What is the time complexity of Set Intersection Size At Least Two?

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

What data structures are used to solve Set Intersection Size At Least Two?

This problem uses Array, Greedy, Sorting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Set Intersection Size At Least Two?

Insight 1: Sorting intervals by their end points is crucial for a greedy approach to work effectively. This allows prioritizing intervals that end sooner, increasing the chance to cover them with a small number of elements. Insight 2: Choosing the two largest possible numbers within an interval (close to the end point) when needing to add elements is a key greedy strategy for minimizing the size of the containing set.

Which companies ask LeetCode 757. Set Intersection Size At Least Two in interviews?

This problem is frequently asked at DP world.

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Set Intersection Size At Least Two - LeetCode 757 Solution

Set Intersection Size At Least Two - Complete Solution Guide

Set Intersection Size At Least Two is LeetCode problem 757, 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] = [start i , end i ] represents all the integers from start i to end i inclusively. A containing set is an array nums where each interval from intervals has at least two integers in nums . For example, if intervals = [[1,3], [3,7], [8,9]] , then [1,2,4,7,8,9] and [2,3,4,8,9] are containing sets . Return the minimum possible size of a containing set . Example 1: Input: intervals = [[1,3],[3,7],[8,9]] Output: 5 Explanation: let nums = [

Detailed Explanation

The problem asks us to find the smallest set of integers such that each interval in a given list of intervals contains at least two numbers from the set. Each interval is defined by a start and end point, inclusive. The goal is to minimize the size of the 'containing set' while ensuring every interval has at least two elements in common with it. For example, if we have the intervals [[1,3], [1,4], [2,5], [3,5]], we need to find the smallest set of numbers such that each interval has at least two numbers from the set. A possible solution is {2, 3, 4}, which has a size of 3.

Solution Approach

The provided solution uses a greedy approach. It first sorts the intervals based on their end points in ascending order and, in case of ties, sorts them in descending order of start points. This sorting strategy is the key to making the greedy approach work. Then, it iterates through the sorted intervals, maintaining the two largest numbers ('p1' and 'p2') currently in the containing set. For each interval, it checks how many of these numbers are already present in the interval. If none are present, it adds two new numbers (end-1 and end). If only one is present, it adds one new number (end). If both are present, it does nothing. The 'end-1' and 'end' choices aim to maximize the coverage of future intervals, therefore minimizing the overall size of the set.

Step-by-Step Algorithm

Step 1: Sort the intervals based on their end points in ascending order. If two intervals have the same end point, sort them in descending order of their start points.
Step 2: Initialize two variables, `p1` and `p2`, to keep track of the two largest elements in the containing set. Initialize them to -1 (or any value smaller than the interval start values). Initialize the answer variable `ans` to 0.
Step 3: Iterate through the sorted intervals.
Step 4: For each interval [start, end], check if `start > p2`. If true, it means that the interval doesn't contain `p1` or `p2`. Add 2 to the `ans`. Set `p1 = end - 1` and `p2 = end`.
Step 5: If `start <= p2` but `start > p1`, it means the interval only contains `p2`. Add 1 to the `ans`. Set `p1 = p2` and `p2 = end`.
Step 6: If `start <= p1`, it means the interval contains both `p1` and `p2`. Do nothing.
Step 7: Return the value of `ans`.

Key Insights

Insight 1: Sorting intervals by their end points is crucial for a greedy approach to work effectively. This allows prioritizing intervals that end sooner, increasing the chance to cover them with a small number of elements.
Insight 2: Choosing the two largest possible numbers within an interval (close to the end point) when needing to add elements is a key greedy strategy for minimizing the size of the containing set.
Insight 3: Sorting intervals in ascending order of end points and then descending order of start points is the correct approach to optimize the greedy algorithm.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Sorting.

Companies

Asked at: DP world.

Set Intersection Size At Least Two | LeetCode 757

Hard

Array Greedy Sorting

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;
import java.util.Comparator;

class Solution {
    /**
     * Finds the minimum size of a set S such that for every interval A in intervals,
     * the size of the intersection of S and A is at least 2.
     *
     * This solution uses a greedy approach. The intervals are sorted first by their
     * end points in ascending order, and then by their start points in descending
     * order. We iterate through the sorted intervals, keeping track of the two
     * largest numbers added to our set S, let's call them p1 and p2 (p1 < p2).
     *
     * For each interval [start, end]:
     * - If start > p2, neither of our tracked points are in the interval. We must
     *   add two new points. The greedy choice is to add `end-1` and `end` to maximize
     *   their utility for future intervals.
     * - If start > p1, only p2 is in the interval. We need one more point. The greedy
     *   choice is `end`. Our new pair of largest points becomes the old p2 and `end`.
     * - If start <= p1, both p1 and p2 are in the interval, so the condition is met.
     *
     * The sorting strategy is key to the correctness of this simple greedy logic.
     */
    public int intersectionSizeTwo(int[][] intervals) {
        Arrays.sort(intervals, new Comparator<int[]>() {
            @Override
            public int compare(int[] a, int[] b) {
                if (a[1] != b[1]) {
                    return a[1] - b[1];
                } else {
                    return b[0] - a[0];
                }
            }
        });

        int p1 = -1;
        int p2 = -1;
        int ans = 0;

        for (int[] interval : intervals) {
            int start = interval[0];
            int end = interval[1];

            if (start > p2) {
                ans += 2;
                p1 = end - 1;
                p2 = end;
            } else if (start > p1) {
                ans += 1;
                p1 = p2;
                p2 = end;
            }
        }

        return ans;
    }
}

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Problem Info

Description

Problem 757: Set Intersection Size At Least Two - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting

Companies

DP world

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:757

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

7 steps

1
Step 1: Sort the intervals based on their end points in ascending order. If two intervals have the same end point, sort them in descending order of their start points.
2
Step 2: Initialize two variables, `p1` and `p2`, to keep track of the two largest elements in the containing set. Initialize them to -1 (or any value smaller than the interval start values). Initialize the answer variable `ans` to 0.
3
Step 3: Iterate through the sorted intervals.
4
Step 4: For each interval [start, end], check if `start > p2`. If true, it means that the interval doesn't contain `p1` or `p2`. Add 2 to the `ans`. Set `p1 = end - 1` and `p2 = end`.
5
Step 5: If `start <= p2` but `start > p1`, it means the interval only contains `p2`. Add 1 to the `ans`. Set `p1 = p2` and `p2 = end`.
6
Step 6: If `start <= p1`, it means the interval contains both `p1` and `p2`. Do nothing.
7
Step 7: Return the value of `ans`.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is primarily determined by the sorting algorithm used. Common sorting algorithms have a time complexity of O(n log n). The rest of the algorithm involves a single loop through the sorted intervals, which takes O(n) time. The space complexity is constant because the algorithm only uses a few extra variables, irrespective of the number of intervals.

📖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

Dynamic Programming

Dynamic programming could potentially be used to solve this problem. However, it would likely be more complex and less efficient than the greedy approach. The state would need to represent which elements are already in the containing set, which can be very large, making the time and space complexity high.

Time: Potentially exponentialSpace: Potentially exponential

Trade-offs: The greedy approach is much simpler and more efficient than dynamic programming for this problem.

✨

Key Insights

1Insight 1: Sorting intervals by their end points is crucial for a greedy approach to work effectively. This allows prioritizing intervals that end sooner, increasing the chance to cover them with a small number of elements.
2Insight 2: Choosing the two largest possible numbers within an interval (close to the end point) when needing to add elements is a key greedy strategy for minimizing the size of the containing set.
3Insight 3: Sorting intervals in ascending order of end points and then descending order of start points is the correct approach to optimize the greedy algorithm.

📚

Key Terms Explained

For Students

Array

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

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

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

Edge Cases

Edge case 1: Empty input (intervals is empty). The algorithm correctly returns 0 because there are no intervals to cover.
Edge case 2: Overlapping intervals. The greedy approach handles overlapping intervals correctly due to the sorting criteria. The algorithm efficiently reuses elements in the containing set for overlapping intervals, minimizing the total size.
Edge case 3: Intervals with the same end points but different start points. The algorithm prioritizes intervals with larger start points, which is correct because it delays adding new elements as much as possible.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect sorting criteria. Sorting only by end points in ascending order is not sufficient. Sorting by end points ascending and start points descending is critical to the solution.

✗Common mistake 2: Not considering the case where the interval contains only one element from the existing set. Failing to handle this case will lead to an incorrect answer.

✗Common mistake 3: Prematurely adding numbers to the result. The elements `end-1` and `end` must only be added when absolutely needed (when the interval contains 0 or 1 element of the current coverage).

✅

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: Resource allocation with constraints. Imagine allocating servers to tasks, where each task requires at least two servers within a specific time window. This problem could help determine the minimum number of servers required to fulfill all task requirements.

🏦Real-world scenario 2: Data compression. Certain data compression algorithms involve finding common subsequences or patterns in data streams. This problem relates to finding the smallest set of data points that satisfy certain coverage criteria.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: July 8, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

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

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Set Intersection Size At Least Two"?

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 greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?