Count of subarrays having sum equal to its length

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

Examples:

Input: N = 3, arr[] = {1, 0, 2}
Output: 3
Explanation:
Total number of subarrays are 6 i.e., {1}, {0}, {2}, {1, 0}, {0, 2}, {1, 0, 2}.
Out of 6 only three subarrays have the number of elements equals to sum of its elements i.e., 
1) {1}, sum = 1, length = 1.
2) {0, 2}, sum = 2, length = 2.
3) {1, 0, 2}, sum = 3, length = 3.

Input: N = 3, arr[] = {1, 1, 0}
Output: 3
Explanation:
Total number of subarrays are 6 i.e. {1}, {1}, {0}, {1, 1}, {1, 0}, {1, 1, 0}.
Out of 6 only three subarrays have the number of elements equals to sum of its elements i.e., 
1) {1}, sum = 1, length = 1.
2) {1}, sum = 1, length = 1.
3) {1, 1}, sum = 2, length = 2.

Naive Approach: The idea is to generate all the subarrays of the array and if the sum of elements of the subarray is equal to the number of elements in it then count this subarray. Print the count after checking all the subarrays.

Time Complexity: O(N²)
Auxiliary Space: O(N)

Efficient Approach: This problem can be converted into a simpler problem by using observation. If all the elements of the array are decremented by 1, then all the subarrays of array arr[] with a sum equal to its number of elements are same as finding the number of subarrays with sum 0 in the new array(formed by decrementing all the elements of arr[ ] by 1). Below are the steps:

1. Decrement all the array elements by 1.
2. Initialize a prefix array with prefix[0] = arr[0].
3. Traverse the given array arr[] from left to right, starting from index 1 and update a prefix sum array as pref[i] = pref[i-1] + arr[i].
4. Initialize the answer to 0.
5. Iterate the prefix array pref[] from left to right and increment the answer by the value of the current element in the map.
6. Increment the value of the current element in the map.
7. Print the value of answer after the above steps.

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)
{
    // Decrementing all the elements
    // of the array by 1
    for (int i = 0; i < N; i++)
        arr[i]--;

    // Making prefix sum array
    int pref[N];
    pref[0] = arr[0];

    for (int i = 1; i < N; i++)
        pref[i] = pref[i - 1] + arr[i];

    // Declare map to store count of
    // elements upto current element
    map<int, int> mp;
    int answer = 0;

    // To count all the subarrays
    // whose prefix sum is 0
    mp[0]++;

    // Iterate the array
    for (int i = 0; i < N; i++) {

        // Increment answer by count of
        // current element of prefix array
        answer += mp[pref[i]];
        mp[pref[i]]++;
    }

    // Return the answer
    return answer;
}

// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 1, 0 };
    int N = sizeof arr / sizeof arr[0];

    // Function call
    cout << 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 int countOfSubarray(int arr[], int N)
{
    // Decrementing all the elements
    // of the array by 1
    for (int i = 0; i < N; i++)
        arr[i]--;

    // Making prefix sum array
    int []pref = new int[N];
    pref[0] = arr[0];

    for (int i = 1; i < N; i++)
        pref[i] = pref[i - 1] + arr[i];

    // Declare map to store count of
    // elements upto current element
    HashMap<Integer,
              Integer> mp = new HashMap<Integer,
                                        Integer>();
    int answer = 0;

    // To count all the subarrays
    // whose prefix sum is 0
    mp.put(0, 1);

    // Iterate the array
    for (int i = 0; i < N; i++)
    {

        // Increment answer by count of
        // current element of prefix array

        if(mp.containsKey(pref[i]))
        {
            answer += mp.get(pref[i]);
            mp.put(pref[i], mp.get(pref[i]) + 1);

        }
        else
        {
            mp.put(pref[i], 1);
        }
    }

    // Return the answer
    return answer;
}

// Driver Code
public static void main(String[] args)
{
    // Given array arr[]
    int arr[] = { 1, 1, 0 };
    int N = arr.length;

    // Function call
    System.out.print(countOfSubarray(arr, N));
}
}

// This code is contributed by sapnasingh4991

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):

    # Decrementing all the elements
    # of the array by 1
    for i in range(N):
        arr[i] -= 1

    # Making prefix sum array
    pref = [0] * N
    pref[0] = arr[0]

    for i in range(1, N):
        pref[i] = pref[i - 1] + arr[i]

    # Declare map to store count of
    # elements upto current element
    mp = defaultdict(lambda : 0)
    answer = 0

    # To count all the subarrays
    # whose prefix sum is 0
    mp[0] += 1

    # Iterate the array
    for i in range(N):

        # Increment answer by count of
        # current element of prefix array
        answer += mp[pref[i]]
        mp[pref[i]] += 1

    # Return the answer
    return answer

# Driver Code

# Given array arr[]
arr = [ 1, 1, 0 ]
N = len(arr)

# Function call
print(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 int countOfSubarray(int []arr, int N)
{
    // Decrementing all the elements
    // of the array by 1
    for (int i = 0; i < N; i++)
        arr[i]--;

    // Making prefix sum array
    int []pref = new int[N];
    pref[0] = arr[0];

    for (int i = 1; i < N; i++)
        pref[i] = pref[i - 1] + arr[i];

    // Declare map to store count of
    // elements upto current element
    Dictionary<int,
               int> mp = new Dictionary<int,
                                        int>();
    int answer = 0;

    // To count all the subarrays
    // whose prefix sum is 0
    mp.Add(0, 1);

    // Iterate the array
    for (int i = 0; i < N; i++)
    {

        // Increment answer by count of
        // current element of prefix array

        if(mp.ContainsKey(pref[i]))
        {
            answer += mp[pref[i]];
            mp[pref[i]]= mp[pref[i]] + 1;
        }
        else
        {
            mp.Add(pref[i], 1);
        }
    }

    // Return the answer
    return answer;
}

// Driver Code
public static void Main(String[] args)
{
    // Given array []arr
    int []arr = { 1, 1, 0 };
    int N = arr.Length;

    // Function call
    Console.Write(countOfSubarray(arr, N));
}
}

// This code is contributed by sapnasingh4991

Javascript

<script>
// Js program for the above approach

// Function that counts the subarrays
// with sum of its elements as its length
function countOfSubarray( arr, N){
    // Decrementing all the elements
    // of the array by 1
    for (let i = 0; i < N; i++)
        arr[i]--;

    // Making prefix sum array
    let pref = [];
    pref[0] = arr[0];

    for (let i = 1; i < N; i++)
        pref[i] = pref[i - 1] + arr[i];

    // Declare map to store count of
    // elements upto current element
    let mp = new Map;
    let answer = 0;

    // To count all the subarrays
    // whose prefix sum is 0
    if(mp[0])
    mp[0]++;
    else
    mp[0] = 1;

    // Iterate the array
    for (let i = 0; i < N; i++) {
        // Increment answer by count of
        // current element of prefix array
        if(mp[pref[i]]){
        answer += mp[pref[i]];
        mp[pref[i]]++;
        }
    }

    // Return the answer
    return answer;
}

// Driver Code
// Given array arr[]
let arr = [ 1, 1, 0 ];
let N = arr.length;
// Function call
document.write(countOfSubarray(arr, N));

</script>

Output:

3

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