Count of Substrings having Sum equal to their Length

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++

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// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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<Integer,
            Integer> mp = new HashMap<Integer,
                                      Integer>();
    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

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Python3

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# 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

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C#

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// 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

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Output: 

6



 

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

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