Prefix Sum Array – Implementation and Applications in Competitive Programming
Given an array arr[] of size n, the task is to 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.
Table of Content
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:
#include <bits/stdc++.h>
using namespace std;
// function to find the prefix sum array
vector<int> findPrefixSum(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 = findPrefixSum(arr);
for(auto i:prefixSum) {
cout<<i<<" ";
}
return 0;
}
// Function to find the prefix sum array
import java.util.*;
class GfG {
// Function to find the prefix sum array
static ArrayList<Integer> findPrefixSum(int[] arr) {
int n = arr.length;
// to store the prefix sum
ArrayList<Integer> prefixSum = new ArrayList<>(n);
for (int i = 0; i < n; i++) {
prefixSum.add(0);
}
// initialize the first element
prefixSum.set(0, arr[0]);
// Adding present element with previous element
for (int i = 1; i < n; i++)
prefixSum.set(i, prefixSum.get(i - 1) + arr[i]);
return prefixSum;
}
public static void main(String[] args) {
int[] arr = {10, 20, 10, 5, 15};
ArrayList<Integer> prefixSum = findPrefixSum(arr);
for (int i : prefixSum) {
System.out.print(i + " ");
}
}
}
# Function to find the prefix sum array
def findPrefixSum(arr):
n = len(arr)
# to store the prefix sum
prefixSum = [0] * n
# initialize the first element
prefixSum[0] = arr[0]
# Adding present element with previous element
for i in range(1, n):
prefixSum[i] = prefixSum[i - 1] + arr[i]
return prefixSum
if __name__ == "__main__":
arr = [10, 20, 10, 5, 15]
prefixSum = findPrefixSum(arr)
for i in prefixSum:
print(i, end=" ")
// Function to find the prefix sum array
using System;
using System.Collections.Generic;
class GfG {
// Function to find the prefix sum array
static List<int> findPrefixSum(int[] arr) {
int n = arr.Length;
// to store the prefix sum
List<int> prefixSum = new List<int>(new int[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;
}
static void Main() {
int[] arr = {10, 20, 10, 5, 15};
List<int> prefixSum = findPrefixSum(arr);
foreach (int i in prefixSum) {
Console.Write(i + " ");
}
}
}
// Function to find the prefix sum array
function findPrefixSum(arr) {
let n = arr.length;
// to store the prefix sum
let prefixSum = new Array(n).fill(0);
// initialize the first element
prefixSum[0] = arr[0];
// Adding present element with previous element
for (let i = 1; i < n; i++)
prefixSum[i] = prefixSum[i - 1] + arr[i];
return prefixSum;
}
let arr = [10, 20, 10, 5, 15];
let prefixSum = findPrefixSum(arr);
for (let i of prefixSum) {
process.stdout.write(i + " ");
}
Output
10 30 40 45 60
Time Complexity: O(n), as we are traversing the array only once.
Auxiliary Space: O(n), to create the array prefxSum[] of size n.
Example Problems based on Prefix Sum
1. Sum of an array between indexes L and R using Prefix Sum:
Given an array arr[] of size n. Given q queries and in each query given i and j, Print the sum of array elements from index i to j.
Input : arr[] = {1, 2, 3, 4, 5}
i = 1, j = 3
i = 2, j = 4
Output : 9
12
Input : arr[] = {1, 2, 3, 4, 5}
i = 0, j = 4
i = 1, j = 2
Output : 15
5Please refer Range sum queries without updates for Naive and Prefix Sum based solutions.
2. Maximum value in an array after m range increment operations:
Consider an array of size n with all initial values as 0. We need to perform the following m range increment operations.
increment(a, b, k) : Increment values from ‘a’ to ‘b’ by ‘k’.
After m operations, we need to calculate the maximum of the values in the array.
Examples :
Input : n = 5 m = 3
a = 0, b = 1, k = 100
a = 1, b = 4, k = 100
a = 2, b = 3, k = 100
Output : 200
Explanation:
Initially array = {0, 0, 0, 0, 0}
After first operation: {100, 100, 0, 0, 0}
After second operation: {100, 200, 100, 100, 100}
After third operation {100, 200, 200, 200, 100}
Maximum element after m operations is 200.Input : n = 4 m = 3
a = 1, b = 2, k = 603
a = 0, b = 0, k = 286
a = 3, b = 3, k = 882
Output : 882
Explanation:
Initially array = {0, 0, 0, 0}
After first operation: {0, 603, 603, 0}
After second operation: {286, 603, 603, 0}
After third operation: {286, 603, 603, 882}
Maximum element after m operations is 882.Please refer Maximum after m range increment operations for details and implementation
Applications of Prefix Sum:
- Equilibrium index of an array: The equilibrium index of an array is an index such that the sum of elements at lower indexes is equal to the sum of elements at higher indexes.
- Find if there is a subarray with 0 sums: Given an array of positive and negative numbers, find if there is a subarray (of size at least one) with 0 sum.
- Maximum subarray size, such that all subarrays of that size have a sum less than k: Given an array of n positive integers and a positive integer k, the task is to find the maximum subarray size such that all subarrays of that size have the sum of elements less than k.
- Find the prime numbers which can be written as sum of most consecutive primes: Given an array of limits. For every limit, find the prime number which can be written as the sum of the most consecutive primes smaller than or equal to the limit.
- Longest Span with same Sum in two Binary arrays: Given two binary arrays, arr1[] and arr2[] of the same size n. Find the length of the longest common span (i, j) where j >= i such that arr1[i] + arr1[i+1] + …. + arr1[j] = arr2[i] + arr2[i+1] + …. + arr2[j].
- Maximum subarray sum modulo m: Given an array of n elements and an integer m. The task is to find the maximum value of the sum of its subarray modulo m i.e find the sum of each subarray mod m and print the maximum value of this modulo operation.
- Maximum occurred integer in n ranges : Given n ranges of the form L and R, the task is to find the maximum occurring integer in all the ranges. If more than one such integer exits, print the smallest one.
- Minimum cost for acquiring all coins with k extra coins allowed with every coin: You are given a list of N coins of different denominations. you can pay an amount equivalent to any 1 coin and can acquire that coin. In addition, once you have paid for a coin, we can choose at most K more coins and can acquire those for free. The task is to find the minimum amount required to acquire all the N coins for a given value of K.
- Random number generator in arbitrary probability distribution fashion: Given n numbers, each with some frequency of occurrence. Return a random number with a probability proportional to its frequency of occurrence.
