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Data Structures & Algorithms May 27, 2020 2 min read

Leetcode 56 Merge Intervals (Java)

// 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 collection of intervals, merge all overlapping intervals.

Example 1:

Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Solution:

We can solve this by comparing the start of the next interval with the end of the current interval and merged if the start of the next interval is less than the end of the current interval.
For this we need to sort the interval based on the start of the intervals.

Note: Please try to solve it yourself before looking into the solution below

class Solution {
    public int[][] merge(int[][] intervals) {
        
        if(intervals.length<=1) return intervals;
        
        List<int[]> mergedIntervals = new ArrayList<>();
        
        Arrays.sort(intervals, (arr1, arr2) -> Integer.compare(arr1[0], arr2[0]));
        
        int[] currentInterval = intervals[0];
        
        mergedIntervals.add(currentInterval);
        
        for(int[] interval: intervals){
            int currentStart = currentInterval[0];
            int currentEnd = currentInterval[1];
            int newStart = interval[0];
            int newEnd = interval[1];
            
            if(currentEnd>=newStart){
                currentInterval[1] = Math.max(currentEnd, newEnd);
            }else{
                currentInterval = interval;
                mergedIntervals.add(currentInterval);
            }
            
        }
        
        
        return mergedIntervals.toArray(new int[mergedIntervals.size()][]);
    }
}

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