Problems
September 5, 2022
Course schedule IV
We will first create an adjacency list from the prerequisites. Then for each queires, we will start DFS from course 1 to course 2, if we can reach to course 2, then we append true in our result, else...
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House robber III
We will traverse the tree in DFS and take both with and without root result. Then we calculate the max from both of them and return the result.
Time Complexity: O(n)
Space...
September 5, 2022
Linked list cycle II
We can use Floyd's cycle detection algorithm to find the position. We will create a fast and slow pointer, fast pointer is twice as fast as the slow pointer. If the two pointer doesn't meet, that...
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Longest absolute file path
Use stack to represent structure of a directory where the root of a directory is at index 0. Directory level can be infer from the number of tabs. ie "" is level 0, "\n" is level 1 and so on. Start...
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Number of enclaves
We will run DFS from every bordering position of the grid and turn the 1's to 0's. Then we will just count the number of position where we still have 1's left, and return that count as result....
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Path sum II
We will use DFS along with backtracking to get the paths of every root to leaf path that adds upto the target.
Time Complexity: O(n)
Space Complexity: O(n), for...
September 5, 2022
Remove duplicate letters
We will create a hashmap to keep track of last occurance of the letters. Then we will have a visited hashset and a stack to keep track of the letters. We traverse sequentially on the string, for each...
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Restore ip addresses
We will create a decision tree based on the rule that each segment of the ip address can be between 0 and 255. Then we run our DFS to get all possible options and return the result. Time...
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Search in rotated sorted array II
To solve this problem we have to follow the folllowing steps: Calculate the mid index. Check if the mid element == target, return True else move to next step. Else if the mid element >= left. If...
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Single number II
We will have 2 elements, one will be store all the values that store all the elements that occurs once and two will store all the elements that occurs twice. Then we make sure that the values that is...
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