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

I got 100% with this code.

function solution(N) {
    let binary = parseInt(N).toString(2);
    let remove_last_zero = binary.slice(0, binary.lastIndexOf("1") + 1);
    let split_1 = remove_last_zero.split("1");
    let remove_empty = split_1.filter(i => i != "");
    if (typeof remove_empty !== 'undefined' && remove_empty.length == 0) {
        return 0;
    }
    let zero_count = remove_empty.map(b => b.length);
    return Math.max.apply(null, zero_count);
}

I did it like this

`function solution(N) {
const binary = N.toString(2)
let finalMax = 0, currentMax = 0

for(let i = 0; i < binary.length; i++) {
if(binary[i] == 1) {
i++
while(binary[i] == 0 && i < binary.length) {
currentMax++
i++
}
if (i == binary.length && binary[i] != 1) {
currentMax = 0
}
if(finalMax < currentMax) {
finalMax = currentMax
currentMax = 0
}
i--
}
currentMax = 0
}
return finalMax
}`

I actually used this approach. How more ugly could it be? lol
function solution(N){
return Math.max(...countZeroesLengthInArray(splitNum(ignoreZeroesAfterLastOne(changeToBin(N)))))
}

const changeToBin = (num) => {
const toChange = num;
return toChange.toString(2);
}

const ignoreZeroesAfterLastOne = (numInString) => {
return numInString.slice(0, numInString.lastIndexOf("1") + 1);
}

const splitNum = (num) => {
return num.split("1");
}

const countZeroesLengthInArray = (arr) => {
return arr.map((curr) => {
return curr.length;
},0);
}

I never saw so many solutions, which are kind of missing the point. Well actually the solution to transform to a "binary" string and splitting is actually funny, but I think most of you are just writing garbage.

The point of the task is to write the most efficient way to calculate the binary gap. This means to use less memory (=> use as less variables as possible), use less branching and use less calculations as possible.

My Solution:

function solution(N) {

    let start = null;
    let max = 0;

    for (let pos = 31; pos >= 0; pos--) {
        if ((N >> pos & 1) === 1) {
            if (start === null) {
                start = pos;
            } else {
                max = max < (start - pos) ? start - pos - 1 : max;
                start = pos;
            }
        }
    }

    return max;
}

Or even less by doing less subtractions:

function solution(N) {

    let start = null;
    let max = 1;

    for (let pos = 31; pos >= 0; pos--) {
        if ((N >> pos & 1) === 1) {
            if (start === null) {
                start = pos;
            } else {
                max = max < (start - pos ) ? start - pos : max;
                start = pos;
            }
        }
    }

    return --max;
}

Further improvements could be e.g. to first ignore all 0 at the beginning with a while loop and set the start then go through the for loop avoiding the branching if (start === null).

But really guys... some should be ashamed...

function solution(N) {
    const number = N.toString(2)
    let max = 0;
    let currentMax = 0;
    for (let index = 0; index < number.length; index++) {
      if (number[index] === "0") currentMax++
      if (number[index] === "1") {
        max = Math.max(currentMax, max)
        currentMax = 0
      }
    }
    return max
}

The easiest solution

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

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;
  }