doyedele1

'''
    Explanation: Recursive DFS (Backtracking) Solution
        * We can't reuse a character that we've visited before
        So, let's do the backtracking brute-force as a human would do

        word = "ABCCED"
        We check if we have an A in the board since that's the first letter in the word
            Yes, we do. Then we check all possible neighbors of A and check if we have a B

        TC: O(rc * dfs) = O(rc * 4 ^ len(word))

doyedele1

'''
    Explanation:    
        [[1,3], [8, 10], [15, 18], [2,6]]
        1-----3
                        8------10
                                                15--------18
                    2------6
        - Sort intervals based on the start times
        [[1,3], [2,6], [8, 10], [15, 18]]
        - Add the first interval to the res nested array

doyedele1

'''
    Explanation I: Naive Solution
        - Enumerate the string and check all the substrings. Check the longest substring that have no repeating characters
        - Update the result to the count of the longest substring with no repeating characters
            abcabcbb
            - From a, we check the substring starting from a
            - From b, we check the substring starting from b
        
            abcabcbb
            abca --> has repeating characters. We do not need to check the next substring, which is abcab because we already have duplicates from the previous iteration. 

doyedele1

class Solution:
    def shiftingLetters(self, s: str, shifts: List[List[int]]) -> str:
        n = len(s)
        arr = [0] * (n + 1)

        for shift in shifts:
            start, end, direction = shift[0], shift[1], shift[2]
            if direction == 1:
                arr[start] += 1
                arr[end + 1] -= 1