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++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to count the// number of leading zerosint 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'sint 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 codeint main(){ int x = 35; int n = 2; cout << FindStringof1s(x, n);} |
Java
// Java implementation of above approachimport java.io.*;class GFG {// Function to count the// number of leading zerosstatic 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'sstatic 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 approachusing System;public class GFG{ // Function to count the// number of leading zerosstatic 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'sstatic 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 zerosfunction 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'sfunction 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 codelet x = 35;let n = 2;document.write(FindStringof1s(x, n));// This code is contributed by gfgking.</script> |
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