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Data Structures & Algorithms June 27, 2020 2 min read

Leetcode Perfect Squares Java Solution

Leetcode Perfect Squares 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:

  1. 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?”
  2. 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.”
  3. Ask for the upgrade. “Now show me the optimal solution. What insight makes it possible? What pattern is this an instance of?”
  4. 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 positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.

Example 1:

Input: n = 12
Output: 3 
Explanation: 12 = 4 + 4 + 4.

Example 2:

Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.

Solution: We will solve this problem using Dynamic Programming, the sum of perfect squares for the number would be the number of perfect squares for previous number +1 whichever is minimum Math.min(dp[i], dp[i-j*j]+1)

class HackerHeap {
    public int numSquares(int n) {
        int max = (int) Math.sqrt(n);
        int[] dp = new int[n+1];
        
        Arrays.fill(dp, Integer.MAX_VALUE);
        
        for(int i=1; i<=n;i++){
            for(int j=1; j<=max;j++){
                if(i==j*j) {
                    dp[i] = 1;
                }else if(i>j*j){
                    dp[i] = Math.min(dp[i], dp[i-j*j]+1);
                }else if(i<j*j) {
                    break;
                }
            }
        }
        return dp[n];
    }
}

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