Leetcode Single Number II Java Solution

// SOLVING THIS WITH AN AI ASSISTANT (2026)

If you are working through this problem with an AI coding assistant — Claude, ChatGPT, Cursor chat, Gemini, GitHub Copilot, Aider, or any agent — the goal isn’t to ask for the answer. It is to use the tool to understand the pattern. The prompt sequence I’d run:

Spec it back to me first. “In your own words, what is this problem actually testing? What’s the smallest example that fails the naive approach?”
Brute-force first, optimize after. “Write the simplest correct solution, even if it’s O(n²). Don’t optimize. Just make it correct, with comments explaining each step.”
Ask for the upgrade. “Now show me the optimal solution. What insight makes it possible? What pattern is this an instance of?”
Stress-test it. “Generate 10 edge cases — empty input, single element, duplicates, max size, sorted, reverse-sorted. Run my solution against each.”

The pattern matters more than the answer. If the agent just hands you optimized code, you’ve trained yourself to lose interviews.

Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Example 1:

Input: [2,2,3,2]
Output: 3

Example 2:

Input: [0,1,0,1,0,1,99]
Output: 99

Solution 1:

We will use a HashMap and store the count of each number, and we will parse through the map and return the key with value 1. This approach will take O(N) time complexity and O(N) space for the Map.

class HackerHeap {
    public int singleNumber(int[] nums) {
        Map<Integer, Integer> count = new HashMap<Integer,Integer>();
        
        for(int i=0;i<nums.length;i++){
            count.put(nums[i],count.getOrDefault(nums[i],0)+1);
        }
        
        for(Integer key:count.keySet()){
            if(count.get(key) == 1) return key;
        }
        
        return -1;
    }
}

Solution 2:

We will use Bitwise operators to find the non repeating number, this approach will take O(N) time complexity but constant space as we are not taking any extra space to store the count like above.

class HackerHeap {
    public int singleNumber(int[] nums) {
        int ones = 0, twos = 0, threes = 0;
    for (int i = 0; i < nums.length; i++) {
        twos |= ones & nums[i];
        ones ^= nums[i];
        threes = ones & twos;
        ones &= ~threes;
        twos &= ~threes;
    }
    return ones;
    }
}

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rajendra