This is how I solved the binary gap problem: Find longest sequence of zeros, bounded by ones, in binary representation of an integer

I wonder how does it fare against solutions such as ones appeared here? Codility binary gap solution using regex

function solution(N) {
    const bin = N.toString(2);

    let currentGap = 0;
    let gaps = [];

    for (i=0; i<bin.length; i++){

      if (bin[i]==="0"){

        if (bin[i+1]==="1"){
          currentGap = 0;

    if (gaps.length===1){
      return gaps[0];
    } else if (gaps.length>1){
      return Math.max(...gaps)
    } else {
      return 0

Here's my approach:

function solution(n) { 
    var maxZeros = 0; 
    while(n !== 0 && n % 2 === 0) {
        n >>>= 1;
    for(var curr=0; n !== 0; n>>>=1) { 
        if(n % 2 === 0) { 
        } else { 
            curr = 0; 
        maxZeros = Math.max(maxZeros, curr); 
    return maxZeros; 

Some notable differences from your solution:

  • Number isn't converted to binary string. This is half for optimization purposes and half for lack of necessity.
  • Approach to handling the "zero gap must be bound by 1s" requisite. The first 1 is automatically handled because it wouldn't be zero if there were still a 1 to handle. However the lower 1 bound is simply handled by shifting until the first 1 is in the lowest digit, eliminating the need to add flags or extra handling.
  • Notice that no memory is required to hold gap information. It is irrelevant as you can save only the information required as you move along.
  • Value n is checked against being 0 as opposed to being greater than zero just because a negative number should not be disregarded just because it is negative.

Hope that helps! If you have any questions, just ask!

  • \$\begingroup\$ Isn't that O(log n)? \$\endgroup\$ – Kruga Jun 6 '17 at 12:38
  • \$\begingroup\$ @Kruga You're right, and OP's would be O(n log n). Nicely spotted. \$\endgroup\$ – Neil Jun 6 '17 at 12:39
  • \$\begingroup\$ What am I saying? OP wasn't calling Math.max inside the loop. That just makes both O(log n). My bad. Removing. \$\endgroup\$ – Neil Jun 6 '17 at 12:42
  • \$\begingroup\$ Hi Neil, interesting solution there! I'll surely inspect it later for actual performance differences using codility out of curiosity. \$\endgroup\$ – kdenz Jun 7 '17 at 19:30

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