doyedele1

'''
  Explanation:
    - Intuitvely, we can find the possible areas for each pair of heights
    - For each pair, we check the area by multiplying the width to the height without spill over. We need to consider the minimum height between the pair.
    - However, can we reduce the number of times we find the possible areas for every pair of the heights which is caused by the minimum height in each pair?
      - Yes, we can. We can have two pointers at both endpoints and perform the area of the ractangle. 
      - When the height is minimum for a left pointer, we can move to the pointer to the right and when the height is minimum for a right pointer, we move the pointer to the left to the right. This is done to look for the next possible maximum height. 
    - TC: O(n), SC: O(1)
'''

doyedele1

'''
    Explanation:
        - n = 19 --> 82 --> 68 --> 100 --> 1
        n // 10 = firstDigit
        n % 10 = lastDigit
        
        - n = 2 --> 4 --> 16 --> 37 --> 58 --> 89 --> 145 --> 42 --> 20 --> 4
        Since we've seen 4 before, then we know that the number is not happy.
        
        Time Complexity: O(log n) - finding the sum square for a given number has a cost of O(logn) because we are processing each digit in the number, and the number of digits in a number is given by (log n): For n > 243, O(243 * 3 + log n + log log n + ....) = O(log n)

doyedele1

'''
    Explanation:
        - Note: Knowing which representation is the correct one for a particular number can be difficult. There are many possible ways of representing the number 120 for instance.
        
        - So if num = 9, X and I are needed. Rule of roman numeral goes from largest to smallest. However, there are some special rules where the symbol with a smaller value can go before the symbol with a larger value.
        
        - If there are no special rules, when num = 1500,
            - We can divide 1500/M. i.e. 1500//1000 = 1. 1 tells us how many Ms in our final result.
            - Then 1500 % 1000 = 500
            - 500 // 500 = 1. 1D in the final result

doyedele1

class Solution:
    def medianSlidingWindow(self, nums: List[int], k: int) -> List[float]:
         # Create an empty array to store your answer
        res = []
        
        # Get the length of the nums array
        n = len(nums)
        
        # Slice the input array to get data from index 0 to index k
        win_arr = nums[:k-1]