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> |
Output
31
Approach: Using predefined functions
x can be converted to a binary string using in-built methods, such as bitset<32>(x).to_string() in C++ STL, bin() in Python3, Integer.toBinary() in Java . And, a string of n 1s can be constructed, whose index in the binary string can be located using predefined functions such as find in C++ STL, index in Python3 and indexOf() in Java.
This approach can be summarized to the following steps:
1. Form the binary string equivalent of the number using in-built methods, such as bitset<32>(x).to_string() in C++ STL, bin() in Python3, Integer.toBinary() in Java.
2. Then, build a string consisting of n 1s, using in-built methods, such as string (n, ‘1’) in C++ STL, ‘1’ * n in Python3, and “1”.repeat(n) in Java.
3. Then, find the index of the string of n 1s in the binary string using in-built methods such as .find() in C++ STL, index() in Python3 and indexOf() in Java.
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to find the string// of n consecutive 1'sint FindStringof1s(unsigned x, int n){ // converting x to a binary string string bin = bitset<32>(x).to_string(); // constructing a string of n 1s string ones(n, '1'); // locating the ones string in bin auto pos = bin.find(ones); if (pos != string::npos) return pos + 1; // if no string is found return -1;}// Driver codeint main(){ int x = 35; int n = 2; // Function call cout << FindStringof1s(x, n);}// This code is contributed by phasing17 |
Java
import java.util.Arrays;public class GFG { // Function to find the string of n consecutive 1's public static int FindStringof1s(int x, int n) { // converting x to a binary string String bin = String.format("%32s", Integer.toBinaryString(x)).replace(' ', '0'); // constructing a string of n 1s char[] onesArray = new char[n]; Arrays.fill(onesArray, '1'); String ones = new String(onesArray); // locating the ones string in bin int pos = bin.indexOf(ones); if (pos != -1) return pos + 1; // if no string is found return -1; } // Driver code public static void main(String[] args) { int x = 35; int n = 2; // Function call System.out.println(FindStringof1s(x, n)); }} |
Python3
# Python3 implementation of above approach# Function to find the string# of n consecutive 1'sdef FindStringof1s(x, n): # converting x to a binary string bin_ = bin(x).zfill(32) # constructing a string of n 1s ones = n * "1" # locating the ones string in bin if ones in bin_: return bin_.index(ones) + 1 # if no string is found return -1;# Driver codex = 35n = 2# Function callprint(FindStringof1s(x, n))# This code is contributed by phasing17 |
C#
using System;public class GFG{// Function to find the string// of n consecutive 1'spublic static int FindStringof1s(uint x, int n){// converting x to a binary stringstring bin = Convert.ToString(x, 2).PadLeft(32, '0'); // constructing a string of n 1s string ones = new string('1', n); // locating the ones string in bin int pos = bin.IndexOf(ones); if (pos != -1) return pos + 1; // if no string is found return -1;}// Driver codepublic static void Main(){ uint x = 35; int n = 2; // Function call Console.WriteLine(FindStringof1s(x, n));}} |
Javascript
function FindStringof1s(x, n) { // converting x to a binary string let bin = x.toString(2).padStart(32, '0'); // constructing a string of n 1s let ones = '1'.repeat(n); // locating the ones string in bin let pos = bin.indexOf(ones); if (pos != -1) return pos + 1; // if no string is found return -1;}// Driver codelet x = 35;let n = 2;// Function callconsole.log(FindStringof1s(x, n)); |
Output
31
Time Complexity: O(log n)
Auxiliary Space: O(1)
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