Print indices of array elements whose removal makes the sum of odd and even-indexed elements equal

Last Updated : 03 Jun, 2021

Given an array arr[] of size N, the task is to find indices of array elements whose removal makes the sum of odd and even indexed elements equal. If no such element exists, then print -1.

Examples:

Input: arr[] = {4, 1, 6, 2}
Output: 1
Explanation:
Removing arr[1] modifies arr[] to {4, 6, 2}
Sum of even-indexed array elements = arr[0] + arr[2] = 4 + 2 = 6
Sum of odd-indexed array elements = arr[1] = 6
Therefore, the required output is 1.

Input:arr = {3, 2, 1}
Output: -1

Naive Approach: The simplest approach to solve this problem to traverse the array and remove the i^th element from the array and check if the sum of odd-indexed array elements equal to the sum of even-indexed array elements or not. If found to be true then print the index.

Time Complexity: O(N²)
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized using the Prefix Sum technique. The idea is to use the fact that If an array element removed from the index, then after that index, the even-indexed array elements become the odd-indexed and vice-versa. Follow the steps below to solve the problem:

Initialize two arrays, say odd[] and even[], to store the prefix sum of odd and even-indexed array elements and prefix sum of even-indexed array elements respectively.
Traverse the array arr[] and compute prefix sum of odd and even-indexed array elements.
Initialize two variables, say P = 0 and Q = 0, to store the sum of even and odd-indexed array elements respectively after removal of an array element.
Traverse the array and remove array elements one by one and update prefix sums accordingly. Check if the sum of even-indexed array elements, P, is equal to the sum of odd-indexed array elements, Q or not. If found to be true, then print the current index.
Otherwise, print -1.

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 find indices of array elements
// whose removal makes the sum of odd and
// even indexed array elements equal
void removeIndicesToMakeSumEqual(vector<int>& arr)
{
    // Stores size of array
    int N = arr.size();
 
    // Store prefix sum of odd
    // index array elements
    vector<int> odd(N, 0);
 
    // Store prefix sum of even
    // index array elements
    vector<int> even(N, 0);
 
    // Update even[0]
    even[0] = arr[0];
 
    // Traverse the given array
    for (int i = 1; i < N; i++) {
 
        // Update odd[i]
        odd[i] = odd[i - 1];
 
        // Update even[i]
        even[i] = even[i - 1];
 
        // If the current index
        // is an even number
        if (i % 2 == 0) {
 
            // Update even[i]
            even[i] += arr[i];
        }
 
        // If the current index
        // is an odd number
        else {
 
            // Update odd[i]
            odd[i] += arr[i];
        }
    }
 
    // Check if at least one
    // index found or not that
    // satisfies the condition
    bool find = 0;
 
    // Store odd indices sum by
    // removing 0-th index
    int p = odd[N - 1];
 
    // Store even indices sum by
    // removing 0-th index
    int q = even[N - 1] - arr[0];
 
    // If p and q are equal
    if (p == q) {
        cout << "0 ";
        find = 1;
    }
 
    // Traverse the array arr[]
    for (int i = 1; i < N; i++) {
 
        // If i is an even number
        if (i % 2 == 0) {
 
            // Update p by removing
            // the i-th element
            p = even[N - 1] - even[i - 1]
                - arr[i] + odd[i - 1];
 
            // Update q by removing
            // the i-th element
            q = odd[N - 1] - odd[i - 1]
                + even[i - 1];
        }
        else {
 
            // Update q by removing
            // the i-th element
            q = odd[N - 1] - odd[i - 1]
                - arr[i] + even[i - 1];
 
            // Update p by removing
            // the i-th element
            p = even[N - 1] - even[i - 1]
                + odd[i - 1];
        }
 
        // If odd index values sum is equal
        // to even index values sum
        if (p == q) {
 
            // Set the find variable
            find = 1;
 
            // Print the current index
            cout << i << " ";
        }
    }
 
    // If no index found
    if (!find) {
 
        // Print not possible
        cout << -1;
    }
}
 
// Driver Code
int main()
{
    vector<int> arr = { 4, 1, 6, 2 };
    removeIndicesToMakeSumEqual(arr);
 
    return 0;
}
 
  

Java

   // Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to find indices of array elements
// whose removal makes the sum of odd and
// even indexed array elements equal
static void removeIndicesToMakeSumEqual(int []arr)
{
     
    // Stores size of array
    int N = arr.length;
 
    // Store prefix sum of odd
    // index array elements
    int []odd = new int[N];
 
    // Store prefix sum of even
    // index array elements
    int []even = new int[N];
 
    // Update even[0]
    even[0] = arr[0];
 
    // Traverse the given array
    for(int i = 1; i < N; i++)
    {
         
        // Update odd[i]
        odd[i] = odd[i - 1];
 
        // Update even[i]
        even[i] = even[i - 1];
 
        // If the current index
        // is an even number
        if (i % 2 == 0) 
        {
             
            // Update even[i]
            even[i] += arr[i];
        }
 
        // If the current index
        // is an odd number
        else
        {
             
            // Update odd[i]
            odd[i] += arr[i];
        }
    }
 
    // Check if at least one
    // index found or not that
    // satisfies the condition
    boolean find = false;
 
    // Store odd indices sum by
    // removing 0-th index
    int p = odd[N - 1];
 
    // Store even indices sum by
    // removing 0-th index
    int q = even[N - 1] - arr[0];
 
    // If p and q are equal
    if (p == q) 
    {
        System.out.print("0 ");
        find = true;
    }
 
    // Traverse the array arr[]
    for(int i = 1; i < N; i++)
    {
         
        // If i is an even number
        if (i % 2 == 0) 
        {
             
            // Update p by removing
            // the i-th element
            p = even[N - 1] - even[i - 1] - 
                     arr[i] + odd[i - 1];
 
            // Update q by removing
            // the i-th element
            q = odd[N - 1] - odd[i - 1] + 
               even[i - 1];
        }
        else
        {
             
            // Update q by removing
            // the i-th element
            q = odd[N - 1] - odd[i - 1] - 
                    arr[i] + even[i - 1];
 
            // Update p by removing
            // the i-th element
            p = even[N - 1] - even[i - 1] + 
                 odd[i - 1];
        }
 
        // If odd index values sum is equal
        // to even index values sum
        if (p == q) 
        {
 
            // Set the find variable
            find = true;
 
            // Print the current index
            System.out.print(i + " ");
        }
    }
 
    // If no index found
    if (!find)
    {
         
        // Print not possible
        System.out.print(-1);
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 4, 1, 6, 2 };
     
    removeIndicesToMakeSumEqual(arr);
}
}
 
// This code is contributed by Amit Katiyar
 
  

Python3

   # Python program to implement
# the above approach
 
# Function to find indices of array elements
# whose removal makes the sum of odd and
# even indexed array elements equal
def removeIndicesToMakeSumEqual(arr):
   
    # Stores size of array
    N = len(arr);
 
    # Store prefix sum of odd
    # index array elements
    odd = [0] * N;
 
    # Store prefix sum of even
    # index array elements
    even = [0] * N;
 
    # Update even[0]
    even[0] = arr[0];
 
    # Traverse the given array
    for i in range(1, N):
 
        # Update odd[i]
        odd[i] = odd[i - 1];
 
        # Update even[i]
        even[i] = even[i - 1];
 
        # If the current index
        # is an even number
        if (i % 2 == 0):
 
            # Update even[i]
            even[i] += arr[i];
 
        # If the current index
        # is an odd number
        else:
 
            # Update odd[i]
            odd[i] += arr[i];
 
    # Check if at least one
    # index found or not that
    # satisfies the condition
    find = False;
 
    # Store odd indices sum by
    # removing 0-th index
    p = odd[N - 1];
 
    # Store even indices sum by
    # removing 0-th index
    q = even[N - 1] - arr[0];
 
    # If p and q are equal
    if (p == q):
        print("0 ");
        find = True;
 
    # Traverse the array arr
    for i in range(1, N):
 
        # If i is an even number
        if (i % 2 == 0):
 
            # Update p by removing
            # the i-th element
            p = even[N - 1] - even[i - 1] - arr[i] + odd[i - 1];
 
            # Update q by removing
            # the i-th element
            q = odd[N - 1] - odd[i - 1] + even[i - 1];
        else:
 
            # Update q by removing
            # the i-th element
            q = odd[N - 1] - odd[i - 1] - arr[i] + even[i - 1];
 
            # Update p by removing
            # the i-th element
            p = even[N - 1] - even[i - 1] + odd[i - 1];
 
        # If odd index values sum is equal
        # to even index values sum
        if (p == q):
           
            # Set the find variable
            find = True;
 
            # Print the current index
            print(i, end = "");
 
    # If no index found
    if (find == False):
       
        # Print not possible
        print(-1);
 
# Driver Code
if __name__ == '__main__':
    arr = [4, 1, 6, 2];
 
    removeIndicesToMakeSumEqual(arr);
 
    # This code is contributed by shikhasingrajput
 
  

C#

   // C# program to implement
// the above approach
using System;
 
class GFG{
     
// Function to find indices of array elements
// whose removal makes the sum of odd and
// even indexed array elements equal
static void removeIndicesToMakeSumEqual(int []arr)
{
     
    // Stores size of array
    int N = arr.Length;
 
    // Store prefix sum of odd
    // index array elements
    int []odd = new int[N];
 
    // Store prefix sum of even
    // index array elements
    int []even = new int[N];
 
    // Update even[0]
    even[0] = arr[0];
 
    // Traverse the given array
    for(int i = 1; i < N; i++)
    {
         
        // Update odd[i]
        odd[i] = odd[i - 1];
 
        // Update even[i]
        even[i] = even[i - 1];
 
        // If the current index
        // is an even number
        if (i % 2 == 0) 
        {
             
            // Update even[i]
            even[i] += arr[i];
        }
 
        // If the current index
        // is an odd number
        else
        {
             
            // Update odd[i]
            odd[i] += arr[i];
        }
    }
 
    // Check if at least one
    // index found or not that
    // satisfies the condition
    bool find = false;
 
    // Store odd indices sum by
    // removing 0-th index
    int p = odd[N - 1];
 
    // Store even indices sum by
    // removing 0-th index
    int q = even[N - 1] - arr[0];
 
    // If p and q are equal
    if (p == q) 
    {
        Console.Write("0 ");
        find = true;
    }
 
    // Traverse the array arr[]
    for(int i = 1; i < N; i++)
    {
         
        // If i is an even number
        if (i % 2 == 0) 
        {
             
            // Update p by removing
            // the i-th element
            p = even[N - 1] - even[i - 1] - 
                     arr[i] + odd[i - 1];
 
            // Update q by removing
            // the i-th element
            q = odd[N - 1] - odd[i - 1] + 
               even[i - 1];
        }
        else
        {
             
            // Update q by removing
            // the i-th element
            q = odd[N - 1] - odd[i - 1] - 
                    arr[i] + even[i - 1];
 
            // Update p by removing
            // the i-th element
            p = even[N - 1] - even[i - 1] + 
                 odd[i - 1];
        }
 
        // If odd index values sum is equal
        // to even index values sum
        if (p == q) 
        {
 
            // Set the find variable
            find = true;
 
            // Print the current index
            Console.Write(i + " ");
        }
    }
 
    // If no index found
    if (!find)
    {
         
        // Print not possible
        Console.Write(-1);
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int[] arr = { 4, 1, 6, 2 };
     
    removeIndicesToMakeSumEqual(arr);
}
}
 
// This code is contributed by AnkThon
 
  

Javascript

   <script>
// javascript program to implement
// the above approach
 
    // Function to find indices of array elements
    // whose removal makes the sum of odd and
    // even indexed array elements equal
    function removeIndicesToMakeSumEqual(arr) {
 
        // Stores size of array
        var N = arr.length;
 
        // Store prefix sum of odd
        // index array elements
        var odd = Array(N).fill(0);
 
        // Store prefix sum of even
        // index array elements
        var even = Array(N).fill(0);
 
        // Update even[0]
        even[0] = arr[0];
 
        // Traverse the given array
        for (i = 1; i < N; i++) {
 
            // Update odd[i]
            odd[i] = odd[i - 1];
 
            // Update even[i]
            even[i] = even[i - 1];
 
            // If the current index
            // is an even number
            if (i % 2 == 0) {
 
                // Update even[i]
                even[i] += arr[i];
            }
 
            // If the current index
            // is an odd number
            else {
 
                // Update odd[i]
                odd[i] += arr[i];
            }
        }
 
        // Check if at least one
        // index found or not that
        // satisfies the condition
        var find = false;
 
        // Store odd indices sum by
        // removing 0-th index
        var p = odd[N - 1];
 
        // Store even indices sum by
        // removing 0-th index
        var q = even[N - 1] - arr[0];
 
        // If p and q are equal
        if (p == q) {
            document.write("0 ");
            find = true;
        }
 
        // Traverse the array arr
        for (i = 1; i < N; i++) {
 
            // If i is an even number
            if (i % 2 == 0) {
 
                // Update p by removing
                // the i-th element
                p = even[N - 1] - even[i - 1] - arr[i] + odd[i - 1];
 
                // Update q by removing
                // the i-th element
                q = odd[N - 1] - odd[i - 1] + even[i - 1];
            } else {
 
                // Update q by removing
                // the i-th element
                q = odd[N - 1] - odd[i - 1] - arr[i] + even[i - 1];
 
                // Update p by removing
                // the i-th element
                p = even[N - 1] - even[i - 1] + odd[i - 1];
            }
 
            // If odd index values sum is equal
            // to even index values sum
            if (p == q) {
 
                // Set the find variable
                find = true;
 
                // Print the current index
                document.write(i + " ");
            }
        }
 
        // If no index found
        if (!find) {
 
            // Print not possible
            document.write(-1);
        }
    }
 
    // Driver Code
     
        var arr = [ 4, 1, 6, 2 ];
 
        removeIndicesToMakeSumEqual(arr);
 
// This code is contributed by Rajput-Ji
</script>
 
  

Output:

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

Count ways to make Bitwise XOR of odd and even indexed elements equal by removing an array element

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Print indices of array elements whose removal makes the sum of odd and even-indexed elements equal

C++

Java

Python3

C#

Javascript

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