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# Count of Substrings having Sum equal to their Length

• Difficulty Level : Easy
• Last Updated : 19 May, 2021

Given a numeric string str, the task is to calculate the number of substrings with the sum of digits equal to their length.

Examples:

Input: str = “112112”
Output:
Explanation:
Substrings “1”, “1”, “11”, “1”, “1”, “11” satisfy the given condition.
Input: str = “1101112”
Output: 12

Naive Approach: The simplest solution is to generate all substrings of the given string and for each substring, check if its sum is equal to its length or not. For each substring found to be true, increase count.
Time Complexity: O(N3
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using a Hashmap and keep updating the count of substrings in the Hashmap and print the required count at the end.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to count the number of``// substrings with sum equal to length``int` `countSubstrings(string s, ``int` `n)``{` `    ``int` `count = 0, sum = 0;` `    ``// Stores the count of substrings``    ``unordered_map<``int``, ``int``> mp;``    ``mp[0]++;` `    ``for` `(``int` `i = 0; i < n; ++i) {` `        ``// Add character to sum``        ``sum += (s[i] - ``'0'``);` `        ``// Add count of substrings to result``        ``count += mp[sum - (i + 1)];` `        ``// Increase count of subarrays``        ``++mp[sum - (i + 1)];``    ``}` `    ``// Return count``    ``return` `count;``}` `// Driver Code``int` `main()``{``    ``string str = ``"112112"``;``    ``int` `n = str.length();``    ``cout << countSubstrings(str, n) << endl;` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG{` `// Function to count the number of``// subStrings with sum equal to length``static` `int` `countSubStrings(String s, ``int` `n)``{``    ``int` `count = ``0``, sum = ``0``;` `    ``// Stores the count of subStrings``    ``HashMap mp = ``new` `HashMap();``    ``mp.put(``0``, ``1``);` `    ``for``(``int` `i = ``0``; i < n; ++i)``    ``{``        ` `        ``// Add character to sum``        ``sum += (s.charAt(i)- ``'0'``);` `        ``// Add count of subStrings to result``        ``count += mp.containsKey(sum - (i + ``1``)) == ``true` `?``                         ``mp.get(sum - (i + ``1``)) : ``0``;` `        ``// Increase count of subarrays``        ``if``(!mp.containsKey(sum - (i + ``1``)))``                    ``mp.put(sum - (i + ``1``), ``1``);``        ``else``            ``mp.put(sum - (i + ``1``),``            ``mp.get(sum - (i + ``1``)) + ``1``);``    ``}` `    ``// Return count``    ``return` `count;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``String str = ``"112112"``;``    ``int` `n = str.length();``    ` `    ``System.out.print(countSubStrings(str, n) + ``"\n"``);``}``}` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program to implement``# the above approach``from` `collections ``import` `defaultdict` `# Function to count the number of``# substrings with sum equal to length``def` `countSubstrings(s, n):``    ` `    ``count, ``sum` `=` `0``, ``0``    ` `    ``# Stores the count of substrings``    ``mp ``=` `defaultdict(``lambda` `: ``0``)``    ``mp[``0``] ``+``=` `1``    ` `    ``for` `i ``in` `range``(n):``        ` `        ``# Add character to sum``        ``sum` `+``=` `ord``(s[i]) ``-` `ord``(``'0'``)``        ` `        ``# Add count of substrings to result``        ``count ``+``=` `mp[``sum` `-` `(i ``+` `1``)]``        ` `        ``# Increase count of subarrays``        ``mp[``sum` `-` `(i ``+` `1``)] ``+``=` `1``        ` `    ``# Return count``    ``return` `count` `# Driver code``str` `=` `'112112'``n ``=` `len``(``str``)` `print``(countSubstrings(``str``, n))` `# This code is contributed by Stuti Pathak`

## C#

 `// C# program to implement``// the above approach``using` `System;``using` `System.Collections.Generic;``class` `GFG{` `// Function to count the number of``// subStrings with sum equal to length``static` `int` `countSubStrings(String s, ``int` `n)``{``    ``int` `count = 0, sum = 0;` `    ``// Stores the count of subStrings``    ``Dictionary<``int``,``               ``int``> mp = ``new` `Dictionary<``int``,``                                        ``int``>();``    ``mp.Add(0, 1);` `    ``for``(``int` `i = 0; i < n; ++i)``    ``{``        ` `        ``// Add character to sum``        ``sum += (s[i]- ``'0'``);` `        ``// Add count of subStrings to result``        ``count += mp.ContainsKey(sum - (i + 1)) == ``true` `?``                             ``mp[sum - (i + 1)] : 0;` `        ``// Increase count of subarrays``        ``if``(!mp.ContainsKey(sum - (i + 1)))``                    ``mp.Add(sum - (i + 1), 1);``        ``else``            ``mp[sum - (i + 1)] = mp[sum - (i + 1)] + 1;``    ``}` `    ``// Return count``    ``return` `count;``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``String str = ``"112112"``;``    ``int` `n = str.Length;``    ` `    ``Console.Write(countSubStrings(str, n) + ``"\n"``);``}``}` `// This code is contributed by Rohit_ranjan`

## Javascript

 ``
Output:
`6`

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

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