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++
// 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
// 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 |
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 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#
// 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 |
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:
1
Time Complexity: O(N)
Auxiliary Space: O(N)