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: 6
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 <bits/stdc++.h>using namespace std;// Function to count the number of// substrings with sum equal to lengthint 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 Codeint main(){ string str = "112112"; int n = str.length(); cout << countSubstrings(str, n) << endl; return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to count the number of// subStrings with sum equal to lengthstatic 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 Codepublic 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 approachfrom collections import defaultdict# Function to count the number of # substrings with sum equal to lengthdef 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 codestr = '112112'n = len(str)print(countSubstrings(str, n))# This code is contributed by Stuti Pathak |
C#
// C# program to implement// the above approachusing System;using System.Collections.Generic;class GFG{// Function to count the number of// subStrings with sum equal to lengthstatic 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 Codepublic 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 |
6
Time Complexity: O(N)
Auxiliary Space: O(N)
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