Number of Even and Odd Bits

Number of Even and Odd Bits - Problem

Bit Manipulation Easy

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: Bits are indexed from right to left in the binary representation of a number (0-indexed).

Return the array [even, odd].

Input & Output

Example 1 — Basic Case

$ Input: n = 17

› Output: [2,0]

💡 Note: 17 in binary is 10001. Even indices (0,2,4): bits are 1,0,1 → count = 2. Odd indices (1,3): bits are 0,0 → count = 0.

Example 2 — Mixed Positions

$ Input: n = 2

› Output: [0,1]

💡 Note: 2 in binary is 10. Even indices (0): bit is 0 → count = 0. Odd indices (1): bit is 1 → count = 1.

Example 3 — Larger Number

$ Input: n = 50

› Output: [1,2]

💡 Note: 50 in binary is 110010. Even indices (0,2,4): bits are 0,0,1 → count = 1. Odd indices (1,3,5): bits are 1,0,1 → count = 2.

Constraints

1 ≤ n ≤ 1000

Visualization

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Asked in

M Microsoft 15 a Amazon 12

The key insight is to process each bit position while tracking whether the position index is even or odd. The optimal bit manipulation approach uses bitwise operations to check each bit directly without string conversion. Time: O(log n), Space: O(1).

Common Approaches

✓ Binary String Conversion

⏱️ Time: O(log n) Space: O(log n)

Convert the integer to its binary string representation, then iterate through each bit position from right to left, counting 1-bits at even and odd indices separately.

Bit Operations

⏱️ Time: O(log n) Space: O(1)

Instead of converting to string, use bit operations to check each bit of the number while maintaining a position counter. Use bitwise AND to check if each bit is set, and bit shifting to move through all positions.

Binary String Conversion — Algorithm Steps

Convert integer n to binary string
Iterate from right to left (index 0, 1, 2...)
Count 1-bits at even indices (0, 2, 4...) and odd indices (1, 3, 5...)

Visualization

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Step-by-Step Walkthrough

Convert to Binary

Convert integer to binary string representation

Index from Right

Index positions from right to left (0, 1, 2, ...)

Count by Parity

Count 1-bits at even indices vs odd indices

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void solution(int n, int* result) {
    char binary[32];
    int len = 0;
    int temp = n;
    
    // Convert to binary string
    while (temp > 0) {
        binary[len++] = (temp % 2) + '0';
        temp /= 2;
    }
    
    int evenCount = 0;
    int oddCount = 0;
    
    // Iterate through bits (already in reverse order)
    for (int i = 0; i < len; i++) {
        if (binary[i] == '1') {
            if (i % 2 == 0) { // Even index
                evenCount++;
            } else { // Odd index
                oddCount++;
            }
        }
    }
    
    result[0] = evenCount;
    result[1] = oddCount;
}

int main() {
    int n;
    scanf("%d", &n);
    
    int result[2];
    solution(n, result);
    
    printf("[%d,%d]\n", result[0], result[1]);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

Converting to binary and iterating through log n bits

⚡ Linearithmic

Space Complexity

O(log n)

Binary string storage requires log n space

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Binary String Conversion — Algorithm Steps

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Code -

Time & Space Complexity

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