# Maximum Subarray Sum possible by replacing an Array element by its Square

• Difficulty Level : Hard
• Last Updated : 09 Jul, 2021

Given an array a[] consisting of N integers, the task is to find the maximum subarray sum that can be obtained by replacing a single array element by its square.

Examples:

Input: a[] = {1, -5, 8, 12, -8}
Output: 152
Explanation:
Replacing 12 by 144, the subarray {8, 144} generates the maximum possible subarray sum in the array.
Input: a[] = {-1, -2, -3}
Output:
Explanation:

Naive Approach: The simplest approach to solve the problem is to replace every element with its square and for each of them, find the maximum subarray sum using Kadane’s algorithm. Finally, print the maximum possible subarray sum obtained.
Time Complexity: O(N2
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using Dynamic Programming. Follow the steps below to solve the problem:

• Initialize memorization table dp[][] where:
• dp[i][0]: Stores the maximum subarray sum that can be obtained including ith element and without squaring any array element.
• dp[i][1]: Stores the maximum subarray sum that can be including ith element and squaring one of the array elements
• Therefore, the recurrence relations are:

dp[i][0] = max(dp[i-1][0] + a[i], a[i]), that is, either extend the previous subarray ending at i – 1th index or start a new subarray from ith index.
dp[i][1] = max(a[i]2, dp[i-1][0] + a[i]2, dp[i-1][1] + a[i]), that is, either start new subarray from ith index or extend previous subarray by adding a[i]2 to dp[i – 1][0] or add a[i] to dp[i – 1][1]

Below is the implementation of the above approach:

## C++

 `// C++ program to implement``// the above approach``#include  ``using` `namespace` `std; ` `// Function to find the maximum subarray``// sum possible``int` `getMaxSum(``int` `a[], ``int` `n)``{``    ``int` `dp[n][2];` `    ``// Stores sum without squaring``    ``dp[0][0] = a[0];` `    ``// Stores sum squaring``    ``dp[0][1] = a[0] * a[0];` `    ``// Stores the maximum subarray sum``    ``int` `max_sum = max(dp[0][0], dp[0][1]);``    ``for``(``int` `i = 1; i < n; i++)``    ``{``        ` `        ``// Either extend the subarray``        ``// or start a new subarray``        ``dp[i][0] = max(a[i],``                      ``dp[i - 1][0] + a[i]);` `        ``// Either extend previous squared``        ``// subarray or start a new subarray``        ``// by squaring the current element``        ``dp[i][1] = max(dp[i - 1][1] + a[i],``                               ``a[i] * a[i]);` `        ``dp[i][1] = max(dp[i][1],``                       ``dp[i - 1][0] +``                       ``a[i] * a[i]);` `        ``// Update maximum subarray sum``        ``max_sum = max(max_sum, dp[i][1]);``        ``max_sum = max(max_sum, dp[i][0]);``    ``}``    ` `    ``// Return answer``    ``return` `max_sum;``}``    ` `// Driver Code``int32_t main()``{``    ``int` `n = 5;``    ``int` `a[] = { 1, -5, 8, 12, -8 };` `    ``// Function call``    ``cout << getMaxSum(a, n) << endl;` `    ``return` `0;``}` `// This code is contributed by rutvik_56`

## Java

 `// Java Program to implement``// the above approach``import` `java.io.*;` `class` `GFG {` `    ``// Function to find the maximum subarray``    ``// sum possible``    ``public` `static` `int` `getMaxSum(``int` `a[], ``int` `n)``    ``{``        ``int` `dp[][] = ``new` `int``[n][``2``];` `        ``// Stores sum without squaring``        ``dp[``0``][``0``] = a[``0``];` `        ``// Stores sum squaring``        ``dp[``0``][``1``] = a[``0``] * a[``0``];` `        ``// Stores the maximum subarray sum``        ``int` `max_sum = Math.max(dp[``0``][``0``], dp[``0``][``1``]);``        ``for` `(``int` `i = ``1``; i < n; i++) {` `            ``// Either extend the subarray``            ``// or start a new subarray``            ``dp[i][``0``] = Math.max(a[i],``                                ``dp[i - ``1``][``0``] + a[i]);` `            ``// Either extend previous squared``            ``// subarray or start a new subarray``            ``// by squaring the current element``            ``dp[i][``1``] = Math.max(dp[i - ``1``][``1``] + a[i],``                                ``a[i] * a[i]);` `            ``dp[i][``1``]``                ``= Math.max(dp[i][``1``],``                        ``dp[i - ``1``][``0``] + a[i] * a[i]);` `            ``// Update maximum subarray sum``            ``max_sum = Math.max(max_sum, dp[i][``1``]);``            ``max_sum = Math.max(max_sum, dp[i][``0``]);``        ``}` `        ``// Return answer``        ``return` `max_sum;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``5``;``        ``int` `a[] = { ``1``, -``5``, ``8``, ``12``, -``8` `};` `        ``// Function call``        ``System.out.println(getMaxSum(a, n));``    ``}``}`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to find the maximum subarray``# sum possible``def` `getMaxSum(a, n):` `    ``dp ``=` `[[``0` `for` `x ``in` `range``(``2``)]``            ``for` `y ``in` `range``(n)]` `    ``# Stores sum without squaring``    ``dp[``0``][``0``] ``=` `a[``0``]` `    ``# Stores sum squaring``    ``dp[``0``][``1``] ``=` `a[``0``] ``*` `a[``0``]` `    ``# Stores the maximum subarray sum``    ``max_sum ``=` `max``(dp[``0``][``0``], dp[``0``][``1``])` `    ``for` `i ``in` `range``(``1``, n):` `        ``# Either extend the subarray``        ``# or start a new subarray``        ``dp[i][``0``] ``=` `max``(a[i],``                    ``dp[i ``-` `1``][``0``] ``+` `a[i])` `        ``# Either extend previous squared``        ``# subarray or start a new subarray``        ``# by squaring the current element``        ``dp[i][``1``] ``=` `max``(dp[i ``-` `1``][``1``] ``+` `a[i],``                        ``a[i] ``*` `a[i])` `        ``dp[i][``1``] ``=` `max``(dp[i][``1``],``                    ``dp[i ``-` `1``][``0``] ``+``                        ``a[i] ``*` `a[i])` `        ``# Update maximum subarray sum``        ``max_sum ``=` `max``(max_sum, dp[i][``1``])``        ``max_sum ``=` `max``(max_sum, dp[i][``0``])` `    ``# Return answer``    ``return` `max_sum` `# Driver Code``n ``=` `5``a ``=` `[ ``1``, ``-``5``, ``8``, ``12``, ``-``8` `]` `# Function call``print``(getMaxSum(a, n))` `# This code is contributed by Shivam Singh`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG{` `// Function to find the maximum subarray``// sum possible``public` `static` `int` `getMaxSum(``int` `[]a, ``int` `n)``{``    ``int` `[,]dp = ``new` `int``[n, 2];` `    ``// Stores sum without squaring``    ``dp[0, 0] = a[0];` `    ``// Stores sum squaring``    ``dp[0, 1] = a[0] * a[0];` `    ``// Stores the maximum subarray sum``    ``int` `max_sum = Math.Max(dp[0, 0], dp[0, 1]);``    ``for``(``int` `i = 1; i < n; i++)``    ``{``        ` `        ``// Either extend the subarray``        ``// or start a new subarray``        ``dp[i, 0] = Math.Max(a[i],``                        ``dp[i - 1, 0] + a[i]);` `        ``// Either extend previous squared``        ``// subarray or start a new subarray``        ``// by squaring the current element``        ``dp[i, 1] = Math.Max(dp[i - 1, 1] + a[i],``                            ``a[i] * a[i]);` `        ``dp[i, 1] = Math.Max(dp[i, 1],``                            ``dp[i - 1, 0] +``                            ``a[i] * a[i]);` `        ``// Update maximum subarray sum``        ``max_sum = Math.Max(max_sum, dp[i, 1]);``        ``max_sum = Math.Max(max_sum, dp[i, 0]);``    ``}` `    ``// Return answer``    ``return` `max_sum;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 5;``    ``int` `[]a = { 1, -5, 8, 12, -8 };` `    ``// Function call``    ``Console.WriteLine(getMaxSum(a, n));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``
Output:
`152`

Time Complexity: O(N)
Auxiliary Space: O(N)

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