Sorting

16. 3Sum Closest

1. Problem Description: Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Problem Link: https://leetcode.com/problems/3sum-closest/

1. Optimized Java Code for "3Sum Closest" Problem:

import java.util.Arrays;

class Solution {
    public int threeSumClosest(int[] nums, int target) {
        // Sort the input array
        Arrays.sort(nums);
        
        // Setting initial closestSum as the sum of first 3 elements
        int closestSum = nums[0] + nums[1] + nums[2];
        
        // Iterate from the beginning of the array
        for (int i = 0; i < nums.length - 2; i++) {
            // Check for duplicate values, skip if the current element is the same as the previous one
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            
            int left = i + 1; // Pointer to the element on the right of current element
            int right = nums.length - 1; // Pointer to the last element
            
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                
                // Check if current sum is closer to target than the existing closestSum
                if (Math.abs(sum - target) < Math.abs(closestSum - target)) {
                    closestSum = sum;
                }
                
                // Move pointers based on the difference between sum and target
                if (sum < target) {
                    left++;
                } else if (sum > target) {
                    right--;
                } else {
                    // If current sum is equal to target, return the sum as we have found the closest possible sum
                    return sum;
                }
            }
        }
        
        return closestSum;
    }
}

1. Explanation and Thought Process:

• The problem asks for finding three integers in a given array whose sum is closest to a given target.

• To approach this problem, we can start by sorting the array in ascending order. Sorting will help us identify the closest sum more efficiently.

• We set the initial closestSum as the sum of the first three elements, assuming it to be the closest sum.

• We then iterate through the array from the beginning, considering each element as a potential candidate for the first number in the triplet.

• To skip duplicate numbers, we check if the current element is the same as the previous one. If true, we continue to the next iteration.

• For each potential first number, we use two pointers: left and right.

• We compare the sum of the current triplet with the closestSum. If the current sum is closer to the target, we update the closestSum.

• Then, we adjust the pointers based on the difference between the sum and the target. If the sum is smaller than the target, we increment the left pointer. If the sum is larger than the target, we decrement the right pointer.

• We break the loop and return if we find a triplet with a sum equal to the target.

• Finally, we return the closestSum as the result.

• This solution has a time complexity of O(n^2), where n is the length of the input array.

17. Sort Colors

1. Problem Description: The problem "Sort Colors" on LeetCode asks us to sort an array containing only 0s, 1s, and 2s in ascending order.

Link to problem: https://leetcode.com/problems/sort-colors/

1. Optimized Java Solution with Detailed Comments:

class Solution {
    public void sortColors(int[] nums) {
        int low = 0; // Pointer to track low (0) values
        int high = nums.length - 1; // Pointer to track high (2) values
        int curr = 0; // Current position pointer
        
        while (curr <= high) {
            if (nums[curr] == 0) {
                swap(nums, curr, low);
                curr++;
                low++;
            }
            else if (nums[curr] == 2) {
                swap(nums, curr, high);
                high--;
            }
            else {
                curr++;
            }
        }
    }
    
    // Helper function to perform array element swapping
    private void swap(int[] nums, int i, int j) {
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}

1. Detailed Description and Algorithm Walkthrough:

The problem requires us to sort an array consisting of only 0s, 1s, and 2s. The main idea behind a solution is to implement a modified version of the Dutch National Flag algorithm. Here's how the algorithm works:

• We can use 3 pointers: low, high, and curr.

• The low pointer points to the next available position for 0s.

• The high pointer points to the next available position for 2s.

• The curr pointer is used to traverse the array.

We initialize the low and curr pointers to 0, and the high pointer to the last index of the array. We start traversing the array from the beginning using the curr pointer.

While the curr pointer is less than or equal to the high pointer, we perform the following steps:

• If the current element is 0, we swap it with the element at the low pointer, increment the curr and low pointers by 1.

• If the current element is 2, we swap it with the element at the high pointer, decrement the high pointer by 1 (as we have sorted a 2 at the right position), and move the curr pointer to the next position without incrementing it (since the swapped element may be 0 or 1, which needs further consideration).

• If the current element is 1, we simply move the curr pointer to the next position.

This process continues until the curr pointer exceeds the high pointer. At this point, the array will be sorted in ascending order of colors.

During an interview, it is important to explain the algorithm and how it solves the problem efficiently. Highlight the key points such as the use of pointers to track positions and swapping elements to maintain the sorting. Also, emphasize the time complexity of the solution, which is O(n) where n is the length of the input array.

It's a good practice to run through an example input array during the explanation to showcase the algorithm's steps clearly.

18. Squares of a Sorted Array

1. Problem Description:

Problem link: https://leetcode.com/problems/squares-of-a-sorted-array/

Given an array of integers sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order.

Example: Input: [-4, -1, 0, 3, 10] Output: [0, 1, 9, 16, 100]

1. Optimized Java Code with detailed comments:

public int[] sortedSquares(int[] nums) {
    int n = nums.length;
    int[] result = new int[n];
    int left = 0; // pointer for the left end of the array
    int right = n - 1; // pointer for the right end of the array

    for (int i = n - 1; i >= 0; i--) {
        if (Math.abs(nums[left]) > Math.abs(nums[right])) { // compare the absolute values of the numbers
            result[i] = nums[left] * nums[left];
            left++;
        } else {
            result[i] = nums[right] * nums[right];
            right--;
        }
    }
    
    return result;
}

1. Detailed Description and Algorithm:

In this problem, we need to square each number in the input array and return a new array sorted in non-decreasing order.

The algorithm for this problem makes use of the two-pointer approach. We can observe that the largest absolute value will be on the ends of the input array. So, we can start by comparing the absolute values of the numbers at the ends and fill the result array from the rightmost position to the left.

The two pointers, left and right, initially point to the ends of the input array. We iterate from the last index of the result array to the first index. In each iteration, we compare the absolute values of nums[left] and nums[right]. If the absolute value at left is greater, we square it and add it to the result array at the current index. Otherwise, we square the value at right and add it to the result array.

By doing this, we guarantee that the elements in the result array are sorted in non-decreasing order because we compare the absolute values and add them to the result array in non-decreasing order.

After filling the result array, we return it as the output.

Thinking through this algorithm during an interview process:

• Understand the problem statement and ask clarifying questions if necessary.

• Consider edge cases such as an empty array or negative elements.

• Analyze the problem and identify a possible approach. The two-pointer approach can be a good choice for this problem.

• Start writing the code with the initial approach and explain the logic step by step.

• Optimize the code by eliminating unnecessary steps or reducing time complexity if possible. In this case, the code is already optimized and straightforward.

• Test the code with different test cases, including the given examples and additional edge cases, to ensure it works correctly.

• If time permits, think about alternative approaches or possible optimizations and discuss them with the interviewer.

19. Merge Intervals

1. Problem Description: Given a collection of intervals, merge all overlapping intervals.

Example: Input: [[1,3],[2,6],[8,10],[15,18]] Output: [[1,6],[8,10],[15,18]] Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].

Problem Link: https://leetcode.com/problems/merge-intervals/

1. Optimized Java Code with Detailed Comments:

import java.util.*;

class Solution {
    public int[][] merge(int[][] intervals) {
        // Sort the intervals based on the start time
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);

        // Create a list to store the merged intervals
        List<int[]> mergedIntervals = new ArrayList<>();

        // Loop through each interval and merge them if necessary
        for (int[] interval : intervals) {
            // If the mergedIntervals list is empty or the current interval doesn't overlap with the last merged interval,
            // add the current interval directly to the mergedIntervals list
            if (mergedIntervals.isEmpty() || interval[0] > mergedIntervals.get(mergedIntervals.size() - 1)[1]) {
                mergedIntervals.add(interval);
            } else {
                // Otherwise, merge the current interval with the last merged interval
                mergedIntervals.get(mergedIntervals.size() - 1)[1] = Math.max(mergedIntervals.get(mergedIntervals.size() - 1)[1], interval[1]);
            }
        }

        // Convert the list of merged intervals to a 2D array and return it
        return mergedIntervals.toArray(new int[mergedIntervals.size()][]);
    }
}

1. Explanation of the Algorithm and Approach: The problem requires merging overlapping intervals in a given collection. The key idea is to sort the intervals based on their start time and then merge them accordingly.

The algorithm starts by sorting the intervals in ascending order based on the start time. This ensures that we can process the intervals in a linear fashion.

Next, we create an ArrayList called mergedIntervals to store the merged intervals. We iterate through each interval in the sorted order.

For each interval, we compare its start time (interval[0]) with the end time of the last merged interval (mergedIntervals.get(mergedIntervals.size() - 1)[1]). If there is no overlap, we add the interval directly to the mergedIntervals list. Otherwise, we update the end time of the last merged interval to be the maximum of its current end time and the end time of the current interval. This ensures that the merged interval encompasses both intervals.

Finally, we convert the mergedIntervals list to a 2D array and return it as the result.

During an interview, it is essential to explain the algorithm step by step, including the sorting approach and the logic behind merging intervals. Talking through each step and providing clear examples will demonstrate your understanding of the problem and your ability to articulate your thought process.

20. Intersection of Two Arrays II

1. Problem Description and Link: The problem "Intersection of Two Arrays II" on LeetCode asks us to find the intersection of two given arrays. We need to return a new array containing the common elements between the two arrays, considering the count of each element in both arrays.

Link to the problem on LeetCode: https://leetcode.com/problems/intersection-of-two-arrays-ii/

1. Optimized Java Code with Detailed Comments:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class IntersectionOfTwoArraysII {

    public int[] intersect(int[] nums1, int[] nums2) {
        // Create a map to store the count of each element in nums1
        Map<Integer, Integer> countMap = new HashMap<>();
        
        // Iterate through nums1 and update the count of each element in countMap
        for (int num : nums1) {
            countMap.put(num, countMap.getOrDefault(num, 0) + 1);
        }
        
        // Create a list to store the common elements
        List<Integer> commonElements = new ArrayList<>();
        
        // Iterate through nums2
        for (int num : nums2) {
            // Check if the element exists in the countMap
            if (countMap.containsKey(num) && countMap.get(num) > 0) {
                // Found a common element, add it to commonElements list
                commonElements.add(num);
                
                // Update the count in countMap
                countMap.put(num, countMap.get(num) - 1);
            }
        }
        
        // Create an array to store the result
        int[] result = new int[commonElements.size()];
        
        // Convert commonElements list to array
        for (int i = 0; i < commonElements.size(); i++) {
            result[i] = commonElements.get(i);
        }
        
        return result;
    }
}

1. Detailed Description of the Algorithm and Interview Thought Process:

The problem asks us to find the intersection of two arrays, and return an array containing the common elements.

To solve this problem efficiently, we can use a hashmap to store the count of each element in the first array (nums1). Then, we need to iterate through the second array (nums2) and check if each element exists in the hashmap. If it does, we add that element to the list of common elements and decrement its count in the hashmap.

This approach ensures that we consider the count of each element in both arrays. By utilizing a hashmap, we can achieve a time complexity of O(n+m), where n and m are the lengths of the input arrays.

During an interview, it's essential to think through the problem and follow these steps:

1. Understand the requirements: Read the problem statement carefully and understand what the problem is asking for. Identify the key inputs and outputs required.

2. Clarify doubts: If you have any confusion or doubts about the problem statement, ask the interviewer for clarification.

3. Identify data structures: Analyze the problem requirements and select the appropriate data structure(s) to solve the problem efficiently. In this case, the hashmap provides an efficient way to store and update the count of each element.

4. Plan the algorithm: Come up with a step-by-step plan to solve the problem using the selected data structure(s). Think about how to iterate through the arrays, update the hashmap, and find the common elements.

5. Write code with comments: Implement the algorithm in your preferred programming language, and add comments to explain each section of the code. Clear and well-commented code showcases your programming skills and thought process.

6. Test your code: After writing the code, test it with various test cases, including edge cases, to ensure its correctness and efficiency.

By following these steps, you can efficiently solve the "Intersection of Two Arrays II" problem on LeetCode and provide a clear and detailed solution during an interview.