) ` `        ``S1 += arr[i]; ` ` `  `    ``// Calculating max_difference ` `    ``max_difference = S1 - (S - S1); ` `    ``return` `max_difference; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `arr[] = { ``8``, ``4``, ``5``, ``2``, ``10` `}; ` `    ``int` `N = arr.length; ` `    ``int` `k = ``2``; ` `    ``System.out.println(maxDifference(arr, N, k)); ` `} ` `} ` ` `  `// This code is contributed by ` `// Surendra_Gangwar `

 `# Python3 code to calculate max_difference ` `# between the sum of two subarrays of ` `# length k and N - k ` ` `  `# Function to calculate max_difference ` `def` `maxDifference(arr, N, k ): ` `    ``S ``=` `0` `    ``S1 ``=` `0` `    ``max_difference ``=` `0` `     `  `    ``# Sum of the array ` `    ``for` `i ``in` `range``(N): ` `        ``S ``+``=` `arr[i] ` `     `  `    ``# Sort the array in descending order ` `    ``arr.sort(reverse``=``True``) ` `    ``M ``=` `max``(k, N ``-` `k) ` `    ``for` `i ``in` `range``( M): ` `        ``S1 ``+``=` `arr[i] ` `     `  `    ``# Calculating max_difference ` `    ``max_difference ``=` `S1 ``-` `(S ``-` `S1) ` `    ``return` `max_difference ` `     `  `# Driver Code ` `arr ``=` `[ ``8``, ``4``, ``5``, ``2``, ``10` `] ` `N ``=` `len``(arr) ` `k ``=` `2` `print``(maxDifference(arr, N, k)) ` ` `  `# This code is contributed by "Sharad_Bhardwaj". `

 `// C# program to calculate max_difference between ` `// the sum of two subarrays of length k and N - k ` `using` `System;  ` ` `  `class` `GFG ` `{ ` ` `  `// Function to calculate max_difference ` `static` `int` `maxDifference(``int` `[]arr, ``int` `N, ``int` `k) ` `{ ` `    ``int` `M, S = 0, S1 = 0, max_difference = 0; ` ` `  `    ``// Sum of the array ` `    ``for` `(``int` `i = 0; i < N; i++) ` `        ``S += arr[i]; ` `    ``int` `temp; ` `     `  `    ``// Sort the array in descending order ` `    ``for` `(``int` `i = 0; i < N; i++)  ` `    ``{ ` `        ``for` `(``int` `j = i + 1; j < N; j++)  ` `        ``{ ` `            ``if` `(arr[i] < arr[j])  ` `            ``{ ` `                ``temp = arr[i]; ` `                ``arr[i] = arr[j]; ` `                ``arr[j] = temp; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``M = Math.Max(k, N - k); ` `    ``for` `(``int` `i = 0; i < M; i++) ` `        ``S1 += arr[i]; ` ` `  `    ``// Calculating max_difference ` `    ``max_difference = S1 - (S - S1); ` `    ``return` `max_difference; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = { 8, 4, 5, 2, 10 }; ` `    ``int` `N = arr.Length; ` `    ``int` `k = 2; ` `    ``Console.Write(maxDifference(arr, N, k)); ` `} ` `} ` ` `  `// This code is contributed by mohit kumar 29 ` ` `

 ` `

Output :
```17
```

Further Optimizations : We can use Heap (or priority queue) to find M largest elements efficiently. Refer k largest(or smallest) elements in an array for details.

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