Largest Binary Gap (JavaScript)

largestBinaryGap.js

function largestBinaryGap(num) {
  var bin = Math.abs(num).toString(2),
      finalMax = 0,
      currentMax;
  
  for (var i = 0; i < bin.length; i++) {
    currentMax = 0;
    while (bin[i] === "0") {
      ++currentMax && ++i;
    }
    finalMax = Math.max(finalMax, currentMax);
  }
  return finalMax;
}

console.log(largestBinaryGap(1)); // 1 //=> 0
console.log(largestBinaryGap(5)); // 101 //=> 1
console.log(largestBinaryGap(6)); // 110 //=> 1
console.log(largestBinaryGap(19)); // 10011 //=> 2
console.log(largestBinaryGap(1041)); // 10000010001 //=> 5
console.log(largestBinaryGap(6291457)); // 11000000000000000000001 //=> 20
console.log(largestBinaryGap(1376796946)); // 1010010000100000100000100010010 //=> 5
console.log(largestBinaryGap(1610612737)); // 1100000000000000000000000000001 //=> 28

Simple solution with 100% score:

function solution(N) {
  let newN = N.toString(2);
  let max = current = 0;
  for (char of newN) {
    if (char == "1") {
      max = Math.max(current, max);
      current = 0;
    } else {
      current++;
    }
  }
  return max;
}

 function solution(N) {
  return N.toString(2)
    .split("1")
    .slice(1, -1)
    .reduce((a, b) => (a > b.length ? a : b.length), 0);
}

This was the one that made the most sense to me and helped debug my own code, thank you so much!

After reading all these solutions, I came up with something that beginners can really easily understand:

const highestBinaryGap = (n)  => {
    let binary_number = n.toString(2)
    splited_binary_number = binary_number.toString().split("1")
    let maxCharacterArray = []
    for (let i = 0; i < splited_binary_number.length; i++) {
        if (splited_binary_number[i] !== "") {
            let length = splited_binary_number[i].length
            maxCharacterArray.push(length)
        }
    }
    let theHigestOccurance = Math.max(...maxCharacterArray)
    return theHigestOccurance
}
// Usage
theHigestBinaryGap = highestBinaryGap(593)
console.log(theHigestBinaryGap)

This is my solution with very less code and no direct loops

  function solution(N) {
      const binnum = Number(N).toString(2);
      let str = binnum.split('1');
      const newarr = str.filter(function(item,index){
          if (item === ''){
          }else if (index === str.length -1) {
          }else {
              return item;
          }
      })
      return newarr.length > 0 ? newarr.sort()[newarr.length -1].length : 0;
  }