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# Range sum queries without updates

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.

Below is the implementation of the above approach:

## C++14

 `// C++ program to find sum between two indexes``// when there is no update.``#include ``using` `namespace` `std;` `// Function to compute sum in given range``int` `rangeSum(``int` `arr[], ``int` `n, ``int` `i, ``int` `j) {``    ``int` `sum = 0;` `    ``// Compute sum from i to j``    ``for` `(``int` `k = i; k <= j; k++) {``        ``sum += arr[k];``    ``}` `    ``return` `sum;``}` `// Driver code``int` `main() {``    ``int` `arr[] = { 1, 2, 3, 4, 5 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``// Function call with queries``    ``cout << rangeSum(arr, n, 1, 3) << endl;``    ``cout << rangeSum(arr, n, 2, 4) << endl;` `    ``return` `0;``}`

Output

```9
12```

Time Complexity: For each query, the time complexity will be O(j-i+1). The overall time complexity is O((j-i+1) * q), where q is the number of queries.
Space Complexity: O(1), not using any extra space.

An Efficient Solution is to precompute prefix sum. Let pre[i] stores sum of elements from arr to arr[i]. To answer a query (i, j), we return pre[j] – pre[i-1].

Below is the implementation of the above approach:

## C++

 `// CPP program to find sum between two indexes``// when there is no update.``#include ``using` `namespace` `std;` `void` `preCompute(``int` `arr[], ``int` `n, ``int` `pre[])``{``    ``pre = arr;``    ``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``int` `rangeSum(``int` `i, ``int` `j, ``int` `pre[])``{``    ``if` `(i == 0)``        ``return` `pre[j];` `    ``return` `pre[j] - pre[i - 1];``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 4, 5 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``int` `pre[n];``  ` `    ``// Function call``    ``preCompute(arr, n, pre);``    ``cout << rangeSum(1, 3, pre) << endl;``    ``cout << rangeSum(2, 4, pre) << endl;` `    ``return` `0;``}`

## 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)>`

## Python3

 `# Python program to find sum between two indexes``# when there is no update.` `# Function to compute prefix sum``def` `preCompute(arr, n, prefix):``  ``prefix[``0``] ``=` `arr[``0``]``  ``for` `i ``in` `range``(``1``, n):``    ``prefix[i] ``=` `prefix[i ``-` `1``] ``+` `arr[i]` `# Returns sum of elements in arr[i..j]``# It is assumed that i <= j``def` `rangeSum(l, r):``  ``if` `l ``=``=` `0``:``    ``print``(prefix[r])``    ``return``  ` `  ``print``(prefix[r] ``-` `prefix[l ``-` `1``])``    ` `# Driver code``arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``]``n ``=` `len``(arr)``prefix ``=` `[``0` `for` `i ``in` `range``(n)]` `# preComputation``preCompute(arr, n, prefix)` `# Range Queries``rangeSum(``1``, ``3``)``rangeSum(``2``, ``4``)` `# This code is contributed by dineshdkda31.`

## 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 = arr;``        ``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.`

## Javascript

 ``

Output

```9
12```

Here time complexity of every range sum query is O(1) and the overall time complexity is O(n).

Auxiliary Space required = O(n), where n is the size of the given array.

The question becomes complicated when updates are also allowed. In such situations when using advanced data structures like Segment Tree or Binary Indexed Tree.

This article is contributed by Rahul Chawla. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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