Interval

35. Insert Interval

Problem Description and Leetcode Link: The problem is to insert a new interval into a list of non-overlapping intervals. The intervals are sorted by their start time and may have overlapping intervals. The task is to insert the new interval and merge any overlapping intervals if necessary. Here is the Leetcode link for the problem:
Optimized Java code with detailed comments:

class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        List<int[]> merged = new ArrayList<>();
        int i = 0;
        
        // Add all intervals that come before the new interval without overlap
        while (i < intervals.length && intervals[i][1] < newInterval[0]) {
            merged.add(intervals[i]);
            i++;
        }
        
        // Merge all overlapping intervals with the new interval
        while (i < intervals.length && intervals[i][0] <= newInterval[1]) {
            newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
            newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
            i++;
        }
        merged.add(newInterval);
        
        // Add all remaining intervals after merging
        while (i < intervals.length) {
            merged.add(intervals[i]);
            i++;
        }
        
        // Convert the merged list into an array and return
        return merged.toArray(new int[merged.size()][2]);
    }
}

Algorithm Description and Interview Insights: The algorithm involves iterating through the given intervals and merging any overlapping intervals with the new interval. The key steps are:

Initialize an empty list merged to store the merged intervals.
Iterate through the intervals until we find an interval that comes after the new interval. Add all intervals that come before the new interval without any overlap.
Merge any overlapping intervals with the new interval. Update the start and end times of the new interval based on the overlapping intervals.
Add the merged new interval to the merged list.
Add all remaining intervals after merging.
Convert the merged list into an array and return.

During an interview, it is important to think through the problem step by step. Understand the problem requirements, constraints, and potential edge cases. Discuss the overall approach, and then dive into the detailed algorithm implementation. Remember to communicate your thought process, clarify any doubts, test your code with sample inputs, and analyze the time and space complexities.

Similar Problems and Abstractions: The algorithm used for inserting an interval and merging overlapping intervals can be abstracted for similar problems that involve interval manipulations, such as:

Similar Problems with Java Code and Descriptions:

a) Merge Intervals:

class Solution {
    public int[][] merge(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        List<int[]> merged = new ArrayList<>();
        int[] currentInterval = intervals[0];
        
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] <= currentInterval[1]) {
                currentInterval[1] = Math.max(currentInterval[1], intervals[i][1]);
            } else {
                merged.add(currentInterval);
                currentInterval = intervals[i];
            }
        }
        
        merged.add(currentInterval);
        return merged.toArray(new int[merged.size()][2]);
    }
}

This problem is similar to "Insert Interval" as it involves merging overlapping intervals. The only difference is that we are given a list of intervals, and we need to merge them.

b) Meeting Rooms II:

class Solution {
    public int minMeetingRooms(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        minHeap.offer(intervals[0][1]);
        
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] >= minHeap.peek()) {
                minHeap.poll();
            }
            minHeap.offer(intervals[i][1]);
        }
        
        return minHeap.size();
    }
}

This problem is similar to "Insert Interval" as it involves scheduling meetings (intervals) and finding the minimum number of rooms required to accommodate all the meetings.

c) Non-overlapping Intervals:

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        if (intervals.length == 0) return 0;
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        int end = intervals[0][1], count = 1;
        
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] >= end) {
                end = intervals[i][1];
                count++;
            } else {
                end = Math.min(end, intervals[i][1]);
            }
        }
        
        return intervals.length - count;
    }
}

This problem is similar to "Insert Interval" as it involves removing the minimum number of intervals to make the remaining intervals non-overlapping.

36. Merge Intervals

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Problem Description: The problem "Merge Intervals" on LeetCode can be found at the following link: https://leetcode.com/problems/merge-intervals/

The problem requires merging overlapping intervals. Given a list of intervals, each interval contains a start time and an end time. The task is to merge all overlapping intervals and return a new list of non-overlapping intervals.

Optimized Java Code for "Merge Intervals" problem:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;

class Solution {
    public int[][] merge(int[][] intervals) {
        // Sort the intervals based on their start time
        Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));
        
        List<int[]> merged = new ArrayList<>();
        int[] currentInterval = intervals[0];
        merged.add(currentInterval);
        
        for (int[] interval : intervals) {
            int currentEnd = currentInterval[1];
            int nextStart = interval[0];
            int nextEnd = interval[1];
            
            // If the intervals overlap, merge them by updating the current interval's end time
            if (currentEnd >= nextStart) {
                currentInterval[1] = Math.max(currentEnd, nextEnd);
            } else {
                // If the intervals don't overlap, add the next interval to the merged list
                currentInterval = interval;
                merged.add(currentInterval);
            }
        }
        
        return merged.toArray(new int[merged.size()][]);
    }
}

Algorithm and Interview Process: The algorithm for merging intervals involves sorting the intervals based on their start times. After sorting, we iterate through the intervals and check for overlap with the current interval. If there is overlap, we merge the intervals by updating the end time of the current interval. If there is no overlap, we add the next interval to the merged list.

During an interview for this problem, you can start by clarifying the problem statement and discussing possible edge cases. Then explain your approach and the reasoning behind it. Implement the algorithm and test it with different test cases. Finally, analyze the time complexity and space complexity of your solution.

Similar Algorithms and Problems: The algorithm used for merging intervals can be applied to several similar problems, such as:

Meeting Rooms: Determine if a person can attend all the meetings without overlapping.
Insert Interval: Given a set of non-overlapping intervals, insert a new interval and merge if necessary.
Non-overlapping Intervals: Find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Similar Problems with Java Code and Descriptions: Here are some similar problems with their Java code and descriptions:

Meeting Rooms:

Problem Description: https://leetcode.com/problems/meeting-rooms/

Java Code:

import java.util.Arrays;

class Solution {
    public boolean canAttendMeetings(int[][] intervals) {
        Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));
        
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] < intervals[i-1][1]) {
                return false;
            }
        }
        
        return true;
    }
}

Description: This problem checks if a person can attend all the meetings without overlapping. We sort the meetings based on their start time and check if the next meeting starts before the previous meeting ends.

Insert Interval:

Problem Description: https://leetcode.com/problems/insert-interval/

Java Code:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        List<int[]> merged = new ArrayList<>();
        int i = 0;
        int n = intervals.length;
        
        // Add all intervals before the new interval
        while (i < n && intervals[i][1] < newInterval[0]) {
            merged.add(intervals[i]);
            i++;
        }
        
        // Merge overlapping intervals
        while (i < n && intervals[i][0] <= newInterval[1]) {
            newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
            newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
            i++;
        }
        
        merged.add(newInterval);
        
        // Add remaining intervals
        while (i < n) {
            merged.add(intervals[i]);
            i++;
        }
        
        return merged.toArray(new int[merged.size()][]);
    }
}

Description: This problem involves inserting a new interval into a set of non-overlapping intervals. We iterate through the intervals and merge overlapping intervals with the new interval as needed.

Non-overlapping Intervals:

Problem Description: https://leetcode.com/problems/non-overlapping-intervals/

Java Code:

import java.util.Arrays;

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        if (intervals.length == 0) {
            return 0;
        }
        
        Arrays.sort(intervals, Comparator.comparingInt(a -> a[1]));
        
        int count = 1;
        int end = intervals[0][1];
        
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] >= end) {
                count++;
                end = intervals[i][1];
            }
        }
        
        return intervals.length - count;
    }
}

Description: This problem finds the minimum number of intervals that need to be removed to make the rest of the intervals non-overlapping. We sort the intervals based on their end times and keep track of the current end time to determine overlap.

37. Non-overlapping Intervals

Problem: The "Non-overlapping Intervals" problem on LeetCode involves finding the minimum number of intervals that need to be removed in order to make all the remaining intervals non-overlapping. The problem can be found on LeetCode at the following link: https://leetcode.com/problems/non-overlapping-intervals/
Optimized Java Code: Here is the optimized Java code for the "Non-overlapping Intervals" problem:

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        if (intervals.length == 0) {
            return 0;
        }
        
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        
        int end = intervals[0][1];
        int count = 1; // count of non-overlapping intervals
        
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] >= end) {
                end = intervals[i][1];
                count++;
            } else {
                end = Math.min(end, intervals[i][1]);
            }
        }
        
        return intervals.length - count;
    }
}

Algorithm Explanation: To solve this problem efficiently, we can follow the greedy approach. First, we sort the intervals based on their start times. Next, we initialize a variable 'end' to store the end time of the current selected interval. We also initialize a variable 'count' to keep track of the number of non-overlapping intervals.

Starting from the second interval, we compare its start time with 'end'. If the start time of the current interval is greater than or equal to 'end', it means there is no overlap and we update 'end' with the end time of the current interval and increment 'count' by 1. If there is an overlap, we need to choose the interval with the smaller end time to keep the remaining intervals as non-overlapping as possible. Thus, we update 'end' with the minimum of its current value and the end time of the current interval.

Finally, we return the total number of intervals minus the count of non-overlapping intervals, which corresponds to the minimum number of intervals that need to be removed.

Related Problems: The algorithm used in the "Non-overlapping Intervals" problem can be applied to similar problems that require finding non-overlapping intervals or merging overlapping intervals. Some of the related problems are:

Merge Intervals: https://leetcode.com/problems/merge-intervals/
Insert Interval: https://leetcode.com/problems/insert-interval/
Meeting Rooms: https://leetcode.com/problems/meeting-rooms/
Meeting Rooms II: https://leetcode.com/problems/meeting-rooms-ii/

Similar Problems with Detailed Java Code and Descriptions:

a) Merge Intervals: Problem Link: https://leetcode.com/problems/merge-intervals/

class Solution {
    public int[][] merge(int[][] intervals) {
        if (intervals.length <= 1) {
            return intervals;
        }
        
        List<int[]> result = new ArrayList<>();
        
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        
        int[] currentInterval = intervals[0];
        result.add(currentInterval);
        
        for (int[] interval : intervals) {
            int currentEnd = currentInterval[1];
            int nextStart = interval[0];
            int nextEnd = interval[1];
            
            if (currentEnd >= nextStart) {
                currentInterval[1] = Math.max(currentEnd, nextEnd);
            } else {
                currentInterval = interval;
                result.add(currentInterval);
            }
        }
        
        return result.toArray(new int[result.size()][]);
    }
}

b) Insert Interval: Problem Link: https://leetcode.com/problems/insert-interval/

class Solution {
    public int[][] insert(int[][] intervals, int[] newInterval) {
        List<int[]> result = new ArrayList<>();
        int i = 0, n = intervals.length;
        
        while (i < n && intervals[i][1] < newInterval[0]) {
            result.add(intervals[i]);
            i++;
        }
        
        int[] mergedInterval = newInterval;
        
        while (i < n && intervals[i][0] <= newInterval[1]) {
            mergedInterval[0] = Math.min(mergedInterval[0], intervals[i][0]);
            mergedInterval[1] = Math.max(mergedInterval[1], intervals[i][1]);
            i++;
        }
        
        result.add(mergedInterval);
        
        while (i < n) {
            result.add(intervals[i]);
            i++;
        }
        
        return result.toArray(new int[result.size()][]);
    }
}

c) Meeting Rooms: Problem Link: https://leetcode.com/problems/meeting-rooms/

class Solution {
    public boolean canAttendMeetings(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] < intervals[i - 1][1]) {
                return false;
            }
        }
        
        return true;
    }
}

d) Meeting Rooms II: Problem Link: https://leetcode.com/problems/meeting-rooms-ii/

class Solution {
    public int minMeetingRooms(int[][] intervals) {
        if (intervals == null || intervals.length == 0) {
            return 0;
        }
        
        int n = intervals.length;
        int[] startTimes = new int[n];
        int[] endTimes = new int[n];
        
        for (int i = 0; i < n; i++) {
            startTimes[i] = intervals[i][0];
            endTimes[i] = intervals[i][1];
        }
        
        Arrays.sort(startTimes);
        Arrays.sort(endTimes);
        
        int rooms = 0;
        int endIndex = 0;
        
        for (int i = 0; i < n; i++) {
            if (startTimes[i] < endTimes[endIndex]) {
                rooms++;
            } else {
                endIndex++;
            }
        }
        
        return rooms;
    }
}

These examples should help you understand how to apply the same algorithm in different scenarios and solve similar problems effectively.

38. Meeting Rooms (Leetcode Premium)

Problem Description and Link: The problem is called "Meeting Rooms" and the link to the leetcode problem is: https://leetcode.com/problems/meeting-rooms/
Optimized Java Code with Detailed Comments:

import java.util.Arrays;

class Interval {
    int start;
    int end;

    Interval(int start, int end) {
        this.start = start;
        this.end = end;
    }
}

class Solution {
    public boolean canAttendMeetings(Interval[] intervals) {
        // Sort the intervals in ascending order of start times
        Arrays.sort(intervals, (a, b) -> a.start - b.start);
        
        // Check if there is any overlap between adjacent intervals
        for (int i = 0; i < intervals.length - 1; i++) {
            if (intervals[i].end > intervals[i + 1].start) {
                return false;
            }
        }
        
        // No overlap found
        return true;
    }
}

Algorithm and Interview Thought Process: The algorithm used to solve this problem is based on sorting the intervals and then checking for any overlap between adjacent intervals. Here is the step-by-step thought process to approach this problem in an interview:

First, we need to understand the problem statement clearly. Given a list of meeting intervals, we need to determine if a person can attend all the meetings without any time overlap.
Next, we can consider different approaches to solve the problem. Sorting the intervals based on start times is an intuitive approach because it allows us to easily detect any overlaps.
One important detail is whether the intervals are inclusive or exclusive. In this problem, the intervals are considered inclusive, i.e., if a meeting ends at time x, another meeting can start at time x.
We can sort the intervals based on their start times, either using a custom comparator or by implementing the Comparable interface in the Interval class.
After sorting the intervals, we iterate through them and check if the end time of the current interval is greater than the start time of the next interval. If this condition is true, it means that there is an overlap, and we return false.
If we reach the end of the loop without finding any overlap, it means that all the meetings can be attended, and we return true.
Finally, we need to consider the time complexity of the solution. Sorting the intervals takes O(n log n) time, and the subsequent iteration takes O(n) time. So, the overall time complexity is O(n log n).

Abstracted Problems: The algorithm used in this problem can be applied to any scenario where we need to check for overlaps between intervals. Some abstracted problems that can be solved using a similar algorithm include:

Scheduling tasks: Given a list of tasks with their start and end times, determine if it is possible to schedule all the tasks without any overlap.
Resource allocation: Given a list of resource requests with their start and end times, determine if it is possible to allocate the resources to the requests without any overlap.
Interval merging: Given a list of intervals, find a way to merge overlapping intervals into a consolidated list.
Exam scheduling: Given a list of exams with their start and end times, determine if it is possible to schedule all the exams without any overlap.

Similar Problems: Here are some similar problems with detailed Java code and descriptions:

Meeting Rooms II: https://leetcode.com/problems/meeting-rooms-ii/

import java.util.Arrays;
import java.util.PriorityQueue;

class Interval {
    int start;
    int end;

    Interval(int start, int end) {
        this.start = start;
        this.end = end;
    }
}

class Solution {
    public int minMeetingRooms(Interval[] intervals) {
        // Sort the intervals in ascending order of start times
        Arrays.sort(intervals, (a, b) -> a.start - b.start);
        
        // Create a min heap to store the end times of the meetings
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        
        // Add the end time of the first meeting to the min heap
        minHeap.add(intervals[0].end);
        
        // Iterate through the remaining meetings
        for (int i = 1; i < intervals.length; i++) {
            // If the current meeting starts after the minimum end time in the heap, then we can reuse that room
            if (intervals[i].start >= minHeap.peek()) {
                minHeap.poll();
            }
            
            // Add the end time of the current meeting to the heap
            minHeap.add(intervals[i].end);
        }
        
        // The size of the min heap is the minimum number of meeting rooms required
        return minHeap.size();
    }
}

This problem is a variation of the original problem where we need to determine the minimum number of meeting rooms required to schedule all the meetings.

Merge Intervals: https://leetcode.com/problems/merge-intervals/

import java.util.Arrays;
import java.util.LinkedList;

class Interval {
    int start;
    int end;

    Interval(int start, int end) {
        this.start = start;
        this.end = end;
    }
}

class Solution {
    public int[][] merge(int[][] intervals) {
        // Sort the intervals based on their start times
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        
        // Create a linked list to store the merged intervals
        LinkedList<int[]> mergedIntervals = new LinkedList<>();
        
        // Iterate through the intervals
        for (int[] interval : intervals) {
            // If the current interval overlaps with the last interval in the merged list, merge them
            if (mergedIntervals.isEmpty() || mergedIntervals.getLast()[1] < interval[0]) {
                mergedIntervals.add(interval);
            } else {
                mergedIntervals.getLast()[1] = Math.max(mergedIntervals.getLast()[1], interval[1]);
            }
        }
        
        // Convert the linked list to a 2D array
        return mergedIntervals.toArray(new int[mergedIntervals.size()][2]);
    }
}

This problem focuses on merging overlapping intervals into a consolidated list of non-overlapping intervals.

These examples demonstrate the versatility of the underlying algorithm and how it can be applied to different problems by making minor modifications.

39. Meeting Rooms II (Leetcode Premium)

Problem Description: You are given an array of meeting time intervals consisting of start and end times. Your task is to find the minimum number of conference rooms required to hold all the meetings.

Java Code for Meeting Rooms II:

import java.util.Arrays;
import java.util.PriorityQueue;

class Solution {
    public int minMeetingRooms(int[][] intervals) {
        if (intervals == null || intervals.length == 0) {
            return 0;
        }
        
        // Sort the intervals based on the start time
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        
        // Create a min heap to store the end times of the meetings
        PriorityQueue<Integer> minHeap = new PriorityQueue<>();
        
        // Add the first meeting end time to the min heap
        minHeap.offer(intervals[0][1]);
        
        // Traverse through the rest of the meetings
        for (int i = 1; i < intervals.length; i++) {
            // If the earliest ending meeting has ended, remove it from the min heap
            if (intervals[i][0] >= minHeap.peek()) {
                minHeap.poll();
            }
            
            // Add the current meeting end time to the min heap
            minHeap.offer(intervals[i][1]);
        }
        
        // The size of the min heap at the end represents the minimum number of rooms required
        return minHeap.size();
    }
}

Algorithm Explanation: The algorithm sorts the intervals based on the start time and uses a min heap to store the end times of the meetings. The min heap keeps track of the earliest ending meeting. When a new meeting starts, if it starts after the earliest ending meeting has ended, we can reuse the room and remove the earliest ending meeting from the min heap. Otherwise, we need to allocate a new room. The size of the min heap at the end represents the minimum number of rooms required.

During the interview process, you can explain the algorithm by breaking it down into the following steps:

Sort the intervals based on the start time.
Create a min heap to store the end times of the meetings.
Add the end time of the first meeting to the min heap.
Traverse through the rest of the meetings.
- If the current meeting starts after the earliest ending meeting has ended, remove the earliest ending meeting from the min heap (as we can reuse the room).
- Add the end time of the current meeting to the min heap.
Return the size of the min heap as the minimum number of rooms required.

Similar Algorithm: This algorithm is a Greedy Algorithm called "Interval Scheduling" or "Activity Selection". It aims to maximize the number of non-overlapping intervals by selecting intervals with earliest finish times.
Similar Problems: Here are some similar problems that can be solved using a similar algorithm:

Please note that the above problems may have slight variations in the input format or problem statement, but the core algorithm remains the same.

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