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

Given an array a[] consisting of N integers, the task is to find the maximum subarray sum that can be pbtained 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 without squaring 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++

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// C++ program to implement
// the above approach
#include <bits/stdc++.h>  
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

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Java

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// 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)); 
    

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Python3

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# 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 

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C#

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// 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

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Output: 

152

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

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