What is the brute force approach and complexity for Single Number II?

For each value, scan the array and return the one that appears exactly once with time complexity O(N²) and space complexity O(1).

What is the optimal approach and complexity for Single Number II?

For each of the 32 bit positions, count the 1s; set the answer's bit when the count % 3 is 1 with optimal time complexity O(32 · N) and space complexity O(1).

How do you identify the pattern for Single Number II?

Look for key signals in the prompt: "every element appears three times except one" indicates Triples do not cancel under XOR. Count the bits instead.; "linear time, constant extra space" indicates 32 counters (or two variables), not a hash map.; "return the single one" indicates The answer is rebuilt from its bits, one column at a time..

Single Number II

Medium·LeetCode #137 Pro OnlyMedium

LeetCode

Problem Statement

You get an array nums. Every value in it appears three times, except for one value that appears exactly once. Return that single value.

Your solution must run in linear time and use constant extra space.

Input

An integer array nums with at least one value. Every value appears three times except one, which appears once.

Output

The one value that appears exactly once.

Example 1

Input: nums = [2, 2, 3, 2]

Output: 3

2 shows up three times. The 3 is the one that shows up once.

Example 2

Input: nums = [0, 1, 0, 1, 0, 1, 99]

Output: 99

0 and 1 each show up three times. The 99 is the single one.

Constraints

• 1 ≤ nums.length ≤ 3 · 10^4
• −2^31 ≤ nums[i] ≤ 2^31 − 1
• Every value appears exactly three times, except one value that appears once.

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“every element appears three times except one”

→ Triples do not cancel under XOR. Count the bits instead.

“linear time, constant extra space”

→ 32 counters (or two variables), not a hash map.

“return the single one”

→ The answer is rebuilt from its bits, one column at a time.

Approach Complexity Ladder

Brute Force — count each value with a loopO(N²)O(1)

For each value, scan the array and return the one that appears exactly once

Better — count with a hash mapO(N)O(N)

Count each value once in a hash map, then return the value with count 1

Optimal — count the bits, then take count % 3O(32 · N)O(1)

For each of the 32 bit positions, count the 1s; set the answer's bit when the count % 3 is 1

Brute Force Intuition

For each value in nums, scan the whole array and count how many times it appears. The first value whose count is 1 is the answer.

Recognition rule for this sub-pattern:

x ^ x = 0 and x ^ 0 = x, so XOR-ing everything cancels the pairs.

Remember this · Bit Manipulation

Trigger

"Every element appears twice except…", power of two, counting set bits, subsets as bitmasks, O(1) space tricks.

Think

What happens to this problem if I look at each bit separately?

Key question

Which bit operation (XOR, AND, OR, shift) cancels or isolates what I need?

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