Maximize the difference of sum of elements at even indices and odd indices by shifting an odd sized subarray to end of given Array.

Last Updated : 31 May, 2022

Given an array arr[] of size N, the task is to maximize the difference of the sum of elements at even indices and elements at odd indices by shifting any subarray of odd length to the end of the array.

Examples:

Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: 3
Explanation: Initially sum of elements at even indices = 1 + 3 + 5 = 9
Sum of elements at odd indices = 2 + 4 + 6 = 12
Difference = 9 -12 = -3
On shifting the subarray from [0, 2] to the end the array becomes {4, 5, 6, 1, 2, 3}
Sum of elements at even indices = 4 + 6 + 2 = 12
Sum of elements at odd indices = 5 + 1 + 3 = 9
Difference = 12 – 9 = 3
This gives the maximum answer out of all such shifts that is equal to 3.

Input: arr[] = {3, 4, 8}
Output: 7

Approach: The task can be solved mathematically, using the prefix sum technique. The problem can be broken down into 4 cases :

Case 1: When length of the array N is even and l (start of subarray) is even {a,b,c,d,e,f} for l = 2, r =4 (one based indexing)
Initial: +a – b + c – d + e -f on shifting sub array of [2,4] to the end the array becomes ={a, e, f, b, c, d}
Final: + a- e +f – b + c-d
On diving the array to 3 subparts [1, l) , [l, r] , (r, N].
The sign of part1 [1, l) remains same, part2 [l, r] remains same and part3 (r, N] becomes negative (opposite sign) after shifting.
If prefix_sum array is computed then sum after shifting becomes
sum1 = prefix_sum(l-1) + ( prefix_sum(r) – prefix_sum(l-1) ) – ((prefix_sum(n) – prefix_sum(r))
sum1 = 2* prefix_sum(r) – prefix_sum(N)

Case 2: When length of the array N is even and l (start of subarray) is odd {a, b, c, d, e, f} for l = 3, r = 5(one based indexing)
Initial: +a – b + c – d + e -f on shifting [3,5] to the end the array becomes = {a, b, f, c, d, e}
Final: + a – b + f – c + d – e
On diving the array to 3 subparts [1, l) , [l, r] , (r, N].
The sign of part1 [1, l) remains same, part2 [l, r] becomes opposite sign and part3 (r, N] becomes opposite sign after shifting.
If prefix_sum array is computed then sum after shifting becomes
sum2 = prefix_sum(l-1) – ( prefix_sum(r) – prefix_sum(l-1)) – ((prefix_sum(n) – prefix_sum(r))
sum2 = 2* prefix_sum(l-1) – prefix_sum(N)

Case 3: When length of the array N is odd and l (start of subarray) is even {a, b, c, d, e} for l = 2, r = 4 (one based indexing)
Initial: +a – b + c – d + e on shifting sub array of [2,4] to the end the array becomes = {a, e, b, c, d}
Final: + a – e + b – c + d
On diving the array to 3 subparts [1, l) , [l, r] , (r, N].
The sign of part1 [1, l) remains same, part2 [l, r] becomes opposite sign and part3 (r, N] becomes negative(opposite sign) after shifting.
If prefix_sum array is computed then sum after shifting becomes
sum3 = prefix_sum(l-1) – (prefix_sum(r) – prefix_sum(l-1)) – ((prefix_sum(n) – prefix_sum(r))
sum3 = 2* prefix_sum(l-1) – prefix_sum(N)

Case 4: When length of the array N is odd and l (start of subarray) is odd {a, b, c, d, e} for l = 3, r =3(one based indexing)
Initial: +a – b + c – d + e on shifting [3,5] to the end the array becomes = {a, b, d, e, c}
Final: + a – b + d – e + c
On diving the array to 3 subparts [1, l) , [l, r] , (r, N].
The sign of part1 [1, l) remains same, part2 [l, r] becomes opposite sign and part3 (r, N] remains same.
If prefix_sum array is computed then sum after shifting becomes
sum4 = prefix_sum(l-1) + (prefix_sum(r) – prefix_sum(l-1)) – ((prefix_sum(n) – prefix_sum(r))
sum4 = 2* prefix_sum(r) – prefix_sum(N)

If n is even result becomes: max(sum1, sum2).

If n is odd result becomes: max(sum3,sum4).

Follow the below steps to solve the problem:

Initialize a vector prefix_sum of size n+1 with 0’s.
Now modify the odd indices of the array by multiplying them with -1.
Iterate through the array in the range[1,n] and fill the prefix_sum vector.
If the size of the array is even
Initialize the variable maxval to -1e8 to store the max sum.
Iterate through the prefix sum vector and check
- If i is even assign maxval as max(maxval, 2 * prefix_sum[i] – prefix_sum[n]).
- Else assign maxval as max(maxval, 2 * prefix_sum[i] – prefix_sum[n]).
If the size of the array is odd.
Initialize the variable maxval to -1e8 to store the max sum.
Iterate through the prefix sum vector and check
- If i is even assign maxval as max(maxval, 2 * prefix_sum[i – 1] – prefix_sum[n])
- Else assign maxval as max(maxval, 2 * prefix_sum[i] – prefix_sum[n])

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum value of
// difference between sum of elements 
// at even indices and sum of elements 
// at odd indices by shifting
// a subarray of odd length to the
// end of the array
int find_maxsum(int arr[], int n)
{
    vector<int> prefix_sum(n + 1, 0);
     
      // Modify the array to keep 
    // alternative +ve and -ve
    for (int i = 0; i < n; i++) {
        if (i % 2 == 1) {
            arr[i] = -arr[i];
        }
    }
   
    // Function to find the prefix sum
    for (int i = 1; i <= n; i++) {
        prefix_sum[i] = prefix_sum[i - 1] 
            + arr[i - 1];
    }
       
    // If n is even
    if (n % 2 == 0) {
         
          // Initialize the maxval to -1e8
        int maxval = -1e8;
        for (int i = 1; i <= n; i++) {
             
              // If i is even (case-1)
            if (i % 2 == 0) {
                maxval = max(maxval, 2 * 
                             prefix_sum[i] 
                             - prefix_sum[n]);
            }
           
            // If i is odd (case-2)
            else {
                maxval = max(maxval, 2 * 
                             prefix_sum[i - 1]
                             - prefix_sum[n]);
            }
        }
       
        // return the maxval
        return maxval;
    }
    else {
        // Initialize the maxval to -1e8
        int maxval = -1e8;
        for (int i = 1; i <= n; i++) {
             
              // If i is even (case 3)
            if (i % 2 == 0) {
                maxval = max(maxval, 2 * 
                             prefix_sum[i - 1]
                             - prefix_sum[n]);
            }
           
            // If i is odd (case 4)
            else {
                maxval = max(maxval, 2 * 
                             prefix_sum[i]
                             - prefix_sum[n]);
            }
        }
       
        // Return the maxval
        return maxval;
    }
}
 
int main()
{
    // Initialize an array
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int N = sizeof(arr) / sizeof(arr[0]);
   
    // Function call
    cout << find_maxsum(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
 
  // Function to find the maximum value of
  // difference between sum of elements 
  // at even indices and sum of elements 
  // at odd indices by shifting
  // a subarray of odd length to the
  // end of the array
  static int find_maxsum(int arr[], int n)
  {
    int prefix_sum[] = new int[n + 1];
    for(int i = 0; i < n + 1; i++) {
      prefix_sum[i] = 0;
    }
 
    // Modify the array to keep 
    // alternative +ve and -ve
    for (int i = 0; i < n; i++) {
      if (i % 2 == 1) {
        arr[i] = -arr[i];
      }
    }
 
    // Function to find the prefix sum
    for (int i = 1; i <= n; i++) {
      prefix_sum[i] = prefix_sum[i - 1] 
        + arr[i - 1];
    }
 
    // If n is even
    if (n % 2 == 0) {
 
      // Initialize the maxval to -1e8
      int maxval = (int)-1e8;
      for (int i = 1; i <= n; i++) {
 
        // If i is even (case-1)
        if (i % 2 == 0) {
          maxval = Math.max(maxval, 2 * 
                            prefix_sum[i] 
                            - prefix_sum[n]);
        }
 
        // If i is odd (case-2)
        else {
          maxval = Math.max(maxval, 2 * 
                            prefix_sum[i - 1]
                            - prefix_sum[n]);
        }
      }
 
      // return the maxval
      return maxval;
    }
    else
    {
       
      // Initialize the maxval to -1e8
      int maxval = (int)-1e8;
      for (int i = 1; i <= n; i++) {
 
        // If i is even (case 3)
        if (i % 2 == 0) {
          maxval = Math.max(maxval, 2 * 
                            prefix_sum[i - 1]
                            - prefix_sum[n]);
        }
 
        // If i is odd (case 4)
        else {
          maxval = Math.max(maxval, 2 * 
                            prefix_sum[i]
                            - prefix_sum[n]);
        }
      }
 
      // Return the maxval
      return maxval;
    }
  }
 
  public static void main(String args[])
  {
     
    // Initialize an array
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int N = arr.length;
 
    // Function call
    System.out.println(find_maxsum(arr, N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python program for the above approach
 
# Function to find the maximum value of
# difference between sum of elements 
# at even indices and sum of elements 
# at odd indices by shifting
# a subarray of odd length to the
# end of the array
def find_maxsum(arr, n):
    prefix_sum = [ 0 for i in range(n + 1)]
     
    # Modify the array to keep 
    # alternative +ve and -ve
    for i in range(n):
        if (i % 2 == 1):
            arr[i] = -arr[i]
             
    # Function to find the prefix sum
    for i in range(1, n+1):
        prefix_sum[i] = prefix_sum[i - 1] + arr[i - 1]
         
    # If n is even
    if (n % 2 == 0):
         
        # Initialize the maxval to -1e8
        maxval = -10**8
        for i in range(1,n+1):
            # If i is even (case-1)
            if (i % 2 == 0):
                maxval = max(maxval, 2 * prefix_sum[i] - prefix_sum[n])
                 
            # If i is odd (case-2)
            else:
                maxval = max(maxval, 2 * prefix_sum[i - 1]- prefix_sum[n])
                 
        # return the maxval
        return maxval
     
    else:
        # Initialize the maxval to -1e8
        maxval = -10**8
        for i in range(1, n+1):
             
            # If i is even (case 3)
            if (i % 2 == 0):
                maxval = max(maxval, 2 * prefix_sum[i - 1] - prefix_sum[n])
             
            # If i is odd (case 4)
            else:
                maxval = max(maxval, 2 * prefix_sum[i] - prefix_sum[n])
                 
        # Return the maxval
        return maxval
         
# Initialize an array
arr = [1, 2, 3, 4, 5, 6]
N = len(arr)
 
# Function call
print(find_maxsum(arr, N))
 
# This code is contributed by Shubham Singh

C#

// C# code to implement the above approach
using System;
public class GFG{
 
  // Function to find the maximum value of
  // difference between sum of elements 
  // at even indices and sum of elements 
  // at odd indices by shifting
  // a subarray of odd length to the
  // end of the array
  static int find_maxsum(int[] arr, int n)
  {
    int[] prefix_sum = new int[n + 1];
    for(int i = 0; i < n + 1; i++) {
      prefix_sum[i] = 0;
    }
 
    // Modify the array to keep 
    // alternative +ve and -ve
    for (int i = 0; i < n; i++) {
      if (i % 2 == 1) {
        arr[i] = -arr[i];
      }
    }
 
    // Function to find the prefix sum
    for (int i = 1; i <= n; i++) {
      prefix_sum[i] = prefix_sum[i - 1] 
        + arr[i - 1];
    }
 
    // If n is even
    if (n % 2 == 0) {
 
      // Initialize the maxval to -1e8
      int maxval = (int)-1e8;
      for (int i = 1; i <= n; i++) {
 
        // If i is even (case-1)
        if (i % 2 == 0) {
          maxval = Math.Max(maxval, 2 * 
                            prefix_sum[i] 
                            - prefix_sum[n]);
        }
 
        // If i is odd (case-2)
        else {
          maxval = Math.Max(maxval, 2 * 
                            prefix_sum[i - 1]
                            - prefix_sum[n]);
        }
      }
 
      // return the maxval
      return maxval;
    }
    else
    {
 
      // Initialize the maxval to -1e8
      int maxval = (int)-1e8;
      for (int i = 1; i <= n; i++) {
 
        // If i is even (case 3)
        if (i % 2 == 0) {
          maxval = Math.Max(maxval, 2 * 
                            prefix_sum[i - 1]
                            - prefix_sum[n]);
        }
 
        // If i is odd (case 4)
        else {
          maxval = Math.Max(maxval, 2 * 
                            prefix_sum[i]
                            - prefix_sum[n]);
        }
      }
 
      // Return the maxval
      return maxval;
    }
  }
 
  // Driver code
  public static void Main()
  {
 
    // Initialize an array
    int[] arr = { 1, 2, 3, 4, 5, 6 };
    int N = arr.Length;
 
    // Function call
    Console.Write(find_maxsum(arr, N));
  }
}
 
// This code is contributed by Shubham Singh

Javascript

<script>
// Javascript program for the above approach
 
 
// Function to find the maximum value of
// difference between sum of elements 
// at even indices and sum of elements 
// at odd indices by shifting
// a subarray of odd length to the
// end of the array
function find_maxsum(arr, n)
{
    let prefix_sum = new Array(n + 1).fill(0);
 
    // Modify the array to keep 
    // alternative +ve and -ve
    for (let i = 0; i < n; i++) {
        if (i % 2 == 1) {
            arr[i] = -arr[i];
        }
    }
 
    // Function to find the prefix sum
    for (let i = 1; i <= n; i++) {
        prefix_sum[i] = prefix_sum[i - 1]
            + arr[i - 1];
    }
 
    // If n is even
    if (n % 2 == 0) {
 
        // Initialize the maxval to -1e8
        let maxval = -1e8;
        for (let i = 1; i <= n; i++) {
 
            // If i is even (case-1)
            if (i % 2 == 0) {
                maxval = Math.max(maxval, 2 *
                    prefix_sum[i]
                    - prefix_sum[n]);
            }
 
            // If i is odd (case-2)
            else {
                maxval = Math.max(maxval, 2 *
                    prefix_sum[i - 1]
                    - prefix_sum[n]);
            }
        }
 
        // return the maxval
        return maxval;
    }
    else {
        // Initialize the maxval to -1e8
        let maxval = -1e8;
        for (let i = 1; i <= n; i++) {
 
            // If i is even (case 3)
            if (i % 2 == 0) {
                maxval = Math.max(maxval, 2 *
                    prefix_sum[i - 1]
                    - prefix_sum[n]);
            }
 
            // If i is odd (case 4)
            else {
                maxval = Math.max(maxval, 2 *
                    prefix_sum[i]
                    - prefix_sum[n]);
            }
        }
 
        // Return the maxval
        return maxval;
    }
}
 
// Initialize an array
let arr = [1, 2, 3, 4, 5, 6];
let N = arr.length;
 
// Function call
document.write(find_maxsum(arr, N));
 
// This code is contributed by gfgking.
</script>

Output

Time Complexity: O(N), as we are using a loop to traverse N times.

Auxiliary Space: O(N), as we are using extra space for prefix_sum.

Maximum difference between sum of even and odd indexed elements of a Subarray

lokeshpotta20

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Practice Tags :

Maximize the difference of sum of elements at even indices and odd indices by shifting an odd sized subarray to end of given Array.

C++

Java

Python3

C#

Javascript

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