LeetCode #1943 — MEDIUM

Describe the Painting

Move from brute-force thinking to an efficient approach using array strategy.

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The Problem

Problem Statement

There is a long and thin painting that can be represented by a number line. The painting was painted with multiple overlapping segments where each segment was painted with a unique color. You are given a 2D integer array segments, where segments[i] = [starti, endi, colori] represents the half-closed segment [starti, endi) with colori as the color.

The colors in the overlapping segments of the painting were mixed when it was painted. When two or more colors mix, they form a new color that can be represented as a set of mixed colors.

  • For example, if colors 2, 4, and 6 are mixed, then the resulting mixed color is {2,4,6}.

For the sake of simplicity, you should only output the sum of the elements in the set rather than the full set.

You want to describe the painting with the minimum number of non-overlapping half-closed segments of these mixed colors. These segments can be represented by the 2D array painting where painting[j] = [leftj, rightj, mixj] describes a half-closed segment [leftj, rightj) with the mixed color sum of mixj.

  • For example, the painting created with segments = [[1,4,5],[1,7,7]] can be described by painting = [[1,4,12],[4,7,7]] because:
    • [1,4) is colored {5,7} (with a sum of 12) from both the first and second segments.
    • [4,7) is colored {7} from only the second segment.

Return the 2D array painting describing the finished painting (excluding any parts that are not painted). You may return the segments in any order.

A half-closed segment [a, b) is the section of the number line between points a and b including point a and not including point b.

Example 1:

Input: segments = [[1,4,5],[4,7,7],[1,7,9]]
Output: [[1,4,14],[4,7,16]]
Explanation: The painting can be described as follows:
- [1,4) is colored {5,9} (with a sum of 14) from the first and third segments.
- [4,7) is colored {7,9} (with a sum of 16) from the second and third segments.

Example 2:

Input: segments = [[1,7,9],[6,8,15],[8,10,7]]
Output: [[1,6,9],[6,7,24],[7,8,15],[8,10,7]]
Explanation: The painting can be described as follows:
- [1,6) is colored 9 from the first segment.
- [6,7) is colored {9,15} (with a sum of 24) from the first and second segments.
- [7,8) is colored 15 from the second segment.
- [8,10) is colored 7 from the third segment.

Example 3:

Input: segments = [[1,4,5],[1,4,7],[4,7,1],[4,7,11]]
Output: [[1,4,12],[4,7,12]]
Explanation: The painting can be described as follows:
- [1,4) is colored {5,7} (with a sum of 12) from the first and second segments.
- [4,7) is colored {1,11} (with a sum of 12) from the third and fourth segments.
Note that returning a single segment [1,7) is incorrect because the mixed color sets are different.

Constraints:

  • 1 <= segments.length <= 2 * 104
  • segments[i].length == 3
  • 1 <= starti < endi <= 105
  • 1 <= colori <= 109
  • Each colori is distinct.

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: There is a long and thin painting that can be represented by a number line. The painting was painted with multiple overlapping segments where each segment was painted with a unique color. You are given a 2D integer array segments, where segments[i] = [starti, endi, colori] represents the half-closed segment [starti, endi) with colori as the color. The colors in the overlapping segments of the painting were mixed when it was painted. When two or more colors mix, they form a new color that can be represented as a set of mixed colors. For example, if colors 2, 4, and 6 are mixed, then the resulting mixed color is {2,4,6}. For the sake of simplicity, you should only output the sum of the elements in the set rather than the full set. You want to describe the painting with the minimum number of non-overlapping half-closed segments of these mixed colors. These segments can be represented by

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[[1,4,5],[4,7,7],[1,7,9]]

Example 2

[[1,7,9],[6,8,15],[8,10,7]]

Example 3

[[1,4,5],[1,4,7],[4,7,1],[4,7,11]]

Related Problems

  • Average Height of Buildings in Each Segment (average-height-of-buildings-in-each-segment)
  • Amount of New Area Painted Each Day (amount-of-new-area-painted-each-day)
  • Shifting Letters II (shifting-letters-ii)
Step 02

Core Insight

What unlocks the optimal approach

  • Can we sort the segments in a way to help solve the problem?
  • How can we dynamically keep track of the sum of the current segment(s)?
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1943: Describe the Painting
class Solution {
    public List<List<Long>> splitPainting(int[][] segments) {
        TreeMap<Integer, Long> d = new TreeMap<>();
        for (int[] e : segments) {
            int l = e[0], r = e[1], c = e[2];
            d.put(l, d.getOrDefault(l, 0L) + c);
            d.put(r, d.getOrDefault(r, 0L) - c);
        }
        List<List<Long>> ans = new ArrayList<>();
        long i = 0, j = 0;
        long cur = 0;
        for (Map.Entry<Integer, Long> e : d.entrySet()) {
            if (Objects.equals(e.getKey(), d.firstKey())) {
                i = e.getKey();
            } else {
                j = e.getKey();
                if (cur > 0) {
                    ans.add(Arrays.asList(i, j, cur));
                }
                i = j;
            }
            cur += e.getValue();
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.