LeetCode #3414 — HARD

Maximum Score of Non-overlapping Intervals

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

You are given a 2D integer array intervals, where intervals[i] = [li, ri, weighti]. Interval i starts at position li and ends at ri, and has a weight of weighti. You can choose up to 4 non-overlapping intervals. The score of the chosen intervals is defined as the total sum of their weights.

Return the lexicographically smallest array of at most 4 indices from intervals with maximum score, representing your choice of non-overlapping intervals.

Two intervals are said to be non-overlapping if they do not share any points. In particular, intervals sharing a left or right boundary are considered overlapping.

Example 1:

Input: intervals = [[1,3,2],[4,5,2],[1,5,5],[6,9,3],[6,7,1],[8,9,1]]

Output: [2,3]

Explanation:

You can choose the intervals with indices 2, and 3 with respective weights of 5, and 3.

Example 2:

Input: intervals = [[5,8,1],[6,7,7],[4,7,3],[9,10,6],[7,8,2],[11,14,3],[3,5,5]]

Output: [1,3,5,6]

Explanation:

You can choose the intervals with indices 1, 3, 5, and 6 with respective weights of 7, 6, 3, and 5.

Constraints:

  • 1 <= intevals.length <= 5 * 104
  • intervals[i].length == 3
  • intervals[i] = [li, ri, weighti]
  • 1 <= li <= ri <= 109
  • 1 <= weighti <= 109
Patterns Used
🔍Modified Binary Search→📊Dynamic Programming→

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: You are given a 2D integer array intervals, where intervals[i] = [li, ri, weighti]. Interval i starts at position li and ends at ri, and has a weight of weighti. You can choose up to 4 non-overlapping intervals. The score of the chosen intervals is defined as the total sum of their weights. Return the lexicographically smallest array of at most 4 indices from intervals with maximum score, representing your choice of non-overlapping intervals. Two intervals are said to be non-overlapping if they do not share any points. In particular, intervals sharing a left or right boundary are considered overlapping.

Baseline thinking

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

Pattern signal: Array · Binary Search · Dynamic Programming

Example 1

[[1,3,2],[4,5,2],[1,5,5],[6,9,3],[6,7,1],[8,9,1]]

Example 2

[[5,8,1],[6,7,7],[4,7,3],[9,10,6],[7,8,2],[11,14,3],[3,5,5]]

Related Problems

  • Two Best Non-Overlapping Events (two-best-non-overlapping-events)
Step 02

Core Insight

What unlocks the optimal approach

  • Use Dynamic Programming.
  • Sort <code>intervals</code> by right boundary.
  • Let <code>dp[r][i]</code> denote the maximum score having picked <code>r</code> intervals from the prefix of <code>intervals</code> ending at index <code>i</code>.
  • <code>dp[r][i] = max(dp[r][i - 1], intervals[i][2] + dp[r][j])</code> where <code>j</code> is the largest index such that <code>intervals[j][1] < intervals[i][0]</code>.
  • Since <code>intervals</code> is sorted by right boundary, we can find index <code>j</code> using binary search.
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
Largest constraint values
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 #3414: Maximum Score of Non-overlapping Intervals
class Solution {
  public int[] maximumWeight(List<List<Integer>> intervals) {
    // Convert input to Interval objects
    List<Interval> indexedIntervals = new ArrayList<>();
    for (int i = 0; i < intervals.size(); ++i) {
      List<Integer> interval = intervals.get(i);
      indexedIntervals.add(new Interval(interval.get(0), interval.get(1), interval.get(2), i));
    }
    indexedIntervals.sort(Comparator.comparingInt(Interval::left));
    T[][] memo = new T[indexedIntervals.size()][5];
    return dp(indexedIntervals, memo, 0, 4).selected.stream().mapToInt(Integer::intValue).toArray();
  }

  private record T(long weight, List<Integer> selected) {}
  private record Interval(int left, int right, int weight, int originalIndex) {}

  private T dp(List<Interval> intervals, T[][] memo, int i, int quota) {
    if (i == intervals.size() || quota == 0)
      return new T(0, List.of());
    if (memo[i][quota] != null)
      return memo[i][quota];

    T skip = dp(intervals, memo, i + 1, quota);

    Interval interval = intervals.get(i);
    final int j = findFirstGreater(intervals, i + 1, interval.right);
    T nextRes = dp(intervals, memo, j, quota - 1);

    List<Integer> newSelected = new ArrayList<>(nextRes.selected);
    newSelected.add(interval.originalIndex);
    Collections.sort(newSelected);
    T pick = new T(interval.weight + nextRes.weight, newSelected);
    return memo[i][quota] =
               (pick.weight > skip.weight ||
                (pick.weight == skip.weight && compareLists(pick.selected, skip.selected) < 0))
                   ? pick
                   : skip;
  }

  // Binary searches the first interval that starts after `rightBoundary`.
  private int findFirstGreater(List<Interval> intervals, int startFrom, int rightBoundary) {
    int l = startFrom;
    int r = intervals.size();
    while (l < r) {
      final int m = (l + r) / 2;
      if (intervals.get(m).left > rightBoundary)
        r = m;
      else
        l = m + 1;
    }
    return l;
  }

  // Compares two lists of integers lexicographically.
  private int compareLists(List<Integer> list1, List<Integer> list2) {
    final int minSize = Math.min(list1.size(), list2.size());
    for (int i = 0; i < minSize; ++i) {
      final int comparison = Integer.compare(list1.get(i), list2.get(i));
      if (comparison != 0)
        return comparison;
    }
    return Integer.compare(list1.size(), list2.size());
  }
}
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(log n)
Space
O(1)

Approach Breakdown

LINEAR SCAN
O(n) time
O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH
O(log n) time
O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.
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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

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