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)

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

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

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