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# Count of groups of consecutive 1s in a given Binary String

• Last Updated : 06 Jul, 2021

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:

1. Group over the range [0, 0] which is equal to “1”.
2. Group over the range [3, 4] which is equal to “11”.
3. 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 ``using` `namespace` `std;` `// Function to find the number of the``// groups of 1s only in the binary``// string``int` `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 code``int` `main()``{``    ``// Input``    ``string S = ``"100110111"``;``    ``int` `N = S.length();` `    ``// Function call``    ``cout << groupsOfOnes(S, N) << endl;``    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.Stack;` `class` `GFG{` `// Function to find the number of the``// groups of 1s only in the binary``// string``static` `int` `groupsOfOnes(String S, ``int` `N)``{``    ` `    ``// Stores number of groups of 1s``    ``int` `count = ``0``;` `    ``// Initialization of the stack``    ``Stack 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 code``public` `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``# string``def` `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 code``if` `__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 approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to find the number of the``// groups of 1s only in the binary``// string``static` `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 code``public` `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

 ``
Output
`3`

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

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