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Prefix Sum Array - Implementation

Last Updated : 13 Jul, 2025
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Given an array arr[], Find the prefix sum of the array. A prefix sum array is another array prefixSum[] of the same size, such that prefixSum[i] is arr[0] + arr[1] + arr[2] . . . arr[i].

Examples: 

Input: arr[] = [10, 20, 10, 5, 15]
Output: [10, 30, 40, 45, 60]
Explanation: For each index i, add all the elements from 0 to i:
prefixSum[0] = 10, 
prefixSum[1] = 10 + 20 = 30, 
prefixSum[2] = 10 + 20 + 10 = 40 and so on.

Input: arr[] = [30, 10, 10, 5, 50]
Output: [30, 40, 50, 55, 105]
Explanation: For each index i, add all the elements from 0 to i:
prefixSum[0] = 30, 
prefixSum[1] = 30 + 10 = 40,
prefixSum[2] = 30 + 10+ 10 = 50 and so on.

Prefix Sum Implementation

The idea is to create an array prefixSum[] of size n, and for each index i in range 1 to n - 1, set prefixSum[i] = prefixSum[i - 1] + arr[i].

To solve the problem follow the given steps:

  • Declare a new array prefixSum[] of the same size as the input array
  • Run a for loop to traverse the input array
  • For each index add the value of the current element and the previous value of the prefix sum array

Below is the implementation of the above approach:

C++
#include <iostream>
#include <vector>
using namespace std;

// function to find the prefix sum array
vector<int> prefSum(vector<int> &arr) {
    int n = arr.size();
    
    // to store the prefix sum
    vector<int> prefixSum(n);

    // initialize the first element
    prefixSum[0] = arr[0];

    // Adding present element with previous element
    for (int i = 1; i < n; i++)
        prefixSum[i] = prefixSum[i - 1] + arr[i];
    
    return prefixSum;
}

int main() {
    vector<int> arr = {10, 20, 10, 5, 15};
    vector<int> prefixSum = prefSum(arr);
    for(auto i: prefixSum) {
        cout << i << " " ;
    }
    return 0;
}
Java Python C# JavaScript

Output
10 30 40 45 60 

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

Please refer Top Problems on Prefix Sum Technique for Interviews for more prefix sum problems.


Prefix Sum Technique (Part 1)
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