Given a binary string S of size N, the task is to find the number of groups of 1s only in the string S.
Examples:
Input: S = “100110111”, N = 9
Output: 3
Explanation:
The following groups are of 1s only:
- Group over the range [0, 0] which is equal to “1”.
- Group over the range [3, 4] which is equal to “11”.
- Group over the range [6, 8] which is equal to “111”.
Therefore, there are a total of 3 groups of 1s only.
Input: S = “0101”
Output: 2
Approach: The problem can be solved by iterating over the characters of the string. Follow the steps below to solve the problem:
- Initialize a variable, say count as 0, which stores the number of substrings of 1s in S.
- Initialize a stack say st to store the substring before an index of 1s only.
- Iterate over the characters of the string S, using the variable i and do the following:
- If the current character is 1, push 1 into stack st.
- Otherwise, If st is not empty, increment count by 1. Else Clear st.
- If st is not empty, increment count by 1, i.e If there is a suffix of 1s.
- Finally, print the total count obtained.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the number of the// groups of 1s only in the binary// stringint groupsOfOnes(string S, int N){ // Stores number of groups of 1s int count = 0; // Initialization of the stack stack<int> st; // Traverse the string S for (int i = 0; i < N; i++) { // If S[i] is '1' if (S[i] == '1') st.push(1); // Otherwise else { // If st is empty if (!st.empty()) { count++; while (!st.empty()) { st.pop(); } } } } // If st is not empty if (!st.empty()) count++; // Return answer return count;}// Driver codeint main(){ // Input string S = "100110111"; int N = S.length(); // Function call cout << groupsOfOnes(S, N) << endl; return 0;} |
Java
// Java program for the above approachimport java.util.Stack;class GFG{// Function to find the number of the// groups of 1s only in the binary// stringstatic int groupsOfOnes(String S, int N){ // Stores number of groups of 1s int count = 0; // Initialization of the stack Stack<Integer> st = new Stack<>(); // Traverse the string S for(int i = 0; i < N; i++) { // If S[i] is '1' if (S.charAt(i) == '1') st.push(1); // Otherwise else { // If st is empty if (!st.empty()) { count++; while (!st.empty()) { st.pop(); } } } } // If st is not empty if (!st.empty()) count++; // Return answer return count;}// Driver codepublic static void main(String[] args){ // Input String S = "100110111"; int N = S.length(); // Function call System.out.println(groupsOfOnes(S, N));}}// This code is contributed by abhinavjain194 |
Python3
# Python3 program for the above approach# Function to find the number of the# groups of 1s only in the binary# stringdef groupsOfOnes(S, N): # Stores number of groups of 1s count = 0 # Initialization of the stack st = [] # Traverse the string S for i in range(N): # If S[i] is '1' if (S[i] == '1'): st.append(1) # Otherwise else: # If st is empty if (len(st) > 0): count += 1 while (len(st) > 0): del st[-1] # If st is not empty if (len(st)): count += 1 # Return answer return count # Driver codeif __name__ == '__main__': # Input S = "100110111" N = len(S) # Function call print(groupsOfOnes(S, N)) # This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the number of the// groups of 1s only in the binary// stringstatic int groupsOfOnes(string S, int N){ // Stores number of groups of 1s int count = 0; // Initialization of the stack Stack<int> st = new Stack<int>(); // Traverse the string S for (int i = 0; i < N; i++) { // If S[i] is '1' if (S[i] == '1') st.Push(1); // Otherwise else { // If st is empty if (st.Count > 0) { count++; while (st.Count > 0) { st.Pop(); } } } } // If st is not empty if (st.Count > 0) count++; // Return answer return count;} // Driver codepublic static void Main(){ // Input string S = "100110111"; int N = S.Length; // Function call Console.Write(groupsOfOnes(S, N));}}// This code is contributed by SURENDRA_GANGWAR. |
Javascript
<script> // JavaScript program for the above approach // Function to find the number of the // groups of 1s only in the binary // string function groupsOfOnes(S, N) { // Stores number of groups of 1s let count = 0; // Initialization of the stack var st = []; // Traverse the string S for (let i = 0; i < N; i++) { // If S[i] is '1' if (S[i] == '1') st.push(1); // Otherwise else { // If st is empty if (st.length != 0) { count++; while (st.length != 0) { st.pop(); } } } } // If st is not empty if (st.length != 0) count++; // Return answer return count; } // Driver code // Input var S = "100110111"; let N = S.length; // Function call document.write(groupsOfOnes(S, N));// This code is contributed by Potta Lokesh</script> |
Output
3
Time Complexity: O(N)
Auxiliary Space: O(N)
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