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

• Last Updated : 04 Jul, 2022

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 ``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

 ``

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 ``using` `namespace` `std;` `// Function to find the string``// of n consecutive 1's``int` `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 code``int` `main()``{``    ``int` `x = 35;``    ``int` `n = 2;` `    ``// Function call``    ``cout << FindStringof1s(x, n);``}` `// This code is contributed by phasing17`

## Python3

 `# Python3 implementation of above approach`  `# Function to find the string``# of n consecutive 1's``def` `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 code``x ``=` `35``n ``=` `2` `# Function call``print``(FindStringof1s(x, n))`   `# This code is contributed by phasing17`

Output

`31`

Time Complexity: O(logn)

Auxiliary Space: O(1)

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