Find all K length subarrays containing only 1s in given Binary String

Last Updated : 29 Dec, 2021

Given a binary string str[], the task is to find all possible K length subarrays containing only 1s and print their starting and ending index.

Examples:

Input: str = "0101000", K=1
Output:
1 1
3 3
Explanation: Substrings at positions 1 and 3 are the substrings with value 1.

Input: str = "11111001", K=3
Output:
0 2
1 3
2 4

Approach: The given problem can be solved with the help of the Sliding Window technique. Create a window of size K initially with a count of 1s from range 0 to K-1. Then traverse the string from index 1 to N-1 and subtract the value of i-1 and add the value of i+K to the current count. If the current count is equal to K, print the value of i and i+k-1 as one of the possible subarrays. Follow the steps below to solve the problem:

Initialize the variable cntOf1s as 0.
Iterate over the range [0, K) using the variable i and perform the following tasks:
- If s[i] equals 1 then increase the value of cntOf1s by 1.
If cntOf1s equals K then print the current substring as one of the answer.
Iterate over the range [1, N) using the variable i and perform the following tasks:
- Reduce the value from the left and add from the right to the variable cntOf1s if the characters at that position are 1.
- If the value of cntOf1s equals K, then print the current substring as the answer.

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 all possible // k length subarrays void get(string s, int k) {     int n = s.length();      int cntOf1s = 0;      // Initial window     for (int i = 0; i < k; i++)         if (s[i] == '1')             cntOf1s++;      if (cntOf1s == k)         cout << 0 << " " << k - 1 << endl;      // Traverse the string     for (int i = 1; i < n; i++) {          // Subtract the value from the left and         // add from the right         cntOf1s = cntOf1s - (s[i - 1] - '0')                   + (s[i + k - 1] - '0');          // Check the condition         if (cntOf1s == k)             cout << i << " " << i + k - 1 << endl;     } }  // Driver Code int main() {     string str = "0110101110";     int K = 2;     get(str, K);     return 0; }

Java

// Java program for the above approach import java.util.*; class GFG{  // Function to find all possible // k length subarrays static void get(char[] s, int k) {     int n = s.length;      int cntOf1s = 0;      // Initial window     for (int i = 0; i < k; i++)         if (s[i] == '1')             cntOf1s++;      if (cntOf1s == k)         System.out.print(0+ " " +  (k - 1 )+"\n");      // Traverse the String     for (int i = 1; i < n && (i + k - 1)<n; i++) {          // Subtract the value from the left and         // add from the right         cntOf1s = cntOf1s - (s[i - 1] - '0')                   + (s[i + k - 1] - '0');          // Check the condition         if (cntOf1s == k)             System.out.print(i+ " " +  (i + k - 1 )+"\n");     } }  // Driver Code public static void main(String[] args) {     String str = "0110101110";     int K = 2;     get(str.toCharArray(), K); } }  // This code is contributed by 29AjayKumar

Python3

# python3 program for the above approach  # Function to find all possible # k length subarrays def get(s, k):      n = len(s)     cntOf1s = 0      # Initial window     for i in range(0, k):         if (s[i] == '1'):             cntOf1s += 1      if (cntOf1s == k):         print(f"{0} {k - 1}")      # Traverse the string     for i in range(1, n - k + 1):          # Subtract the value from the left and         # add from the right         cntOf1s = cntOf1s - (ord(s[i - 1]) - ord('0')) + \             (ord(s[i + k - 1]) - ord('0'))          # Check the condition         if (cntOf1s == k):             print(f"{i} {i + k - 1}")  # Driver Code if __name__ == "__main__":      str = "0110101110"     K = 2     get(str, K)  # This code is contributed by rakeshsahni

// C# program for the above approach using System; class GFG {    // Function to find all possible   // k length subarrays   static void get(char[] s, int k)   {     int n = s.Length;      int cntOf1s = 0;      // Initial window     for (int i = 0; i < k; i++)       if (s[i] == '1')         cntOf1s++;      if (cntOf1s == k)       Console.Write(0 + " " + (k - 1) + "\n");      // Traverse the String     for (int i = 1; i < n && (i + k - 1) < n; i++)     {        // Subtract the value from the left and       // add from the right       cntOf1s = cntOf1s - (s[i - 1] - '0')         + (s[i + k - 1] - '0');        // Check the condition       if (cntOf1s == k)         Console.Write(i + " " + (i + k - 1) + "\n");     }   }    // Driver Code   public static void Main()   {     String str = "0110101110";     int K = 2;     get(str.ToCharArray(), K);   } }  // This code is contributed by Saurabh Jaiswal

JavaScript

  <script>         // JavaScript code for the above approach          // Function to find all possible         // k length subarrays         function get(s, k) {             let n = s.length;              let cntOf1s = 0;              // Initial window             for (let i = 0; i < k; i++)                 if (s[i] == '1')                     cntOf1s++;              if (cntOf1s == k)                 document.write(0 + ' ' + (k - 1) + '<br>');              // Traverse the string             for (let i = 1; i < n; i++) {                  // Subtract the value from the left and                 // add from the right                 cntOf1s = cntOf1s - (s[i - 1].charCodeAt(0) - '0'.charCodeAt(0))                     + (s[i + k - 1].charCodeAt(0) - '0'.charCodeAt(0));                  // Check the condition                 if (cntOf1s == k)                     document.write(i + ' ' + (i + k - 1) + '<br>');             }         }          // Driver Code         let str = "0110101110";          let K = 2;         get(str, K);    // This code is contributed by Potta Lokesh     </script>

Output

1 2  6 7  7 8

Time Complexity: O(N)
Auxiliary Space: O(1)

Count of substrings of a Binary string containing only 1s

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Find all K length subarrays containing only 1s in given Binary String

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