Problems
October 13, 2022
Find positive integer solution for a given equation
Given a function f(x, y) and a value z, return all positive integer pairs x and y where f(x,y) == z. Rephrase the problem...
October 12, 2022
Determine whether matrix can be obtained by rotation
We will rotate the matrix 4 times and check if it is equal to the target matrix. Time complexity: O(n^2) Space complexity: O(n^2)
ReadOctober 12, 2022
Largest perimeter triangle
We will sort the array in descending order and check if the sum of the first three numbers is greater than the sum of the next three numbers. If it is, we return the sum of the first three numbers....
ReadOctober 11, 2022
Number of recent calls
We will use a queue to store the recent calls. We will remove the calls that are not in the range of 3000ms. Then we will return the size of the queue. Time complexity: O(n) Space complexity: O(n)
ReadOctober 11, 2022
Increasing triplet subsequence
We will keep track of the smallest and second smallest number in the array. If we find a number that is greater than both, we return true. Otherwise, we return false. Time Complexity:...
ReadOctober 10, 2022
Break a palindrome
We can check the half of the string and if a character is not a, we can replace it with a. If all characters are a, we can replace the last character with b. If the string is a single character, we...
ReadOctober 9, 2022
Find bottom left tree value
We will traverse the tree with BFS and assign the first node of the each level in our result. At the end of the traversal, we will return the result, which will store the left most value of the last...
ReadOctober 9, 2022
Two sum IV input is a bst
This is very similar to two sum problem. But rather than traversing an array, we will traverse a tree. We will be using DFS to traverse the tree and along with the way, we will store the difference...
ReadOctober 9, 2022
Flip equivalent binary trees
We will start traversing both tree with DFS and check whether the left subtree and right subtree are equal for both tree or left subtree is equal to right subtree and vice-versa both tree. Time...
ReadOctober 9, 2022
Frequency of the most frequent element
First we will sort the array. Then it turns into a sliding window problem, the key is to find out the valid condition k + sum >= size * max. For every new element nums[r] to the sliding window, add...
Read