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Data Structures & Algorithms June 7, 2020 3 min read

Leetcode 406 Queue Reconstruction by Height

// 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.

Suppose you have a random list of people standing in a queue. Each person is described by a pair of integers (h, k), where h is the height of the person and k is the number of people in front of this person who have a height greater than or equal to h. Write an algorithm to reconstruct the queue.

Note:
The number of people is less than 1,100.

Example:

Input:
[[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]]

Output:
[[5,0], [7,0], [5,2], [6,1], [4,4], [7,1]]

Solution:

The idea is to first sort by height which is the first index and then sort by the number of people before them.
For the above input [[7,0], [4,4], [7,1], [5,0], [6,1], [5,2]].

We will first sort by decreasing height which will result in

Arrays.sort(people, (a,b)->a[0]==b[0]?a[1]-b[1]:b[0]-a[0])
[[7,0],[7,1],[6,1],[5,0],[4,4]]

Now we need to place them in the order of the weight, so we loop through the array and depending on the weight we will add to the list at the position of the weight.

Below code adds the people to the location with respect to their weight.

for(int i=0;i<people.length;i++){
    list.add(people[i][1],people[i]);
}

Complete code:

class Solution {
    public int[][] reconstructQueue(int[][] people) {
        Arrays.sort(people, (a,b) -> a[0]==b[0]?a[1]-b[1]:b[0]-a[0]);
        List<int[]> returnVal = new ArrayList<>();
        
        for(int i=0;i<people.length;i++){
            returnVal.add(people[i][1],people[i]);
        }
        return returnVal.toArray(new int[people.length][2]);
     }
}

The code run in O(N*N) as the sort takes N*N comparing both the height and weight.



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