Problems
October 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...
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 8, 2022
Find all anagrams in a string
We will use counter to calculate the number of characters of string p. Then we take a sliding window of lenght p, then compare the character count with the character count of p. If we found a match,...
ReadOctober 8, 2022
Unique length 3 palindromic subsequences
For each palindromes in format of "aba", we enumerate the character on two side. We find its first occurrence and its last occurrence, all the characters in the middle are the candidate for the...
ReadOctober 8, 2022
Find the index of the first occurrence in a string
We will create a substring of length of the needle and compare it with the needle, if we find a match, we return the index as result, otherwise return -1.
Time Complexity: O(n)
Space...
October 8, 2022
Remove element
We will take a pointer at the beginning of the array, then iterate over the whole array. If the value doesn't match the given values, we assing it to the pointer's position of the array and then...
ReadOctober 7, 2022
Search insert position
This one is the classic binary search problem. We will strat looking for the target, if we find the target then we insert on that position. If we don't find the target, then we insert at the last low...
ReadOctober 7, 2022
Single threaded cpu
First we sort the tasks according to start time, remember to keep a reference to the original task index. Set the current time to the first start time in the task list. Push all tasks whose start...
ReadOctober 7, 2022
Reverse linked list II
First if the left and right position of the list is same, we can just return the list. Otherwise, we will take a pointer, traverse till the left position, then reverse the list in place till the...
ReadOctober 6, 2022
Length of last word
We will first start from the end and move the end pointer till we find a character. Then we take another pointer beginning from the end character and move towards the beginning of the string until we...
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