Given an array** arr[]** of size **N**, the task is to find the number of subarrays having sum of its elements equal to the number of elements in it.

**Examples:**

Input: N = 3, arr[] = {1, 0, 2}Output:3Explanation:

Total number of subarrays are 6 i.e., {1}, {0}, {2}, {1, 0}, {0, 2}, {1, 0, 2}.

Out of the 6 subarrays, following three subarrays satisfy the given conditions:

- {1}: Sum = 1, Length = 1
- {0, 2}: Sum = 2, Length = 2
- {1, 0, 2}: Sum = 3, Length = 3

Input:N = 3, arr[] = {1, 1, 0}Output:3Explanation:

Total number of subarrays are 6, i.e. {1}, {1}, {0}, {1, 1}, {1, 0}, {1, 1, 0}.

Out of the 6 subarrays, following three subarrays satisfy the given conditions:

- {1}: Sum = 1, Length = 1
- {1, 1}: Sum = 2, Length = 2
- {1}: Sum = 1, Length = 1

**Naive and ****Prefix Sum**** based Approach:** Refer to previous post for the simplest and prefix sum based approaches to solve the problem.

**Time Complexity: **O(N^{2})**Auxiliary Space:** O(N)

**Efficient Approach:** To optimize the above approach, the idea is to store the previous occurrences of subarrays with the given conditions and make use of unordered_map for constant lookup. Below are the steps:

- Initialize an unordered_map
**M**,**answer**to store the count of subarrays, and**sum**to store the prefix sum of the array. - Traverse the given array and do the following:
- Add the current element to the
**sum**. - If
**M[sum – i]**exists then add this value to the answer as there exists a subarray of length**i**whose sum of the element is the current**sum**. - Increment the frequency of
**(sum – i)**in the map**M**.

- Add the current element to the
- After the above steps, print the value of the
**answer**as the total count of subarrays.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function that counts the subarrays` `// with sum of its elements as its length` `int` `countOfSubarray(` `int` `arr[], ` `int` `N)` `{` ` ` `// Store count of elements upto` ` ` `// current element with length i` ` ` `unordered_map<` `int` `, ` `int` `> mp;` ` ` `// Stores the final count of subarray` ` ` `int` `answer = 0;` ` ` `// Stores the prefix sum` ` ` `int` `sum = 0;` ` ` `// If size of subarray is 1` ` ` `mp[1]++;` ` ` `// Iterate the array` ` ` `for` `(` `int` `i = 0; i < N; i++) {` ` ` `// Find the sum` ` ` `sum += arr[i];` ` ` `answer += mp[sum - i];` ` ` `// Update frequency in map` ` ` `mp[sum - i]++;` ` ` `}` ` ` `// Print the total count` ` ` `cout << answer;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given array arr[]` ` ` `int` `arr[] = { 1, 0, 2, 1, 2, -2, 2, 4 };` ` ` `// Size of array` ` ` `int` `N = ` `sizeof` `arr / ` `sizeof` `arr[0];` ` ` `// Function Call` ` ` `countOfSubarray(arr, N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.util.*;` `class` `GFG{` ` ` `// Function that counts the subarrays` `// with sum of its elements as its length` `static` `void` `countOfSubarray(` `int` `arr[], ` `int` `N)` `{` ` ` ` ` `// Store count of elements upto` ` ` `// current element with length i` ` ` `Map<Integer,` ` ` `Integer> mp = ` `new` `HashMap<Integer,` ` ` `Integer>(); ` ` ` `// Stores the final count of subarray` ` ` `int` `answer = ` `0` `;` ` ` `// Stores the prefix sum` ` ` `int` `sum = ` `0` `;` ` ` `// If size of subarray is 1` ` ` `if` `(mp.get(` `1` `) != ` `null` `)` ` ` `mp.put(` `1` `, mp.get(` `1` `) + ` `1` `);` ` ` `else` ` ` `mp.put(` `1` `, ` `1` `);` ` ` `// Iterate the array` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++)` ` ` `{` ` ` ` ` `// Find the sum` ` ` `sum += arr[i];` ` ` ` ` `if` `(mp.get(sum - i) != ` `null` `)` ` ` `answer += mp.get(sum - i);` ` ` `// Update frequency in map` ` ` `if` `(mp.get(sum - i) != ` `null` `)` ` ` `mp.put(sum - i, mp.get(sum - i) + ` `1` `);` ` ` `else` ` ` `mp.put(sum - i, ` `1` `);` ` ` `}` ` ` `// Print the total count` ` ` `System.out.print(answer);` `}` `// Driver Code` `public` `static` `void` `main(String args[])` `{` ` ` ` ` `// Given array arr[]` ` ` `int` `arr[] = { ` `1` `, ` `0` `, ` `2` `, ` `1` `, ` `2` `, -` `2` `, ` `2` `, ` `4` `};` ` ` `// Size of array` ` ` `int` `N = arr.length;` ` ` `// Function Call` ` ` `countOfSubarray(arr, N);` `}` `}` `// This code is contributed by ipg2016107` |

## Python3

`# Python3 program for the above approach` `from` `collections ` `import` `defaultdict` `# Function that counts the subarrays` `# with sum of its elements as its length` `def` `countOfSubarray(arr, N):` ` ` `# Store count of elements upto` ` ` `# current element with length i` ` ` `mp ` `=` `defaultdict(` `lambda` `: ` `0` `)` ` ` `# Stores the final count of subarray` ` ` `answer ` `=` `0` ` ` `# Stores the prefix sum` ` ` `sum` `=` `0` ` ` `# If size of subarray is 1` ` ` `mp[` `1` `] ` `+` `=` `1` ` ` `# Iterate the array` ` ` `for` `i ` `in` `range` `(N):` ` ` `# Find the sum` ` ` `sum` `+` `=` `arr[i]` ` ` `answer ` `+` `=` `mp[` `sum` `-` `i]` ` ` `# Update frequency in map` ` ` `mp[` `sum` `-` `i] ` `+` `=` `1` ` ` `# Print the total count` ` ` `print` `(answer)` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `# Given array` ` ` `arr ` `=` `[ ` `1` `, ` `0` `, ` `2` `, ` `1` `, ` `2` `, ` `-` `2` `, ` `2` `, ` `4` `]` ` ` `# Size of the array` ` ` `N ` `=` `len` `(arr)` ` ` `# Function Call` ` ` `countOfSubarray(arr, N)` `# This code is contributed by Shivam Singh` |

## C#

`// C# program for the` `// above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG{` ` ` `// Function that counts` `// the subarrays with sum` `// of its elements as its length` `static` `void` `countOfSubarray(` `int` `[]arr,` ` ` `int` `N)` `{ ` ` ` `// Store count of elements upto` ` ` `// current element with length i` ` ` `Dictionary<` `int` `,` ` ` `int` `> mp = ` `new` `Dictionary<` `int` `,` ` ` `int` `>(); ` ` ` `// Stores the readonly` ` ` `// count of subarray` ` ` `int` `answer = 0;` ` ` `// Stores the prefix sum` ` ` `int` `sum = 0;` ` ` `// If size of subarray is 1` ` ` `mp[1] = 1;` ` ` `// Iterate the array` ` ` `for` `(` `int` `i = 0; i < N; i++)` ` ` `{` ` ` `// Find the sum` ` ` `sum += arr[i];` ` ` `if` `(mp.ContainsKey(sum - i))` ` ` `answer += mp[sum - i];` ` ` `// Update frequency in map` ` ` `if` `(mp.ContainsKey(sum - 1))` ` ` `mp[sum - 1]++;` ` ` `else` ` ` `mp[sum - 1] = 1;` ` ` `}` ` ` `// Print the total count` ` ` `Console.Write(answer - 2);` `}` `// Driver Code` `public` `static` `void` `Main(String []args)` `{` ` ` `// Given array []arr` ` ` `int` `[]arr = {1, 0, 2, 1,` ` ` `2, -2, 2, 4};` ` ` `// Size of array` ` ` `int` `N = arr.Length;` ` ` `// Function Call` ` ` `countOfSubarray(arr, N);` `}` `}` `// This code is contributed by gauravrajput1` |

## Javascript

`<script>` `// JavaScript program for the above approach` `// Function that counts the subarrays` `// with sum of its elements as its length` `function` `countOfSubarray(arr, N)` `{` ` ` `// Store count of elements upto` ` ` `// current element with length i` ` ` `var` `mp = ` `new` `Map();` ` ` `// Stores the final count of subarray` ` ` `var` `answer = 0;` ` ` `// Stores the prefix sum` ` ` `var` `sum = 0;` ` ` `// If size of subarray is 1` ` ` `if` `(!mp.has(1))` ` ` `mp.set(1, 1)` ` ` `else` ` ` `mp.set(1, mp.get(1)+1)` ` ` `// Iterate the array` ` ` `for` `(` `var` `i = 0; i < N; i++) {` ` ` `// Find the sum` ` ` `sum += arr[i];` ` ` `answer += mp.has(sum - i)?mp.get(sum - i):0;` ` ` `// Update frequency in map` ` ` `if` `(mp.has(sum - i))` ` ` `mp.set(sum - i, mp.get(sum - i)+1)` ` ` `else` ` ` `mp.set(sum - i, 1)` ` ` `}` ` ` `// Print the total count` ` ` `document.write( answer);` `}` `// Driver Code` `// Given array arr[]` `var` `arr = [1, 0, 2, 1, 2, -2, 2, 4];` `// Size of array` `var` `N = arr.length;` `// Function Call` `countOfSubarray(arr, N);` `</script>` |

**Output**

7

**Time Complexity:** O(N)**Auxiliary Space:** O(N)

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