What is the time complexity of Maximum Score of Non-overlapping Intervals?

The optimal solution has a time complexity of O(n*n*log(n)).

What data structures are used to solve Maximum Score of Non-overlapping Intervals?

This problem uses Array, Binary Search, Dynamic Programming, Sorting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Score of Non-overlapping Intervals?

Insight 1: Sorting intervals based on start and end times simplifies non-overlapping checks and allows for efficient processing. Insight 2: Dynamic programming is suitable to explore all possible combinations of non-overlapping intervals and their weights.

Which companies ask LeetCode 3414. Maximum Score of Non-overlapping Intervals in interviews?

This problem is frequently asked at Sprinklr.

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Maximum Score of Non-overlapping Intervals - LeetCode 3414 Solution

Maximum Score of Non-overlapping Intervals - Complete Solution Guide

Maximum Score of Non-overlapping Intervals is LeetCode problem 3414, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a 2D integer array intervals , where intervals[i] = [l i , r i , weight i ] . Interval i starts at position l i and ends at r i , and has a weight of weight i . 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-overlap

Detailed Explanation

The problem requires us to select at most 4 non-overlapping intervals from a given set of intervals, each having a start time, end time, and weight. The goal is to maximize the total weight of the chosen intervals while also ensuring that the selected intervals' indices, when sorted, form the lexicographically smallest array possible in case of ties in total weight.

Solution Approach

The provided solution uses a dynamic programming approach to find the maximum weight of at most 4 non-overlapping intervals. It first sorts the intervals by their start and end times. Then, it uses a DP table to store the maximum weight achievable using up to `k` intervals, ending at a given index. The DP table is built from the end of the sorted intervals to the beginning. A binary search is employed to find the next non-overlapping interval efficiently. When there's a tie in the total weight, the code compares the lists of interval indices lexicographically to choose the smallest one.

Step-by-Step Algorithm

Step 1: Augment each interval with its original index and store this information.
Step 2: Sort the intervals (with indices) by start time, and then by end time in case of ties.
Step 3: Create a DP table `dp[i][k]` where `dp[i][k]` stores a pair: (maximum weight using up to `k` intervals from index `i` onwards, list of interval indices used to achieve the maximum weight).
Step 4: Iterate through the intervals in reverse order (from `n-1` to `0`).
Step 5: For each interval `i`, iterate through the possible number of intervals to choose (`k` from `1` to `4`).
Step 6: Compute the two options: (a) skip interval `i` and take the result from `dp[i+1][k]`, (b) take interval `i`. To take interval `i`, find the next non-overlapping interval `j` using binary search (bisect_right). The weight will then be weight of interval `i` + value of `dp[j][k-1]`.
Step 7: Compare the weight obtained by skipping or taking the interval, and compare their indices lexicographically in case of weight ties. Store the result.
Step 8: The result `dp[0][4]` stores the maximum weight using up to 4 intervals and the lexicographically smallest set of indices.
Step 9: Return the list of indices.

Key Insights

Insight 1: Sorting intervals based on start and end times simplifies non-overlapping checks and allows for efficient processing.
Insight 2: Dynamic programming is suitable to explore all possible combinations of non-overlapping intervals and their weights.
Insight 3: The lexicographical tie-breaker adds complexity, requiring careful comparison of index lists when weights are equal.

Complexity Analysis

Time Complexity: O(n*n*log(n))

Space Complexity: O(n)

Topics

This problem involves: Array, Binary Search, Dynamic Programming, Sorting.

Companies

Asked at: Sprinklr.

Maximum Score of Non-overlapping Intervals | LeetCode 3414

Hard

Array Binary Search Dynamic Programming Sorting

Time: O(n*n*log(n))Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int[] maximumWeight(List<List<Integer>> intervals) {
        int n = intervals.size();

        List<List<Integer>> intervalsWithIndices = new ArrayList<>();
        for (int i = 0; i < intervals.size(); i++) {
            List<Integer> interval = intervals.get(i);
            List<Integer> intervalWithIndex = new ArrayList<>(interval);
            intervalWithIndex.add(i);
            intervalsWithIndices.add(intervalWithIndex);
        }

        intervalsWithIndices.sort((a, b) -> {
            int startComparison = Integer.compare(a.get(0), b.get(0));
            if (startComparison != 0) {
                return startComparison;
            }
            return Integer.compare(a.get(1), b.get(1));
        });

        List<Integer> startTimes = new ArrayList<>();
        for (List<Integer> iv : intervalsWithIndices) {
            startTimes.add(iv.get(0));
        }

        Pair[][] dp = new Pair[n + 1][5];
        for (int i = 0; i <= n; i++) {
            for (int k = 0; k < 5; k++) {
                dp[i][k] = new Pair(0, new ArrayList<>());
            }
        }

        for (int i = n - 1; i >= 0; i--) {
            for (int k = 1; k < 5; k++) {
                Pair resSkip = dp[i + 1][k];

                int l = intervalsWithIndices.get(i).get(0);
                int r = intervalsWithIndices.get(i).get(1);
                int w = intervalsWithIndices.get(i).get(2);
                int originalIdx = intervalsWithIndices.get(i).get(3);

                int nextI = binarySearchRight(startTimes, r, i + 1);

                Pair scoreTakePair = dp[nextI][k - 1];
                int scoreTake = scoreTakePair.score;
                List<Integer> indicesTakeRest = scoreTakePair.indices;

                int currentScore = w + scoreTake;
                List<Integer> currentIndices = new ArrayList<>(indicesTakeRest);
                currentIndices.add(originalIdx);
                Collections.sort(currentIndices);

                Pair resTake = new Pair(currentScore, currentIndices);

                Pair result;
                if (resTake.score > resSkip.score) {
                    result = resTake;
                } else if (resSkip.score > resTake.score) {
                    result = resSkip;
                } else {
                    if (compareLists(resTake.indices, resSkip.indices) < 0) {
                        result = resTake;
                    } else {
                        result = resSkip;
                    }
                }

                dp[i][k] = result;
            }
        }

        return dp[0][4].indices.stream().mapToInt(Integer::intValue).toArray();
    }

    private int binarySearchRight(List<Integer> list, int target, int lo) {
        int hi = list.size();
        while (lo < hi) {
            int mid = lo + (hi - lo) / 2;
            if (list.get(mid) <= target) {
                lo = mid + 1;
            } else {
                hi = mid;
            }
        }
        return lo;
    }
    
    private int compareLists(List<Integer> list1, List<Integer> list2) {
        int size1 = list1.size();
        int size2 = list2.size();
        int minSize = Math.min(size1, size2);

        for (int i = 0; i < minSize; i++) {
            int comparison = Integer.compare(list1.get(i), list2.get(i));
            if (comparison != 0) {
                return comparison;
            }
        }

        return Integer.compare(size1, size2);
    }


    static class Pair {
        int score;
        List<Integer> indices;

        public Pair(int score, List<Integer> indices) {
            this.score = score;
            this.indices = indices;
        }
    }
}

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Problem Info

Description

Problem 3414: Maximum Score of Non-overlapping Intervals - Detailed explanation and solution approaches.

Topics

Array Binary Search Dynamic Programming Sorting

Companies

Sprinklr

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3414

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Augment each interval with its original index and store this information.
2
Step 2: Sort the intervals (with indices) by start time, and then by end time in case of ties.
3
Step 3: Create a DP table `dp[i][k]` where `dp[i][k]` stores a pair: (maximum weight using up to `k` intervals from index `i` onwards, list of interval indices used to achieve the maximum weight).
4
Step 4: Iterate through the intervals in reverse order (from `n-1` to `0`).
5
Step 5: For each interval `i`, iterate through the possible number of intervals to choose (`k` from `1` to `4`).
6
Step 6: Compute the two options: (a) skip interval `i` and take the result from `dp[i+1][k]`, (b) take interval `i`. To take interval `i`, find the next non-overlapping interval `j` using binary search (bisect_right). The weight will then be weight of interval `i` + value of `dp[j][k-1]`.
7
Step 7: Compare the weight obtained by skipping or taking the interval, and compare their indices lexicographically in case of weight ties. Store the result.
8
Step 8: The result `dp[0][4]` stores the maximum weight using up to 4 intervals and the lexicographically smallest set of indices.
9
Step 9: Return the list of indices.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*n*log(n))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is derived from the nested loops for DP, with the binary search contributing the logarithmic factor. The space complexity is primarily due to the DP table and the array used for storing start times. The size of the result indices list is bounded by a constant, therefore does not contribute to the overall space complexity.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

A recursive approach with memoization could also solve this problem. The recursion would explore taking or skipping each interval. The memoization would store the results to avoid redundant calculations.

Time: O(n * n * log(n))Space: O(n)

Trade-offs: Recursion with memoization may be easier to understand initially, but can lead to stack overflow errors for very large inputs, and might be slightly less efficient in practice than the iterative DP approach due to function call overhead.

✨

Key Insights

1Insight 1: Sorting intervals based on start and end times simplifies non-overlapping checks and allows for efficient processing.
2Insight 2: Dynamic programming is suitable to explore all possible combinations of non-overlapping intervals and their weights.
3Insight 3: The lexicographical tie-breaker adds complexity, requiring careful comparison of index lists when weights are equal.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input `intervals`. The DP initialization handles this, and the final result will be an empty list with score 0.
Edge case 2: Intervals that completely overlap. The sorting and non-overlapping logic handle this correctly, by selecting the appropriate interval based on start/end times.
Edge case 3: Number of intervals is less than 4. The DP logic handles this, finding the optimal solution with the available intervals.
Edge case 4: All intervals overlap. The algorithm will choose the interval with the highest weight and lexicographically smallest index if only one interval can be chosen. If no intervals can be chosen, the algorithm will return an empty list.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly sorting the intervals (e.g., only sorting by start time without considering end time).

✗Common mistake 2: Failing to handle the lexicographical tie-breaker, leading to incorrect results when multiple combinations have the same maximum weight.

✗Common mistake 3: Not correctly implementing the non-overlapping check, potentially including overlapping intervals in the solution.

✗Common mistake 4: Not correctly freeing memory in C, leading to memory leaks.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Scheduling tasks or meetings with maximum value and non-overlapping constraints. For example, a conference scheduling system might want to select a set of talks that maximizes attendee interest while ensuring no attendee has two talks scheduled at the same time.

🏦Real-world scenario 2: Resource allocation with limited resources and conflicting demands. A company might have a set of projects, each requiring certain resources and generating a certain profit. The goal is to select a subset of projects that maximizes profit while respecting resource limitations and avoiding conflicts (e.g., two projects requiring the same machine at the same time).

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 22, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Score of Non-overlapping Intervals"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?