Maximum sum subarray after altering the array

Given an array arr[] of size N. The task is to find the maximum subarray sum possible after performing the given operation at most once. In a single operation, you can choose any index i and either the subarray arr[0…i] or the subarray arr[i…N-1] can be reversed.

Examples:

Input: arr[] = {3, 4, -2, 1, 3}
Output: 11
After reversing arr[0…2], arr[] = {-2, 4, 3, 1, 3}



Input: arr[] = {-3, 5, -1, 2, 3}
Output: 10
Reverse arr[2…4], arr[] = {-3, 5, 3, 2, -1}

Naive approach: Use the Kadane’s algorithm to find the maximum subarray sum for the following cases:

  1. Find the maximum subarray sum for the original array i.e. when no operation is performed.
  2. Find the maximum subarray sum after reversing the subarray arr[0…i] for all possible values of i.
  3. Find the maximum subarray sum after reversing the subarray arr[i…N-1] for all possible values of i.

Print the maximum subarray sum so far in the end.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the maximum subarray sum
int maxSumSubarray(vector<int> arr, int size)
{
  
    int max_so_far = INT_MIN, max_ending_here = 0;
  
    for (int i = 0; i < size; i++) {
        max_ending_here = max_ending_here + arr[i];
        if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
  
        if (max_ending_here < 0)
            max_ending_here = 0;
    }
    return max_so_far;
}
  
// Function to reverse the subarray arr[0...i]
void getUpdatedArray(vector<int>& arr,
                     vector<int>& copy, int i)
{
    for (int j = 0; j <= (i / 2); j++) {
        copy[j] = arr[i - j];
        copy[i - j] = arr[j];
    }
    return;
}
  
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
int maxSum(vector<int> arr, int size)
{
  
    // To store the result
    int resSum = INT_MIN;
  
    // When no operation is performed
    resSum = max(resSum, maxSumSubarray(arr, size));
  
    // Find the maximum subarray sum after
    // reversing the subarray arr[0...i]
    // for all possible values of i
    vector<int> copyArr = arr;
    for (int i = 1; i < size; i++) {
        getUpdatedArray(arr, copyArr, i);
        resSum = max(resSum,
                     maxSumSubarray(copyArr, size));
    }
  
    // Find the maximum subarray sum after
    // reversing the subarray arr[i...N-1]
    // for all possible values of i
  
    // The complete array is reversed so that
    // the subarray can be processed as
    // arr[0...i] instead of arr[i...N-1]
    reverse(arr.begin(), arr.end());
    copyArr = arr;
    for (int i = 1; i < size; i++) {
        getUpdatedArray(arr, copyArr, i);
        resSum = max(resSum,
                     maxSumSubarray(copyArr, size));
    }
  
    return resSum;
}
  
// Driver code
int main()
{
    vector<int> arr{ -9, 21, 24, 24, -51, -6,
                     17, -42, -39, 33 };
    int size = arr.size();
  
    cout << maxSum(arr, size);
  
    return 0;
}

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Output:

102

Efficient approach: In this approach, apply the Kadane’s algorithm to find the subarray with the maximum sum that will be the first solution i.e. no operation is performed yet. Now, do some precomputation to avoid repetition.
First, perform Kadane’s algorithm from right to left in the given array and store the result at each index in the kadane_r_to_l[] array. Basically this array will give the maximum sub_array sum for arr[i…N-1] for each valid i.
Now, perform the preffix_sum of the given array. On the resulting array, perform the following operation.
For each valid i, preffix_sum[i] = max(prefix_sum[i – 1], prefix_sum[i]). We will use this array to get the max prefix sum among all prefixes in the sub_array prefix_sum[0…i].
Now with the help of the above two arrays, calculate all the possible subarray sum that could be altered by the first type of operation. Logic is very simple, find the maximum prefix sum in the arr[0…i] and maximum sub_array sum in arr[i+1…N]. After reversing the first part, max_prefix_sum of arr[i…0] and maximum sub_array sum in arr[i+1…N] will be all together in contiguous manner that will give the subarray with max_sum in arr[0…N].

Now for each i from 0 to N – 2, the summation of prefix_sum[i] + kadane_r_to_l[i + 1] will give the the max subarray sum for each iteration. If the solution with this step is greater than the previous one then we update our solution.

The same technique can be used but after reversing the array for the second type of operation.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns true if all
// the array element are <= 0
bool areAllNegative(vector<int> arr)
{
    for (int i = 0; i < arr.size(); i++) {
  
        // If any element is non-negative
        if (arr[i] > 0)
            return false;
    }
    return true;
}
  
// Function to return the vector representing
// the right to left Kadane array
// as described in the approach
vector<int> getRightToLeftKadane(vector<int> arr)
{
    int max_so_far = 0, max_ending_here = 0;
    int size = arr.size();
    for (int i = size - 1; i >= 0; i--) {
        max_ending_here = max_ending_here + arr[i];
        if (max_ending_here < 0)
            max_ending_here = 0;
        else if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
        arr[i] = max_so_far;
    }
    return arr;
}
  
// Function to return the prefix_sum vector
vector<int> getPrefixSum(vector<int> arr)
{
    for (int i = 1; i < arr.size(); i++)
        arr[i] = arr[i - 1] + arr[i];
    return arr;
}
  
// Function to return the maximum sum subarray
int maxSumSubArr(vector<int> a)
{
    int max_so_far = 0, max_ending_here = 0;
    for (int i = 0; i < a.size(); i++) {
        max_ending_here = max_ending_here + a[i];
        if (max_ending_here < 0)
            max_ending_here = 0;
        else if (max_so_far < max_ending_here)
            max_so_far = max_ending_here;
    }
    return max_so_far;
}
  
// Function to get the maximum sum subarray
// in the modified array
int maxSumSubWithOp(vector<int> arr)
{
  
    // kadane_r_to_l[i] will store the maximum subarray
    // sum for thre subarray arr[i...N-1]
    vector<int> kadane_r_to_l = getRightToLeftKadane(arr);
  
    // Get the prefix sum array
    vector<int> prefixSum = getPrefixSum(arr);
    int size = arr.size();
  
    for (int i = 1; i < size; i++) {
  
        // To get max_prefix_sum_at_any_index
        prefixSum[i] = max(prefixSum[i - 1], prefixSum[i]);
    }
  
    int max_subarray_sum = 0;
  
    for (int i = 0; i < size - 1; i++) {
  
        // Summation of both gives the maximum subarray
        // sum after applying the operation
        max_subarray_sum
            = max(max_subarray_sum,
                  prefixSum[i] + kadane_r_to_l[i + 1]);
    }
    return max_subarray_sum;
}
  
// Function to return the maximum
// subarray sum after performing the
// given operation at most once
int maxSum(vector<int> arr, int size)
{
  
    // If all element are negative then
    // return the maximum element
    if (areAllNegative(arr)) {
        return (*max_element(arr.begin(), arr.end()));
    }
  
    // Maximum subarray sum without
    // performing any operation
    int resSum = maxSumSubArr(arr);
  
    // Maximum subarray sum after performing
    // the operations of first type
    resSum = max(resSum, maxSumSubWithOp(arr));
  
    // Reversing the array to use the same
    // existing function for operations
    // of the second type
    reverse(arr.begin(), arr.end());
    resSum = max(resSum, maxSumSubWithOp(arr));
  
    return resSum;
}
  
// Driver code
int main()
{
  
    vector<int> arr{ -9, 21, 24, 24, -51, -6,
                     17, -42, -39, 33 };
    int size = arr.size();
  
    cout << maxSum(arr, size);
  
    return 0;
}

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Output:

102


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