LeetCode 868: Binary Gap Solution

Problem Explanation

Explanation

To solve this problem, we need to convert the given integer n into its binary representation and then iterate through the binary digits to find the longest distance between any two adjacent 1's.

Convert the integer n to its binary representation.
Iterate through the binary representation and keep track of the distance between adjacent 1's.
Update the maximum distance found so far.
Return the maximum distance as the result.

Time Complexity

The time complexity of this solution is O(log n) because we need to convert the integer n into its binary representation, which requires log n bits.

Space Complexity

The space complexity of this solution is O(log n) to store the binary representation of the integer n.

Solution Code

class Solution {
    public int binaryGap(int n) {
        String binary = Integer.toBinaryString(n);
        int maxDistance = 0;
        int lastOneIndex = -1;
        
        for (int i = 0; i < binary.length(); i++) {
            if (binary.charAt(i) == '1') {
                if (lastOneIndex != -1) {
                    maxDistance = Math.max(maxDistance, i - lastOneIndex);
                }
                lastOneIndex = i;
            }
        }
        
        return maxDistance;
    }
}

Frequently Asked Questions

How to solve LeetCode 868 (Binary Gap)?

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What is the time complexity of LeetCode 868 (Binary Gap)?

The time complexity for LeetCode 868 (Binary Gap) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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868. Binary Gap