Given an array arr of integers of size n. We need to compute the sum of elements from index i to index j. The queries consisting of i and j index values will be executed multiple times.
Examples:
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
5
A Simple Solution is to compute the sum for every query.
An Efficient Solution is to precompute prefix sum. Let pre[i] stores sum of elements from arr[0] to arr[i]. To answer a query (i, j), we return pre[j] – pre[i-1].
Below is the implementation of the above approach:
CPP
// CPP program to find sum between two indexes// when there is no update.#include <bits/stdc++.h>using namespace std;void preCompute(int arr[], int n, int pre[]){ pre[0] = arr[0]; for (int i = 1; i < n; i++) pre[i] = arr[i] + pre[i - 1];}// Returns sum of elements in arr[i..j]// It is assumed that i <= jint rangeSum(int i, int j, int pre[]){ if (i == 0) return pre[j]; return pre[j] - pre[i - 1];}// Driver codeint main(){ int arr[] = { 1, 2, 3, 4, 5 }; int n = sizeof(arr) / sizeof(arr[0]); int pre[n]; // Function call preCompute(arr, n, pre); cout << rangeSum(1, 3, pre) << endl; cout << rangeSum(2, 4, pre) << endl; return 0;} |
Python3
# Python program to find sum between two indexes# when there is no update.def find_ans(ar, j, k): l = len(ar) for i in range(1, l): ar[i] = ar[i] + ar[i-1] print(ar[k] - ar[j-1]) return# Driver codepr = [1, 2, 3, 4, 5]ar = pr[:]find_ans(ar, 1, 3)ar = pr[:]find_ans(ar, 2, 4)# Code Contributed by Mohit Gupta_OMG <(0_o)> |
Java
// Java program to find sum between two indexes// when there is no update.import java.util.*;import java.lang.*;class GFG { public static void preCompute(int arr[], int n, int pre[]) { pre[0] = arr[0]; for (int i = 1; i < n; i++) pre[i] = arr[i] + pre[i - 1]; } // Returns sum of elements in arr[i..j] // It is assumed that i <= j public static int rangeSum(int i, int j, int pre[]) { if (i == 0) return pre[j]; return pre[j] - pre[i - 1]; } // Driver code public static void main(String[] args) { int arr[] = { 1, 2, 3, 4, 5 }; int n = arr.length; int pre[] = new int[n]; preCompute(arr, n, pre); System.out.println(rangeSum(1, 3, pre)); System.out.println(rangeSum(2, 4, pre)); }}// Code Contributed by Mohit Gupta_OMG <(0_o)> |
C#
// Program to find sum between two// indexes when there is no update.using System;class GFG { public static void preCompute(int[] arr, int n, int[] pre) { pre[0] = arr[0]; for (int i = 1; i < n; i++) pre[i] = arr[i] + pre[i - 1]; } // Returns sum of elements in // arr[i..j] // It is assumed that i <= j public static int rangeSum(int i, int j, int[] pre) { if (i == 0) return pre[j]; return pre[j] - pre[i - 1]; } // Driver code public static void Main() { int[] arr = { 1, 2, 3, 4, 5 }; int n = arr.Length; int[] pre = new int[n]; // Function call preCompute(arr, n, pre); Console.WriteLine(rangeSum(1, 3, pre)); Console.WriteLine(rangeSum(2, 4, pre)); }}// Code Contributed by Anant Agarwal. |
9 12
Here time complexity of every range sum query is O(1) and the overall time complexity is O(n).
The question becomes complicated when updates are also allowed. In such situations when using advanced data structures like Segment Tree or Binary Indexed Tree.
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