Here's a step by step solution to solving the binary gap challenge on codility. Hope it helps!

**QUESTION**

A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N.

For example, number 9 has binary representation 1001 and contains a binary gap of length 2. The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3. The number 20 has binary representation 10100 and contains one binary gap of length 1. The number 15 has binary representation 1111 and has no binary gaps. The number 32 has binary representation 100000 and has no binary gaps.

Write a function:

function solution(N);

that, given a positive integer N, returns the length of its longest binary gap. The function should return 0 if N doesn't contain a binary gap.

For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its longest binary gap is of length 5. Given N = 32 the function should return 0, because N has binary representation '100000' and thus no binary gaps.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..2,147,483,647].

**BREAKDOWN**

Our example here is the number 10041

To solve this we need to have pointers i and j which will help us count the number of 0's.

After that we need to :

- Convert the binary number 10041 to string
- Initiate pointers i & j
- Set pointers i & j to 0
- Increment j until it hits a 1
- When j hits a 1, set the value of i to be equal to j (i=j)
- Store the value of (j-i-1) which is the value of the count of the first set of 0's
- Repeat for the entire array

**SOLUTION CODE**

Here's the code solution from the breakdown above:

```
function solution(N){
let binaryN = (N >>>0).toString(2); //Changes the binary number to string
let maxGap = 0; //sets the maximum number of binary gaps to 0
for(let i = 0; i<binaryN.length; i++){ //sets the counter i to zero and loops through the entire binary length while incrementing by 1
for(let j=i; j<binaryN.length; j++){ //sets the counter j to i and loops through the entire binary length while incrementing by 1
if(binaryN[i] == 1 && binaryN[j] && i!=j){ //checks whether both i and j are equal to 1 but not equal to each other
if(maxGap< j-i-1){ //Calculates the maximum gap
maxGap = j-i-1;
}
i=j;
}
}
}
return maxGap;
}
```

## Discussion (1)