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Leetcode 88: Merge Sorted Array

May 14, 2025

Question

In this Leetcode Problem, we are given two already sorted arrays, as well as their expected sizes in the m and n variables. The Nums1 array is also filled with zeros at the end to ensure the Nums2 array can fit inside of Nums1.

Our task is to merge both of these presorted arrays together Inplace inside of Nums1, while maintaining a sorted ordering. Inplace is different from most Leetcode problems, as we are only required to edit the given nums1 array. Instead of checking a specific function return value, the Leetcode Judge instead checks what we have done with the given nums1.

Additional, the Follow Up challenge specifies we want to solve this problem in O(m+n) time. The easy (m+n) log(n) solution would be to concatenate the given arrays together, then sort them using a built-in sort function. However, because most built-in sorting methods are log(n) time complexity, this would break our follow up challenge. Instead, we can use a more optimal two pointer approach.

Solving

To solve this problem, we use an I variable to iterate from right to left on the nums1 array. We then use two pointers named a,b as a way to check values from the two given arrays. The pointer a is used to look up values in nums1, and the pointer b is used for numbers in nums2. So where the I variable is just for iterating through nums1, a and b are actually used for performing our comparisons.

If nums1[a] is greater then nums2[b], then we can keep that value at nums1 the same. Note that we only check this if our a pointer is still at a valid slot in nums1. Otherwise, we can safely assume that the current I th position in nums1 needs to be replaced with nums2[b]. Since nums2 is guaranteed to be the smaller length of the two given arrays, we can exit out of our while loop.

Instead of returning an array, we can just let the code end at the while loop, as the Leetcode Judge is only concerned with what nums1 is equal to.

TLDR:

We must merge two given sorted arrays together inplace. To solve the follow up challenge of O(m+n) time complexity, we can easily do this using a two pointer solutions instead of relying on built in sorting methods.

Leetcode 70: Climbing Stairs

May 12, 2025

DevGod

Elf Vtuber

About

This leetcode problem is about a fictional set of stairs, and finding the total amount of ways you could climb to the Nth step. Whilst climbing the stairs, you can choose to climb at most 1 or 2 steps at a time. Additionally, the problems lets us know that stairs 1,2,3 are equal to 1,2,3 ways respectively. I ended up using Top Down Dynamic Programming alongside Recursion plus Memoization for time optimization.

One way to think about this problem, is to imagine what leads up to your target Nth stair. For example, if you wanted to reach the 100th stair, you have to either climb one step from the 99th stair, or climb two steps from the 98th stair. Since the Nth stair is reachable from the previous two stairs, we can calculate the number of ways to reach the Nth stair by adding together the previous two stairs amount of climbing combinations. This forms a pattern similar to a Fibonacci number problems, where we want to find the answer based on the sum of previous answers we’ve found for the problem.

The sub problem of “Climbing Stairs” is that for each Nth stair, we must find the number of ways the previous two stairs had before adding them together. The base cases of this problem is that for any Nth stair less than or equal to three, the number of ways will be the same value as Nth. It is when stair’s N is greater than three that we must use our DP solution to find the total number of ways to climb up to that stair.

Additionally, since N can get large, I use a hash map named MEMO to store past results to reduce unnecessary recursive function calls, optimizing my solution by relying on cached past results for better speed. If my MEMO has a result for a given step I am looking for, then I just use that. However, if MEMO doesn’t contain the number of ways for the Nth stair I am looking for, then I proceed forward with running my recursive solution.

TLDR

Used Dynamic Programming, Recursion and Memoization to find the each Stairs number of ways to climb. Each Nth stair number of total ways relies on the previous two stairs total ways added together. Stairs 1,2,3 have one way, two ways, and three ways to climb to them respectively.

Leetcode 1: Two Sum

May 2, 2025

DevGod

Elf Vtuber

About

Two Sum is the first ever Leetcode problem published. In this problem, we are given an array of numbers, and a target integer. We must find two points in the given array who’s values added together are equal to the given target value. At the end, we return the indexes of where these two points exist in the given array.

To solve this problem, we can use a data structure known as a Hash Table. Basically, Hash Tables are used for optimally searching, setting, and getting key value pairs. In this problem, using a Hash Table is important for storing past results that we can use later in our for loop. By sacrificing this extra little bit of memory, we can write an O(N) time complexity solution for the problem.

Specifically, we want to find the difference between the target and the current value we’re evaluating in the given array. This difference value is a number we need to find in the future to form a pair that equals the given target. Our Hash Table’s keys will be set to this difference, while our values will be set to our current index for use later. For example, if our first element in the array is 1, and our target is 5, then we would store {4:1}.

We continue scanning through the given array, whilst recording our difference key and our index values along the way. If we ever come across a value in the given array that happens to equal a key stored inside of our hash table, we then know we have found a matching pair that equals target. We can then return the current index we are checking, alongside our previous index already stored inside of our Hash Table. Back to our above example, if we have {4:1} in our Hash Table, and we found a 4 further in the array, we can return our current index along with the past index we’ve stored in the Hash Table.

TLDR:

Using a Hash Table to store the needed future value along with the past Index. Then if we hit the required future value further in the array, we can return our past Index, along with the current index.

Leetcode 3375: Minimum Operations to Make Array Values Equal to K

Apr 7, 2025

DevGod

Elf Vtuber

For today’s LeetCode problem, I honestly had to reread the prompt several times to really grasp what it was asking. It’s an easy problem conceptually, but the wording makes it a bit tricky to follow at first.

The main idea is this: we’re given an array and we’re allowed to perform a specific operation multiple times. We can choose a number from the array — but only if there’s exactly one unique number in the array that’s greater than it. For example, in the array [1, 2, 2, 2, 2, 2], the number 1 is valid because the only number greater than it is 2. But in [1, 2, 3], the number 1 is not valid because both 2 and 3 are greater than it — meaning there’s more than one unique number greater than 1.

Once we’ve identified valid numbers, we can use them to perform the operation: turning all instances of a greater number into that number itself. So in [1, 2, 2, 2, 2], we can transform all the 2s into 1s in a single operation — because we’re allowed to perform this operation any number of times, but it can only decrease values, never increase them.

A key detail is that we can only convert numbers downward. So if we’re given a target value K, we must ensure it’s possible to reduce all the values in the array to K using this method. If K is greater than any number in the array, it’s impossible — since we can’t increase values. For example, if K = 3 and the array is [1, 2, 2, 2, 2], we can’t reach 3 because we can’t increase any of the values.

Our goal is to return the minimum number of operations required to make every value in the array equal to K. To solve this, I used a Set in JavaScript to track all the unique values greater than K since each unique value greater than K will need its own operation to be reduced to K. If at any point we encounter a value less than K, we know the task is impossible and should return -1.

TLDR

Count how many unique values in the array are greater than K — that’s the minimum number of operations needed. If the array contains any value less than K, it’s impossible to solve, so return -1.

Leetcode 763: Partition Labels

Mar 28, 2025

DevGod

Elf Vtuber

For today’s daily leetcode problem, we are given a string and need to split it into as many partitions as possible, ensuring that each letter appears in at most one partition. Additionally, the partitions must preserve the order of letters as they appear in the original string. However, we don’t need to store the actual substrings—our task is simply to return the lengths of these partitions.

Example Walkthrough Consider the string “pizzatime”. The optimal partitions would be [1,6,1,1]. Here’s why:

The letters p, m, and e each form their own single-letter partitions since they don’t share any letter with another partition.
The substring “izzati” forms a partition because it contains multiple occurrences of i and z, and we need to ensure that all instances of a letter belong to a single partition.
Since the letters must remain in order and cannot belong to multiple partitions, “zzati” is grouped together with i to form a partition of length 6.

Another example is “partitionlabels”, which results in [1,13,1]:

The letters p and s form their own separate partitions.
The middle partition, “artitionlabel”, must include all occurrences of a and l.
Since l appears late in the string, we must extend the partition to ensure all ls are included, resulting in a large middle partition.

Implementation Approach Since I was in a hurry and the constraints were small, I used JavaScript’s built-in .lastIndexOf() method to determine the last occurrence of each letter in real time. The key insight is that a letter’s last occurrence dictates where a partition must end. While this approach runs in 𝑂(n^2) time due to repeated .lastIndexOf() calls, an optimized solution could use a hashmap for 𝑂(1) lookups.

To efficiently track when to end a partition, I maintain a max variable, representing the farthest last occurrence seen so far. As I iterate through the string, I update max using:

max = Math.max(max, s.lastIndexOf(s[L]));

This ensures that if a previous letter has a later last occurrence, we extend the current partition accordingly. This is crucial for cases like “partitionlabels”, where encountering l extends the partition beyond a’s last occurrence.

I iterate through the string using a loop and keep track of partitionSize. Whenever I reach max, I add partitionSize to the result array, reset it, and continue. Instead of constructing substrings, I only track partition lengths, as that’s all the problem requires.

This greedy approach ensures an optimal partitioning of the string while keeping the implementation simple and efficient.