Find the array element having equal sum of Prime Numbers on its left and right
Last Updated :
23 Jul, 2025
Given an array arr[] of size N, the task is to find the index in the given array where the sum of the prime numbers present to its left is equal to the sum of the prime numbers present to its right.
Examples:
Input: arr[] = {11, 4, 7, 6, 13, 1, 5}
Output: 3
Explanation: Sum of prime numbers to left of index 3 = 11 + 7 = 18
Sum of prime numbers to right of index 3 = 13 + 5 = 18
Input: arr[] = {5, 2, 1, 7}
Output: 2
Explanation: Sum of prime numbers to left of index 2 = 5 + 2 = 7
Sum of prime numbers to right of index 2 = 7
Naive Approach: The simplest approach is to traverse the array and check for the given condition for every index in the range [0, N - 1]. If the condition is found to be true for any index, then print the value of that index.
Time Complexity: O(N2*?M), where M is the largest element in the array
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized using the Sieve of Eratosthenes and prefix sum technique to pre-store the sum of prime numbers left and right to an array element. Follow the steps below to solve the problem:
- Traverse through the array to find the maximum value present in the array.
- Use Sieve of Eratosthenes to find out the prime numbers which are less than or equal to the maximum value present in the array. Store these elements in a Map.
- Initialize an array, say first_array. Traverse the array and store the sum of all prime numbers up to ith index at first_array[i].
- Initialize an array, say second_array. Traverse the array in reverse and store the sum of all elements up to ith index at second_array[i].
- Traverse the arrays first_array and second_array and check if at any index, both their values are equal or not. If found to be true, return that index.
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 an index in the
// array having sum of prime numbers
// to its left and right equal
int find_index(int arr[], int N)
{
// Stores the maximum value
// present in the array
int max_value = INT_MIN;
for (int i = 0; i < N; i++) {
max_value = max(max_value, arr[i]);
}
// Stores all positive
// elements which are <= max_value
map<int, int> store;
for (int i = 1; i <= max_value; i++) {
store[i]++;
}
// If 1 is present
if (store.find(1) != store.end()) {
// Remove 1
store.erase(1);
}
// Sieve of Eratosthenes to
// store all prime numbers which
// are <= max_value in the Map
for (int i = 2; i <= sqrt(max_value); i++) {
int multiple = 2;
// Erase non-prime numbers
while ((i * multiple) <= max_value) {
if (store.find(i * multiple) != store.end()) {
store.erase(i * multiple);
}
multiple++;
}
}
// Stores the sum of
// prime numbers from left
int prime_sum_from_left = 0;
// Stores the sum of prime numbers
// to the left of each index
int first_array[N];
for (int i = 0; i < N; i++) {
// Stores the sum of prime numbers
// to the left of the current index
first_array[i] = prime_sum_from_left;
if (store.find(arr[i]) != store.end()) {
// Add current value to
// the prime sum if the
// current value is prime
prime_sum_from_left += arr[i];
}
}
// Stores the sum of
// prime numbers from right
int prime_sum_from_right = 0;
// Stores the sum of prime numbers
// to the right of each index
int second_array[N];
for (int i = N - 1; i >= 0; i--) {
// Stores the sum of prime
// numbers to the right of
// the current index
second_array[i] = prime_sum_from_right;
if (store.find(arr[i]) != store.end()) {
// Add current value to the
// prime sum if the
// current value is prime
prime_sum_from_right += arr[i];
}
}
// Traverse through the two
// arrays to find the index
for (int i = 0; i < N; i++) {
// Compare the values present
// at the current index
if (first_array[i] == second_array[i]) {
// Return the index where
// both the values are same
return i;
}
}
// No index is found.
return -1;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 11, 4, 7, 6, 13, 1, 5 };
// Size of Array
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << find_index(arr, N);
return 0;
}
// Java program for the above approach
import java.lang.*;
import java.util.*;
class GFG{
// Function to find an index in the
// array having sum of prime numbers
// to its left and right equal
static int find_index(int arr[], int N)
{
// Stores the maximum value
// present in the array
int max_value = Integer.MIN_VALUE;
for(int i = 0; i < N; i++)
{
max_value = Math.max(max_value, arr[i]);
}
// Stores all positive
// elements which are <= max_value
Map<Integer, Integer> store= new HashMap<>();
for(int i = 1; i <= max_value; i++)
{
store.put(i, store.getOrDefault(i, 0) + 1);
}
// If 1 is present
if (store.containsKey(1))
{
// Remove 1
store.remove(1);
}
// Sieve of Eratosthenes to
// store all prime numbers which
// are <= max_value in the Map
for(int i = 2; i <= Math.sqrt(max_value); i++)
{
int multiple = 2;
// Erase non-prime numbers
while ((i * multiple) <= max_value)
{
if (store.containsKey(i * multiple))
{
store.remove(i * multiple);
}
multiple++;
}
}
// Stores the sum of
// prime numbers from left
int prime_sum_from_left = 0;
// Stores the sum of prime numbers
// to the left of each index
int[] first_array = new int[N];
for(int i = 0; i < N; i++)
{
// Stores the sum of prime numbers
// to the left of the current index
first_array[i] = prime_sum_from_left;
if (store.containsKey(arr[i]))
{
// Add current value to
// the prime sum if the
// current value is prime
prime_sum_from_left += arr[i];
}
}
// Stores the sum of
// prime numbers from right
int prime_sum_from_right = 0;
// Stores the sum of prime numbers
// to the right of each index
int[] second_array= new int[N];
for(int i = N - 1; i >= 0; i--)
{
// Stores the sum of prime
// numbers to the right of
// the current index
second_array[i] = prime_sum_from_right;
if (store.containsKey(arr[i]))
{
// Add current value to the
// prime sum if the
// current value is prime
prime_sum_from_right += arr[i];
}
}
// Traverse through the two
// arrays to find the index
for(int i = 0; i < N; i++)
{
// Compare the values present
// at the current index
if (first_array[i] == second_array[i])
{
// Return the index where
// both the values are same
return i;
}
}
// No index is found.
return -1;
}
// Driver code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 11, 4, 7, 6, 13, 1, 5 };
// Size of Array
int N = arr.length;
// Function Call
System.out.println(find_index(arr, N));
}
}
// This code is contributed by offbeat
# Python3 program for the above approach
from math import sqrt
# Function to find an index in the
# array having sum of prime numbers
# to its left and right equal
def find_index(arr, N):
# Stores the maximum value
# present in the array
max_value = -10**9
for i in range(N):
max_value = max(max_value, arr[i])
# Stores all positive
# elements which are <= max_value
store = {}
for i in range(1, max_value + 1):
store[i] = store.get(i, 0) + 1
# If 1 is present
if (1 in store):
# Remove 1
del store[1]
# Sieve of Eratosthenes to
# store all prime numbers which
# are <= max_value in the Map
for i in range(2, int(sqrt(max_value)) + 1):
multiple = 2
# Erase non-prime numbers
while ((i * multiple) <= max_value):
if (i * multiple in store):
del store[i * multiple]
multiple += 1
# Stores the sum of
# prime numbers from left
prime_sum_from_left = 0
# Stores the sum of prime numbers
# to the left of each index
first_array = [0]*N
for i in range(N):
# Stores the sum of prime numbers
# to the left of the current index
first_array[i] = prime_sum_from_left
if arr[i] in store:
# Add current value to
# the prime sum if the
# current value is prime
prime_sum_from_left += arr[i]
# Stores the sum of
# prime numbers from right
prime_sum_from_right = 0
# Stores the sum of prime numbers
# to the right of each index
second_array = [0]*N
for i in range(N - 1, -1, -1):
# Stores the sum of prime
# numbers to the right of
# the current index
second_array[i] = prime_sum_from_right
if (arr[i] in store):
# Add current value to the
# prime sum if the
# current value is prime
prime_sum_from_right += arr[i]
# Traverse through the two
# arrays to find the index
for i in range(N):
# Compare the values present
# at the current index
if (first_array[i] == second_array[i]):
# Return the index where
# both the values are same
return i
# No index is found.
return -1
# Driver Code
if __name__ == '__main__':
# Given array arr[]
arr= [11, 4, 7, 6, 13, 1, 5]
# Size of Array
N = len(arr)
# Function Call
print (find_index(arr, N))
# This code is contributed by mohit kumar 29.
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find an index in the
// array having sum of prime numbers
// to its left and right equal
static int find_index(int[] arr, int N)
{
// Stores the maximum value
// present in the array
int max_value = Int32.MinValue;
for(int i = 0; i < N; i++)
{
max_value = Math.Max(max_value, arr[i]);
}
// Stores all positive
// elements which are <= max_value
Dictionary<int,
int> store = new Dictionary<int,
int>();
for(int i = 1; i <= max_value; i++)
{
if (!store.ContainsKey(i))
store[i] = 0;
store[i]++;
}
// If 1 is present
if (store.ContainsKey(1))
{
// Remove 1
store.Remove(1);
}
// Sieve of Eratosthenes to
// store all prime numbers which
// are <= max_value in the Map
for(int i = 2; i <= Math.Sqrt(max_value); i++)
{
int multiple = 2;
// Erase non-prime numbers
while ((i * multiple) <= max_value)
{
if (store.ContainsKey(i * multiple))
{
store.Remove(i * multiple);
}
multiple++;
}
}
// Stores the sum of
// prime numbers from left
int prime_sum_from_left = 0;
// Stores the sum of prime numbers
// to the left of each index
int[] first_array = new int[N];
for(int i = 0; i < N; i++)
{
// Stores the sum of prime numbers
// to the left of the current index
first_array[i] = prime_sum_from_left;
if (store.ContainsKey(arr[i]))
{
// Add current value to
// the prime sum if the
// current value is prime
prime_sum_from_left += arr[i];
}
}
// Stores the sum of
// prime numbers from right
int prime_sum_from_right = 0;
// Stores the sum of prime numbers
// to the right of each index
int[] second_array = new int[N];
for(int i = N - 1; i >= 0; i--)
{
// Stores the sum of prime
// numbers to the right of
// the current index
second_array[i] = prime_sum_from_right;
if (store.ContainsKey(arr[i]))
{
// Add current value to the
// prime sum if the
// current value is prime
prime_sum_from_right += arr[i];
}
}
// Traverse through the two
// arrays to find the index
for(int i = 0; i < N; i++)
{
// Compare the values present
// at the current index
if (first_array[i] == second_array[i])
{
// Return the index where
// both the values are same
return i;
}
}
// No index is found.
return -1;
}
// Driver Code
public static void Main()
{
// Given array arr[]
int[] arr = { 11, 4, 7, 6, 13, 1, 5 };
// Size of Array
int N = arr.Length;
// Function Call
Console.WriteLine(find_index(arr, N));
}
}
// This code is contributed by ukasp
<script>
// Javascript program for the above approach
// Function to find an index in the
// array having sum of prime numbers
// to its left and right equal
function find_index(arr,N)
{
// Stores the maximum value
// present in the array
let max_value = Number.MIN_VALUE;
for (let i = 0; i < N; i++) {
max_value = Math.max(max_value, arr[i]);
}
// Stores all positive
// elements which are <= max_value
let store=new Map();
for (let i = 1; i <= max_value; i++) {
if(!store.has(i))
store.set(i,0);
store.set(i,store.get(i)+1);
}
// If 1 is present
if (store.has(1)) {
// Remove 1
store.delete(1);
}
// Sieve of Eratosthenes to
// store all prime numbers which
// are <= max_value in the Map
for (let i = 2; i <= Math.sqrt(max_value); i++) {
let multiple = 2;
// Erase non-prime numbers
while ((i * multiple) <= max_value) {
if (store.has(i * multiple)) {
store.delete(i * multiple);
}
multiple++;
}
}
// Stores the sum of
// prime numbers from left
let prime_sum_from_left = 0;
// Stores the sum of prime numbers
// to the left of each index
let first_array=new Array(N);
for (let i = 0; i < N; i++) {
// Stores the sum of prime numbers
// to the left of the current index
first_array[i] = prime_sum_from_left;
if (store.has(arr[i])) {
// Add current value to
// the prime sum if the
// current value is prime
prime_sum_from_left += arr[i];
}
}
// Stores the sum of
// prime numbers from right
let prime_sum_from_right = 0;
// Stores the sum of prime numbers
// to the right of each index
let second_array=new Array(N);
for (let i = N - 1; i >= 0; i--) {
// Stores the sum of prime
// numbers to the right of
// the current index
second_array[i] = prime_sum_from_right;
if (store.has(arr[i]) ) {
// Add current value to the
// prime sum if the
// current value is prime
prime_sum_from_right += arr[i];
}
}
// Traverse through the two
// arrays to find the index
for (let i = 0; i < N; i++) {
// Compare the values present
// at the current index
if (first_array[i] == second_array[i]) {
// Return the index where
// both the values are same
return i;
}
}
// No index is found.
return -1;
}
// Driver Code
// Given array arr[]
let arr=[11, 4, 7, 6, 13, 1, 5];
// Size of Array
let N=arr.length;
// Function Call
document.write(find_index(arr, N));
// This code is contributed by patel2127
</script>
Time Complexity: O(N + max(arr[])loglog(max(arr[]))
Auxiliary Space: O(max(arr[]) + N)
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