LeetCode #2595 — EASY

Number of Even and Odd Bits

Build confidence with an intuition-first walkthrough focused on bit manipulation fundamentals.

Solve on LeetCode

The Problem

Problem Statement

You are given a positive integer n.

Let even denote the number of even indices in the binary representation of n with value 1.

Let odd denote the number of odd indices in the binary representation of n with value 1.

Note that bits are indexed from right to left in the binary representation of a number.

Return the array [even, odd].

Example 1:

Input: n = 50

Output: [1,2]

Explanation:

The binary representation of 50 is 110010.

It contains 1 on indices 1, 4, and 5.

Example 2:

Input: n = 2

Output: [0,1]

Explanation:

The binary representation of 2 is 10.

It contains 1 only on index 1.

Constraints:

1 <= n <= 1000

Step 01

Brute Force Baseline

Problem summary: You are given a positive integer n. Let even denote the number of even indices in the binary representation of n with value 1. Let odd denote the number of odd indices in the binary representation of n with value 1. Note that bits are indexed from right to left in the binary representation of a number. Return the array [even, odd].

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Bit Manipulation

Example 1

Example 2

Core Insight

What unlocks the optimal approach

Maintain two integer variables, even and odd, to count the number of even and odd indices in the binary representation of integer n.
Divide n by 2 while n is positive, and if n modulo 2 is 1, add 1 to its corresponding variable.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2595: Number of Even and Odd Bits
class Solution {
    public int[] evenOddBit(int n) {
        int[] ans = new int[2];
        for (int i = 0; n > 0; n >>= 1, i ^= 1) {
            ans[i] += n & 1;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2595: Number of Even and Odd Bits
func evenOddBit(n int) []int {
	ans := make([]int, 2)
	for i := 0; n != 0; n, i = n>>1, i^1 {
		ans[i] += n & 1
	}
	return ans
}

# Accepted solution for LeetCode #2595: Number of Even and Odd Bits
class Solution:
    def evenOddBit(self, n: int) -> List[int]:
        ans = [0, 0]
        i = 0
        while n:
            ans[i] += n & 1
            i ^= 1
            n >>= 1
        return ans

// Accepted solution for LeetCode #2595: Number of Even and Odd Bits
impl Solution {
    pub fn even_odd_bit(mut n: i32) -> Vec<i32> {
        let mut ans = vec![0; 2];

        let mut i = 0;
        while n != 0 {
            ans[i] += n & 1;

            n >>= 1;
            i ^= 1;
        }

        ans
    }
}

// Accepted solution for LeetCode #2595: Number of Even and Odd Bits
function evenOddBit(n: number): number[] {
    const ans = Array(2).fill(0);
    for (let i = 0; n > 0; n >>= 1, i ^= 1) {
        ans[i] += n & 1;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.