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

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 ith 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++
Java
Python3
C#
Javascript
// 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 equalvoid 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 Codeint main(){    vector<int> arr = { 4, 1, 6, 2 };    removeIndicesToMakeSumEqual(arr);     return 0;}
// Java program to implement// the above approachimport java.util.*; class GFG{ // Function to find indices of array elements// whose removal makes the sum of odd and// even indexed array elements equalstatic 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 Codepublic static void main(String[] args){    int[] arr = { 4, 1, 6, 2 };         removeIndicesToMakeSumEqual(arr);}} // This code is contributed by Amit Katiyar
# 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 equaldef 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 Codeif __name__ == '__main__':    arr = [4, 1, 6, 2];     removeIndicesToMakeSumEqual(arr);     # This code is contributed by shikhasingrajput
// C# program to implement// the above approachusing System; class GFG{     // Function to find indices of array elements// whose removal makes the sum of odd and// even indexed array elements equalstatic 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 Codepublic static void Main(String[] args){    int[] arr = { 4, 1, 6, 2 };         removeIndicesToMakeSumEqual(arr);}} // This code is contributed by AnkThon
<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>
1
 
Time Complexity: O(N)Auxiliary Space: O(N)
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