LeetCode 2595: Number of Even and Odd Bits Solution

Master LeetCode problem 2595 (Number of Even and Odd Bits), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2595. Number of Even and Odd Bits

Easy

Bit Manipulation

Problem Explanation

Explanation

To solve this problem, we need to count the number of 1s at even and odd positions in the binary representation of the given integer n. We can do this by iterating through the binary representation of n and counting the number of 1s at even and odd positions.

Initialize two variables even and odd to count the number of 1s at even and odd positions.
Iterate through the binary representation of n from right to left.
Keep track of the position (index) of the bit, starting from index 0 for the rightmost bit.
If the current bit is 1 and the position is even, increment the even count. If the position is odd, increment the odd count.
Return the array [even, odd].

Time Complexity

The time complexity of this solution is O(log n), where n is the given integer. This is because we need to iterate through the binary representation of n.

Space Complexity

The space complexity of this solution is O(1) as we are using only a constant amount of extra space.

Solution Code

class Solution {
    public int[] countBits(int n) {
        int even = 0, odd = 0;
        int position = 0;
        while (n > 0) {
            if ((n & 1) == 1) {
                if (position % 2 == 0) {
                    even++;
                } else {
                    odd++;
                }
            }
            n = n >> 1;
            position++;
        }
        return new int[]{even, odd};
    }
}

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