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Leetcode 58: Length Of Last Word

Jul 6, 2025

Question

In this Leetcode question, we are given a string representing a sentence. Each word is separated by the ” ” space character. Our goal is to return the length of the last word in the string. An important observation to note is that our given string could have white space characters before the start, as well as after the end. Additionally, more than one white space character could also separate each word in the string.

Solving

To solve this Leetcode problem, I start by using the built-in JS .trim() method. This trim method is able to remove all white space at the beginning and end of the string. By using trim on the given string, I can proceed with my JS methods knowing the end of the string is just the last word, and no additional space characters.

Next I use the string .split(” ”) method to split in-between space characters. This method transforms our trimmed string into an array of strings, separated by the space characters. As a side note, this method does leave some empty elements, but that is alright since the last element will just be our full last word. Since we are only concerned about the last element of this split array, any empty strings present in this string array can just be ignored. There is a way to use regular expressions to more accurately split the given string to avoid empty strings. However since I was more interested in code brevity, I instead used just a ” ” single space character instead.

After splitting our trimmed string into an array of strings, we can now pop() to return the last element. The pop method is a built-in JS array method, that removes and returns the last word of the split string array. Another way to retrieve the last element would have been with [s.trim().split(” “).length - 1]. While this is the slightly more time efficient way, the .pop() method just took less characters. Given the small constraints regarding the given string’s length, I preferred the least amount of code vs the most time efficient code.

Lastly, now that we have trimmed the original string, split it into an array of strings and popped the last element, we can now just return the .length property to answer the question.

TLDR:

I used built-in Javascript array and string methods to return the length of the last word in the given string.

Leetcode 55: Jump Game

Jul 3, 2025

DevGod

Elf Vtuber

Question

In this Leetcode question, we are given an array of numbers. We start at the first index of the array, and our goal is to reach the end of the array. Each value of the array represents the maximum jump distance to the right. For example, if the first index’s value is 5, you could jump anywhere between 1 to 5 indexs ahead. If the index’s value is 0, then we are NOT allowed to jump anywhere beyond that spot, effectively trapping us. It is important to note that we are never given negative values in the given array. This means that leftward jumps are never allowed, given this constraint of the array Numbers.

Our goal is to determine if we can reach the end using any series of jumps. We return true or false depending on the reachability of the array’s end.

Solving

To solve this Leetcode problem, I used a greedy approach. I used a for loop to iterate left to right through the array. Whilst I traverse the array, the variable named maxJump records the furthest reachable index. In-between iterations, I check whether my current position is ‘impossible’ to reach. If an impossible index is detected, we return false and terminate early. If we go throughout the entire array, from left to right, without hitting an impossible to reach spot, we return true.

TLDR:

Used a Greedy approach to traverse the array, determining if the array’s end is reachable using any series of jumps. I solved this problem by iterating throughout the array, searching for impossible to reach spots.

Leetcode 121: Best Time to Buy and Sell Stock

Jul 2, 2025

DevGod

Elf Vtuber

Question

For this Leetcode problem, we are given an array of numbers representing the fluctuating values of an imaginary stock. Each index of the given array represents a new day. Using this array, we must determine the maximum possible profit that can be achieved. The profit is determined by choosing ONE day to buy the stock, and one future day to sell the stock. In other words, a key constraint of this specific ‘buying stocks’ problem is the fact we are limited to only one sell and one buy operation.

Solving

To solve this problem, we can use Kadane’s algorithm as a template to find the maximum possible profit. This algorithm is used for many problems involving finding the max subarray sum. In our case, we can use it to find the highest profit peak for buying and selling the stock.

We use a for loop to keep track of our stock’s profits for each given day. To keep track of our current possible profit, each iteration we add the difference between the current Ith day’s value minus the previous day’s price. We use a Math.max(0,curProfit) to greedily ensure our current potential profit never dips below zero. Lastly, we use a variable maxProfit to keep track of the largest potential profit we have seen throughout the array.

TLDR:

I used Kadane’s algorithm as a template to create a maxProfit function, for finding the maximum possible profit achieved by buying and selling a stock.

Leetcode 169: Majority Element

Jun 14, 2025

DevGod

Elf Vtuber

Shout out Solaris428 for help

Question

This Leetcode problem asks us to find the “Majority Element” from a given array of numbers. The majority element is the element of the array that appears > N/2 times. Basically, the majority element is the only absolute mode in this given array. Leetcode will only give you input arrays guaranteed to have one majority element. We must find and return the majority element to solve this problem.

Follow up

We must solve this question using linear O(n) time complexity and O(1) constant space complexity. In other words, any solutions that use sorting algorithms would break this follow up’s time complexity rule. This is because sorting algorithms, Wether implemented or using language’s built in sorting methods, will result in at best nlogn time complexity. Secondly, we can not use data structures such as frequency Arrays or Hash tables, as these would be n space complexity at best.

Solving

Since there is only at most one majority element present in our given array, we can use a running frequency of elements. This is because the majority element appears more often then all other values of the array. In other words, if we were to take the frequency of the majority element, subtracted by the sum of all the other frequencies of the non majority elements, the total would be more than zero. This only occurs for the majority element, as that is going to be the only element in the array that appears more often then all of the other elements.

We start by initializing a score variable used for tracking the current frequency, as well as the current “potential answer” we are investigating. We use a for loop to iterate through all of the numbers of the given array. If the num of the for loop is equal to the potential answer we are currently investigating, then we increment the current frequency score, else we decrease the score. If our frequency score ever reaches zero, then we change our current ans we are investigating to the current num of the for loop. This is because if our frequency is zero, at that point in time our current potential answer is <= the frequency of other values.

In other words, whenever our currently investigated potential ans running frequency is equal to zero, we must start investigating a new value at that time. Continuing with this pattern of investigating potential majority elements as they become suspects, and then once they get cleared we investigate the next value.

Lastly, once we have reached the end of our for loop, then our answer that is left should be the majority element we are looking for, so we return the ans. This solution fits our follow up, as we use constant space, and we use liner time complexity.

TLDR:

We use an approach similar to a running prefix sum, where we keep track of the majority element by returning the element who’s frequency at the end of array traversal is greater then zero.

Leetcode 26: Remove Duplicates from Sorted Array

Jun 14, 2025

DevGod

Elf Vtuber

Question

In this Leetcode problem, we are given a pre-sorted, strictly increasing array of numbers and must remove adjacent duplicate values in-place. However, Leetcode includes a custom judge for this problem that verifies whether the array is modified in-place. For example, we do NOT want to initialize a new array to solve this problem, we want to solve this problem in-place using the original array. Another requirement of the custom judge for this question is to return K. The value K represents the amount of unique values we have after removing adjacent duplicates from the given array.

Solving

To solve this problem, I used a two pointer approach to modify the array in-place and keep track of unique values, K. I implement a for-loop that moves my right pointer R across the array. Every time nums[R] and nums[R+1] don’t equal each other, I have found a new unique value. Anytime my left pointer array element updates, I move the L one space. The right pointer, R and R+1 continue to move forwards until they reach the end of the given array.

In other words, my right pointer R is used in a sliding window style traversal to check all adjacent duplicate values in the array. Meanwhile, my left pointer L is strictly used to update positions in the left of the array, as I find unique values. Additionally, my left pointer L also keeps track of the amount of unique values that I have encountered in the given array. This is because every time I modify the position of left pointer L, it increments by one, which effectively keeps a running score of every time I updated the left section of the array.

Lastly, although the two-pointer solution does not remove the remaining “junk” values in the right section of the array, the custom judge only checks the first K elements for correctness.

TLDR

I used a two pointer approach to modify the given array in-place to hold the unique values in the left section of the array, satisfying the custom Leetcode judge requirements.