LeetCode #477 — MEDIUM

Total Hamming Distance

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given an integer array nums, return the sum of Hamming distances between all the pairs of the integers in nums.

Example 1:

Input: nums = [4,14,2]
Output: 6
Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
showing the four bits relevant in this case).
The answer will be:
HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.

Example 2:

Input: nums = [4,14,4]
Output: 4

Constraints:

1 <= nums.length <= 10⁴
0 <= nums[i] <= 10⁹
The answer for the given input will fit in a 32-bit integer.

Step 01

Brute Force Baseline

Problem summary: The Hamming distance between two integers is the number of positions at which the corresponding bits are different. Given an integer array nums, return the sum of Hamming distances between all the pairs of the integers in nums.

Baseline thinking

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

Pattern signal: Array · Math · Bit Manipulation

Example 1

[4,14,2]

Example 2

[4,14,4]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #477: Total Hamming Distance
class Solution {
    public int totalHammingDistance(int[] nums) {
        int ans = 0, n = nums.length;
        for (int i = 0; i < 32; ++i) {
            int a = 0;
            for (int x : nums) {
                a += (x >> i & 1);
            }
            int b = n - a;
            ans += a * b;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #477: Total Hamming Distance
func totalHammingDistance(nums []int) (ans int) {
	for i := 0; i < 32; i++ {
		a := 0
		for _, x := range nums {
			a += x >> i & 1
		}
		b := len(nums) - a
		ans += a * b
	}
	return
}

# Accepted solution for LeetCode #477: Total Hamming Distance
class Solution:
    def totalHammingDistance(self, nums: List[int]) -> int:
        ans, n = 0, len(nums)
        for i in range(32):
            a = sum(x >> i & 1 for x in nums)
            b = n - a
            ans += a * b
        return ans

// Accepted solution for LeetCode #477: Total Hamming Distance
impl Solution {
    pub fn total_hamming_distance(nums: Vec<i32>) -> i32 {
        let mut ans = 0;
        for i in 0..32 {
            let mut a = 0;
            for &x in nums.iter() {
                a += (x >> i) & 1;
            }
            let b = (nums.len() as i32) - a;
            ans += a * b;
        }
        ans
    }
}

// Accepted solution for LeetCode #477: Total Hamming Distance
function totalHammingDistance(nums: number[]): number {
    let ans = 0;
    for (let i = 0; i < 32; ++i) {
        const a = nums.filter(x => (x >> i) & 1).length;
        const b = nums.length - a;
        ans += a * b;
    }
    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(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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.