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

  • Last Updated : 11 Jun, 2021

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++




// 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);
}

Java




// 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));
    }
}

C#




// 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));
    }
}

Javascript




<script>
// Javascript implementation of above approach
 
 
// Function to count the
// number of leading zeros
function countLeadingZeros(x) {
    let y;
    let 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
function FindStringof1s(x, n) {
    let 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
let x = 35;
let n = 2;
document.write(FindStringof1s(x, n));
 
// This code is contributed by gfgking.
</script>
Output: 
31

 

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