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Reverse a subarray to maximize sum of even-indexed elements of given array

  • Difficulty Level : Hard
  • Last Updated : 02 Aug, 2021

Given an array arr[], the task is to maximize the sum of even-indexed elements by reversing a subarray and print the maximum sum obtained.

Examples: 

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Input: arr[] = {1, 2, 1, 2, 1} 
Output:
Explanation: 
Sum of initial even-indexed elements = a[0] + a[2] + a[4] = 1 + 1 + 1 = 3 
Reversing subarray {1, 2, 1, 2} modifies the array to {2, 1, 2, 1, 1}. 
Hence, the maximized sum = 2 + 2 + 1 = 5



Input: arr[] = {7, 8, 4, 5, 7, 6, 8, 9, 7, 3} 
Output: 37 

Naive Approach: 
The simplest approach to solve the problem is to generate all the possible permutations by reversal of elements one by one and calculate the sum at even indices for each permutation. Print the maximum possible sum among all the permutations. 

Time Complexity: O(N3
Auxiliary Space: O(N)

Efficient Approach: 
The above approach can be further optimized to O(N) computational complexity by using Dynamic Programming to check the maximum difference by rotation of arrays.
Follow the steps below to solve the problem: 

  • Compare the elements at odd index with even index and also keep track of them.
  • Initialize two arrays leftDP[] and rightDP[].
  • For every odd index, leftDP[] stores the difference of the element at current index with the element on its left and rightDP[] stores that of the right.
  • If the difference calculated for the previous index is positive, add it to the current difference:

if(dp[i – 1] > 0) 
dp[i] = dp[i-1] + curr_diff 
 

  • Otherwise, store the current difference:

dp[i] = curr_diff; 
 

Below is the implementation of the above approach: 

C++




// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return maximized sum
// at even indices
int maximizeSum(int arr[], int n)
{
    int sum = 0;
    for(int i = 0; i < n; i = i + 2)
        sum += arr[i];
 
    // Stores difference with
    // element on the left
    int leftDP[n / 2];
 
    // Stores difference with
    // element on the right
    int rightDP[n / 2];
 
    int c = 0;
 
    for(int i = 1; i < n; i = i + 2)
    {
         
        // Compute and store
        // left difference
        int leftDiff = arr[i] - arr[i - 1];
 
        // For first index
        if (c - 1 < 0)
            leftDP = leftDiff;
 
        else
        {
             
            // If previous difference
            // is positive
            if (leftDP > 0)
                leftDP = leftDiff + leftDP;
 
            // Otherwise
            else
                leftDP[i] = leftDiff;
        }
 
        int rightDiff;
 
        // For the last index
        if (i + 1 >= n)
            rightDiff = 0;
 
        // Otherwise
        else
            rightDiff = arr[i] - arr[i + 1];
 
        // For first index
        if (c - 1 < 0)
            rightDP = rightDiff;
        else
        {
             
            // If the previous difference
            // is positive
            if (rightDP > 0)
                rightDP = rightDiff +
                             rightDP;
            else
                rightDP = rightDiff;
        }
        c++;
    }
    int maxi = 0;
    for(int i = 0; i < n / 2; i++)
    {
        maxi = max(maxi, max(leftDP[i],
                            rightDP[i]));
    }
    return maxi + sum;
}
 
// Driver Code
int main()
{
    int arr[] = { 7, 8, 4, 5, 7,
                  6, 8, 9, 7, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int ans = maximizeSum(arr, n);
     
    cout << (ans);
}
 
// This code is contributed by chitranayal

Java




// Java Program to implement
// the above approach
import java.io.*;
 
class GFG {
 
    // Function to return maximized sum
    // at even indices
    public static int maximizeSum(int[] arr)
    {
 
        int n = arr.length;
        int sum = 0;
        for (int i = 0; i < n; i = i + 2)
            sum += arr[i];
 
        // Stores difference with
        // element on the left
        int leftDP[] = new int[n / 2];
 
        // Stores difference with
        // element on the right
        int rightDP[] = new int[n / 2];
 
        int c = 0;
 
        for (int i = 1; i < n; i = i + 2) {
 
            // Compute and store
            // left difference
            int leftDiff = arr[i]
                           - arr[i - 1];
 
            // For first index
            if (c - 1 < 0)
                leftDP = leftDiff;
 
            else {
 
                // If previous difference
                // is positive
                if (leftDP > 0)
                    leftDP = leftDiff
                                + leftDP;
 
                // Otherwise
                else
                    leftDP[i] = leftDiff;
            }
 
            int rightDiff;
 
            // For the last index
            if (i + 1 >= arr.length)
                rightDiff = 0;
 
            // Otherwise
            else
                rightDiff = arr[i]
                            - arr[i + 1];
 
            // For first index
            if (c - 1 < 0)
                rightDP = rightDiff;
            else {
 
                // If the previous difference
                // is positive
                if (rightDP > 0)
                    rightDP = rightDiff
                                 + rightDP;
                else
                    rightDP = rightDiff;
            }
            c++;
        }
        int max = 0;
        for (int i = 0; i < n / 2; i++) {
            max = Math.max(max,
                           Math.max(
                               leftDP[i],
                               rightDP[i]));
        }
 
        return max + sum;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = { 7, 8, 4, 5, 7, 6,
                      8, 9, 7, 3 };
        int ans = maximizeSum(arr);
        System.out.println(ans);
    }
}

Python3




# Python3 program to implement
# the above approach
 
# Function to return maximized sum
# at even indices
def maximizeSum(arr):
 
    n = len(arr)
    sum = 0
 
    for i in range(0, n, 2):
        sum += arr[i]
 
    # Stores difference with
    # element on the left
    leftDP = [0] * (n)
 
    # Stores difference with
    # element on the right
    rightDP = [0] * (n)
 
    c = 0
    for i in range(1, n, 2):
 
        # Compute and store
        # left difference
        leftDiff = arr[i] - arr[i - 1]
 
        # For first index
        if (c - 1 < 0):
            leftDP[i] = leftDiff
        else:
 
            # If previous difference
            # is positive
            if (leftDP[i] > 0):
                leftDP[i] = (leftDiff +
                             leftDP[i - 1])
 
            # Otherwise
            else:
                leftDP[i] = leftDiff
 
        rightDiff = 0
 
        # For the last index
        if (i + 1 >= len(arr)):
            rightDiff = 0
             
        # Otherwise
        else:
            rightDiff = arr[i] - arr[i + 1]
 
        # For first index
        if (c - 1 < 0):
            rightDP[i] = rightDiff
        else:
 
            # If the previous difference
            # is positive
            if (rightDP[i] > 0):
                rightDP[i] = (rightDiff +
                              rightDP[i - 1])
            else:
                rightDP[i] = rightDiff
                 
        c += 1
 
    maxm = 0
 
    for i in range(n // 2):
        maxm = max(maxm, max(leftDP[i],
                            rightDP[i]))
 
    return maxm + sum
 
# Driver Code
if __name__ == '__main__':
 
    arr = [ 7, 8, 4, 5, 7,
            6, 8, 9, 7, 3 ]
    ans = maximizeSum(arr)
 
    print(ans)
 
# This code is contributed by mohit kumar 29

C#




// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to return maximized sum
// at even indices
public static int maximizeSum(int[] arr)
{
    int n = arr.Length;
    int sum = 0;
     
    for(int i = 0; i < n; i = i + 2)
        sum += arr[i];
 
    // Stores difference with
    // element on the left
    int []leftDP = new int[n / 2];
 
    // Stores difference with
    // element on the right
    int []rightDP = new int[n / 2];
 
    int c = 0;
 
    for(int i = 1; i < n; i = i + 2)
    {
         
        // Compute and store
        // left difference
        int leftDiff = arr[i] - arr[i - 1];
 
        // For first index
        if (c - 1 < 0)
            leftDP = leftDiff;
             
        else
        {
             
            // If previous difference
            // is positive
            if (leftDP > 0)
                leftDP = leftDiff +
                            leftDP;
                             
            // Otherwise
            else
                leftDP = leftDiff;
        }
 
        int rightDiff;
 
        // For the last index
        if (i + 1 >= arr.Length)
            rightDiff = 0;
 
        // Otherwise
        else
            rightDiff = arr[i] - arr[i + 1];
 
        // For first index
        if (c - 1 < 0)
            rightDP = rightDiff;
             
        else
        {
             
            // If the previous difference
            // is positive
            if (rightDP > 0)
                rightDP = rightDiff +
                             rightDP;
            else
                rightDP = rightDiff;
        }
        c++;
    }
     
    int max = 0;
     
    for(int i = 0; i < n / 2; i++)
    {
        max = Math.Max(max,
                       Math.Max(leftDP[i],
                               rightDP[i]));
    }
    return max + sum;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 7, 8, 4, 5, 7, 6,
                  8, 9, 7, 3 };
    int ans = maximizeSum(arr);
     
    Console.WriteLine(ans);
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
// Javascript Program to implement
// the above approach
 
// Function to return maximized sum
    // at even indices
function maximizeSum(arr)
{
    let n = arr.length;
        let sum = 0;
        for (let i = 0; i < n; i = i + 2)
            sum += arr[i];
  
        // Stores difference with
        // element on the left
        let leftDP = new Array(Math.floor(n / 2));
  
        // Stores difference with
        // element on the right
        let rightDP = new Array(Math.floor(n / 2));
  
         for(let i=0;i<n/2;i++)
        {
            leftDP[i]=0;
            rightDP[i]=0;
        }
          
        let c = 0;
  
        for (let i = 1; i < n; i = i + 2) {
  
            // Compute and store
            // left difference
            let leftDiff = arr[i]
                           - arr[i - 1];
  
            // For first index
            if (c - 1 < 0)
                leftDP[i] = leftDiff;
  
            else {
  
                // If previous difference
                // is positive
                if (leftDP[i] > 0)
                    leftDP[i] = leftDiff
                                + leftDP[i-1];
  
                // Otherwise
                else
                    leftDP[i] = leftDiff;
            }
  
            let rightDiff;
  
            // For the last index
            if (i + 1 >= arr.length)
                rightDiff = 0;
  
            // Otherwise
            else
                rightDiff = arr[i]
                            - arr[i + 1];
  
            // For first index
            if (c - 1 < 0)
                rightDP[i] = rightDiff;
            else {
  
                // If the previous difference
                // is positive
                if (rightDP[i] > 0)
                    rightDP[i] = rightDiff
                                 + rightDP[i-1];
                else
                    rightDP[i] = rightDiff;
            }
            c++;
        }
        let max = 0;
        for (let i = 0; i < n / 2; i++) {
            max = Math.max(max,
                           Math.max(
                               leftDP[i],
                               rightDP[i]));
        }
  
        return max + sum;
}
 
// Driver Code
let arr=[7, 8, 4, 5, 7, 6,
                      8, 9, 7, 3];
let ans = maximizeSum(arr);
document.write(ans);
                 
// This code is contributed by avanitrachhadiya2155
</script>
Output: 
37

 

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




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