steph-crown

Question

How I Solved It?

This was a classic Geometry problem. All I needed to do was in three steps:

• Initialize a maxArea value
• Compare every rectangle with one another (a nested loop)
• For each rectangle pair, Check if they intersect (the minimum of the rights minus the maximum of the lefts > 0)
• If intersects, find the largest possible square (square of the smaller side of the intersection boundary)
• Update maxArea with the maximum of this value in previous step and maxArea

steph-crown

Question

How I Solved It

• After some thinking, it became clear that my focus shouldn't be on the grid, but on those bars hBars and vBars that can be removed from the grid
• I needed to find the maximum number of consecutive bars that can be removed from hBars, such that the same number can be removed from vBars.
• It doesn't mater the position of the bars in the grid. As long as they're consecutive and the same number of horizontal and vertical bars is removed, you'll definitely form a square
• So I got started. I created a reusable function maxConsecutives that finds the max number of consecutive bars in a bars array, ensuring we can't remove the bars at the edges cos they'll be useless to our goal of finding bordered squares
• Then I called this function with hBars and vBars, picked the minimum of both return values, added by 1 and squared.
• I added by 1 because whe

steph-crown

Question

This question is a confirmation that I need to take a separate, dedicated short course on dynamic programming.

With help from solutions on YouTube though, I was able to get this:

function maxDotProduct(nums1: number[], nums2: number[]): number {
    const m = nums1.length;
 const n = nums2.length;

steph-crown

Question

The Hint helped me with this one.

How I Solved It?

• I calculated the sum for each node, and using what was stated in the hint, for each node, I multiplied its sum with the sum of the complement of the node. Then I found the maximum of all such products.
• The dilemma was now how to get the totals of each node in an array. So I created a sumInOrder function to that, within the root function so it could update the totals array in the main function.

steph-crown

Question

This problem was more about understanding the binary tree data structure than it was about thinking of some algorithm to use. I used TypeScript for this one.

How I Solved It

• The idea was simple. Initialize a maxSum to -Infinity
• Then, find the sum, levelSum for each levels and update maxSum to this levelSum if levelSum is greater
• Finally return maxSum
• The only thing I had to think of was how to traverse the tree in a way that gave me all the nodes at a level in a single loop cycle so I could add them.
• I used breath-first search (BFS) to accomplish this

steph-crown

Question

This is one of those problems where the question is C. Ronaldo (or L. Messi, depending on your cup of tea), and the solution is ... J. Lingard.

I knew this was a greedy problem and I just needed to find a pattern to be able to make a valid assumption about the array without having to do individual checks between adjacent elements. It took me a while, but a comment from consistentNotMotivated lit all the lightbulbs in my head. See comment below:

How I Solved It

Well, it's clear from the comment referenced above. In more straightforward terms,

steph-crown

Question

I admit, I checked the hint before getting started, and I saw the hint that You only need to loop to the square root of a number to find its divisors. To be fair, I think I would have figured that out myself after observing the patterns (winks twice).

How I Solved It

• After confirming the square root theory, the solution became so obvious and I wondered if there's was a hidden twist that could have made this question MEDIUM level rather than EASY.
• Anyways, I solved by looping throughn the array, and for each number, I start from 1 up to the its square root to find the divisors.
• Then if the divisor count is more than 4, I add them to the sum
• Finally, I return the sum
• And it worked, wow. So no hidden twists. Boring.

steph-crown

Problem

Did You Know :)?

• This was my easiest of the year so far.

How I Solved It

• It was immediately clear what I needed to do
• Loop through, store the count in a hashmap, and once any count crossed 1, return true, else return false at the end of the loop
• Then I remembered I had seen something similar before, the day before. See here. And I remembered that there was an even sleeker way to do this with a hashset since we don't care about the counts and we just want to know the item that occurs more than once.
• With a hashset, i just need to check if the item already exists before, then return true

steph-crown

Problem

This was one of those problems that you need to take your time to understand and find a pattern

How I Solved It

• I started by trying to find the number of possible combinations for each row. I quickly calculated the first row as having 12 possible combinations, that is 3 uniques in first cell, multiplied by 2 uniques in second cell and 2 uniques in third cell. Second and third cells have 2 uniques each because they mustn't take the color directly to their left. This means there are 2 left for each
• After figuring this out, finding a general pattern for subsequent rows proved more difficult than I envisaged, due to the dependence of each cell on both the pattern in the row above it and the selected color in the cell to its left
• After about 30 minutes of coming close to figuring something out, I conceded and checked YouTube to see if I could find an explanation. Luckily I found

steph-crown

Problem

How I Solved It

• This was very straightforward. It was immediately obvious that I needed to just find the number that occurs most in the list. I just had to use a hashmap to store counts and get the largest
• I observed even further that I didn't need to wait to count everything in the list. As I add the counts to the map, once I encountered a number with count of n/2, that's our guy.
• Then something even more interesting clicked. The question says there are 2 * n numbers and one number occurs n times. This means there is only space for n more numbers. But then it says there are n + 1 unique numbers. This means that, beyond the number that occurs n times, there are n more numbers in the list. It follows therefore that, these other n numbers occur exactly one time each. Therefore if we can find a number occuring more than once, that's our guy.
• I didn't need to