Find consecutive 1s of length >= n in binary representation of a number

Given two integers x and n, the task is to search for the first consecutive stream of 1s (in the x’s 32-bit binary representation) which is greater than or equal to n in length and return its position. If no such string exists then return -1.

Examples:

Input: x = 35, n = 2
Output: 31
Binary representation of 35 is 00000000000000000000000000100011 and two consecutive 1’s are present at position 31.

Input: x = 32, n = 3
Output: -1
32 = 00000000000000000000000000100000 in binary and it does not have a sub-string of 3 consecutive 1’s.

Approach: Use Bitwise operation to calculate the no. of leading zeros in the number and then use it to find the position from where we need to start searching for consecutive 1’s. Skip the search for leading zeros.



Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to count the
// number of leading zeros
int countLeadingZeros(int x)
{
    unsigned y;
    int n;
    n = 32;
    y = x >> 16;
    if (y != 0) {
        n = n - 16;
        x = y;
    }
    y = x >> 8;
    if (y != 0) {
        n = n - 8;
        x = y;
    }
    y = x >> 4;
    if (y != 0) {
        n = n - 4;
        x = y;
    }
    y = x >> 2;
    if (y != 0) {
        n = n - 2;
        x = y;
    }
    y = x >> 1;
    if (y != 0)
        return n - 2;
    return n - x;
}
  
// Function to find the string
// of n consecutive 1's
int FindStringof1s(unsigned x, int n)
{
    int k, p;
  
    // Initialize position to return.
    p = 0;
    while (x != 0) {
  
        // Skip leading 0's
        k = countLeadingZeros(x);
        x = x << k;
  
        // Set position after leading 0's
        p = p + k;
  
        // Count first group of 1's.
        k = countLeadingZeros(~x);
  
        // If length of consecutive 1's
        // is greater than or equal to n
        if (k >= n)
            return p + 1;
  
        // Not enough 1's
        // skip over to next group
        x = x << k;
  
        // Update the position
        p = p + k;
    }
  
    // if no string is found
    return -1;
}
  
// Driver code
int main()
{
    int x = 35;
    int n = 2;
    cout << FindStringof1s(x, n);
}

chevron_right






                        filter_none









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java  implementation of above approach 


  


import java.io.*; 


  


class GFG { 


// Function to count the 


// number of leading zeros 


static int countLeadingZeros(int x) 


{ 


    int y; 


    int n; 


    n = 32; 


    y = x >> 16; 


    if (y != 0) { 


        n = n - 16; 


        x = y; 


    } 


    y = x >> 8; 


    if (y != 0) { 


        n = n - 8; 


        x = y; 


    } 


    y = x >> 4; 


    if (y != 0) { 


        n = n - 4; 


        x = y; 


    } 


    y = x >> 2; 


    if (y != 0) { 


        n = n - 2; 


        x = y; 


    } 


    y = x >> 1; 


    if (y != 0) 


        return n - 2; 


    return n - x; 


} 


  


// Function to find the string 


// of n consecutive 1's 


static int FindStringof1s(int x, int n) 


{ 


    int k, p; 


  


    // Initialize position to return. 


    p = 0; 


    while (x != 0) { 


  


        // Skip leading 0's 


        k = countLeadingZeros(x); 


        x = x << k; 


  


        // Set position after leading 0's 


        p = p + k; 


  


        // Count first group of 1's. 


        k = countLeadingZeros(~x); 


  


        // If length of consecutive 1's 


        // is greater than or equal to n 


        if (k >= n) 


            return p + 1; 


  


        // Not enough 1's 


        // skip over to next group 


        x = x << k; 


  


        // Update the position 


        p = p + k; 


    } 


  


    // if no string is found 


    return -1; 


} 


  


// Driver code 


      


    public static void main (String[] args) { 


  


    int x = 35; 


    int n = 2; 


    System.out.println(FindStringof1s(x, n)); 


    } 


} 



















                        chevron_right







                        filter_none









C#














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// C# implementation of above approach 


  


using System; 


  


public class GFG{ 


      


// Function to count the 


// number of leading zeros 


static int countLeadingZeros(int x) 


{ 


    int y; 


    int n; 


    n = 32; 


    y = x >> 16; 


    if (y != 0) { 


        n = n - 16; 


        x = y; 


    } 


    y = x >> 8; 


    if (y != 0) { 


        n = n - 8; 


        x = y; 


    } 


    y = x >> 4; 


    if (y != 0) { 


        n = n - 4; 


        x = y; 


    } 


    y = x >> 2; 


    if (y != 0) { 


        n = n - 2; 


        x = y; 


    } 


    y = x >> 1; 


    if (y != 0) 


        return n - 2; 


    return n - x; 


} 


  


// Function to find the string 


// of n consecutive 1's 


static int FindStringof1s(int  x, int n) 


{ 


    int k, p; 


  


    // Initialize position to return. 


    p = 0; 


    while (x != 0) { 


  


        // Skip leading 0's 


        k = countLeadingZeros(x); 


        x = x << k; 


  


        // Set position after leading 0's 


        p = p + k; 


  


        // Count first group of 1's. 


        k = countLeadingZeros(~x); 


  


        // If length of consecutive 1's 


        // is greater than or equal to n 


        if (k >= n) 


            return p + 1; 


  


        // Not enough 1's 


        // skip over to next group 


        x = x << k; 


  


        // Update the position 


        p = p + k; 


    } 


  


    // if no string is found 


    return -1; 


} 


  


// Driver code 


      


    static public void Main (){ 


    int x = 35; 


    int n = 2; 


    Console.WriteLine (FindStringof1s(x, n)); 


    } 


} 



















                        chevron_right







                        filter_none











Output:


31





competitive-programming-img





    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:
Shubham__Ranjan
Check out this Author's contributed articles.
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on  the "Improve Article" button below.


Improved By :  Sach_Code