Position of leftmost set bit in given binary string where all 1s appear at end

Last Updated : 15 Feb, 2022

Given a binary string S of length N, such that all 1s appear on the right. The task is to return the index of the first set bit found from the left side else return -1.

Examples:

Input: s = 00011, N = 5
Output: 3
Explanation: The first set bit from the left side is at index 3.

Input: s = 0000, N = 4
Output: -1

Approach: This problem can be solved using Binary Search.

Apply Binary search in the range [1, N] to find the first set bit as follows:
Update mid as (l+r)/2
If s[mid] is set bit, update ans as mid and r as mid-1
else update l as mid + 1
Return the last stored value in ans.

Below is the implementation of the above approach:

C++

// C++ Program to find Position // Of leftmost set bit #include <iostream> using namespace std;  // Function to find // A bit is set or not bool isSetBit(string& s, int i) {     return s[i] == '1'; }  // Function to find // First set bit int firstSetBit(string& s, int n) {     long l = 0, r = n, m, ans = -1;      // Applying binary search     while (l <= r) {          m = (l + r) / 2;         if (isSetBit(s, m)) {              // store the current             // state of m in ans             ans = m;             r = m - 1;         }         else {             l = m + 1;         }     }      // Returning the position     // of first set bit     return ans; }  // Driver code int main() {      string s = "111";     int n = s.length();     cout << firstSetBit(s, n);      return 0; }

Java

// Java program for the above approach class GFG {    // Function to find   // A bit is set or not   static boolean isSetBit(String s, int i)   {     return s.charAt(i) == '1' ? true : false;   }    // Function to find   // First set bit   static int firstSetBit(String s, int n)   {     int l = 0, r = n, m, ans = -1;      // Applying binary search     while (l <= r) {        m = (l + r) / 2;       if (isSetBit(s, m)) {          // store the current         // state of m in ans         ans = m;         r = m - 1;       }       else {         l = m + 1;       }     }      // Returning the position     // of first set bit     return ans;   }    // Driver Code   public static void main(String args[])   {     String s = "111";     int n = s.length();     System.out.print(firstSetBit(s, n));   } }  // This code is contributed by gfgking

// C# program for the above approach using System; using System.Collections.Generic;  class GFG {    // Function to find   // A bit is set or not   static bool isSetBit(string s, int i)   {     return s[i] == '1';   }    // Function to find   // First set bit   static int firstSetBit(string s, int n)   {     int l = 0, r = n, m, ans = -1;      // Applying binary search     while (l <= r) {        m = (l + r) / 2;       if (isSetBit(s, m)) {          // store the current         // state of m in ans         ans = m;         r = m - 1;       }       else {         l = m + 1;       }     }      // Returning the position     // of first set bit     return ans;   }    // Driver Code   public static void Main()   {     string s = "111";     int n = s.Length;     Console.Write(firstSetBit(s, n));   } }  // This code is contributed by sanjoy_62.

JavaScript

  <script>         // JavaScript code for the above approach          // Function to find         // A bit is set or not         function isSetBit(s, i) {             return s[i] == '1';         }          // Function to find         // First set bit         function firstSetBit(s, n) {             let l = 0, r = n, m, ans = -1;              // Applying binary search             while (l <= r) {                  m = Math.floor((l + r) / 2);                 if (isSetBit(s, m)) {                      // store the current                     // state of m in ans                     ans = m;                     r = m - 1;                 }                 else {                     l = m + 1;                 }             }              // Returning the position             // of first set bit             return ans;         }          // Driver code         let s = "111";         let n = s.length;         document.write(firstSetBit(s, n));         // This code is contributed by Potta Lokesh     </script>

Python

# Python Program to find Position # Of leftmost set bit  # Function to find # A bit is set or not def isSetBit(s, i):          return s[i] == '1'  # Function to find # First set bit def firstSetBit(s, n):      l = 0     r = n     m = 0     ans = -1      # Applying binary search     while (l <= r):          m = (l + r) // 2         if (isSetBit(s, m)):              # store the current             # state of m in ans             ans = m             r = m - 1          else:             l = m + 1      # Returning the position     # of first set bit     return ans  # Driver code  s = "111" n = len(s) print(firstSetBit(s, n))  # This code is contributed by Samim Hossain Mondal.

Output:

Time Complexity: O(LogN)
Auxiliary Space: o(1)

Find any permutation of Binary String of given size not present in Array

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Position of leftmost set bit in given binary string where all 1s appear at end

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