To solve the problem of finding the longest binary gap in a given positive integer ( N ), we can follow these steps:

Convert the integer ( N ) into its binary representation. In JavaScript, you can do this using
toString(2)
method. 
Traverse through the binary representation to identify sequences of consecutive zeros (‘0’). Keep track of the length of each sequence.

As we identify these sequences, keep track of the maximum length encountered.

Finally, return the length of the longest binary gap found. If no binary gaps are found, return 0.
Here’s the JavaScript implementation based on the described approach:
function solution(N) {
const binaryString = N.toString(2); // Convert N to binary string
let maxGapLength = 0;
let currentGapLength = 0;
let inGap = false;
for (let i = 0; i < binaryString.length; i++) {
if (binaryString[i] === '1') {
if (inGap) {
// We've encountered the end of a gap
if (currentGapLength > maxGapLength) {
maxGapLength = currentGapLength;
}
currentGapLength = 0;
}
inGap = true;
} else {
if (inGap) {
// We are inside a gap
currentGapLength++;
}
}
}
return maxGapLength;
}
N.toString(2)
converts ( N ) to its binary string representation.
Then we ierate through each character of the binary string
When encountering ‘1’, check if we were already in a gap (i.e., inGap
is true). If so, this signifies the end of a gap, so update maxGapLength
if the current gap is longer than previously recorded.
When encountering ‘0’, if we are in a gap (inGap
is true), increment currentGapLength
.
After the loop, ensure to check if the last recorded gap (if any) is the longest.
This algorithm is efficient with a time complexity of ( O(log N) ), where ( N ) is the integer input, due to the conversion to binary and subsequent traversal of its digits. This ensures it can handle the maximum input size within reasonable time constraints.