amarjitdhillon

kubectl set image pod podname nginx=nginx:1.15-alpine

kubectl edit pod podName

# Create the nginx pod with version 1.17.4 and expose it on port 80
kubectl run nginx --image=nginx:1.17.4 --restart=Never --port=
kubectl run nginx-pod --image=nginx:alpine # will create a pod with name nginx-pod

# add the command in the pod
command: ['sleep', '5000']

amarjitdhillon

class Solution:
    def maxCount(self, m: int, n: int, ops: List[List[int]]) -> int:
        for r,c in ops:
            # expand and shrink the matrix size (bottom-right boundry of matrix) based on the current operation
            m = min(m,r)
            n = min(c,n)
        # total number of elements will be multiplication of bottom and right boundry asssming left and top boundry is 0,0
        return m*n
        

amarjitdhillon

class Solution:
    def getModifiedArray(self, length: int, updates: List[List[int]]) -> List[int]:
        # Initialize an array of length+1 with all values as 0
        arr = [0] * (length+1)
        
        # Iterate over all the given ranges
        for start, end, value in updates:
            '''
            Just update the start index with +val and end+1 with -val as we need to see the effect till end
            We will use the prefix sum to retrive the actual sum, later

amarjitdhillon

class Solution:
    def coinChange(self, coins: List[int], amount: int) -> int:
        # Create a dp array for 0 to amount (inclusive) with the max values
        dp = [float('inf')] * (amount+1)
        
        # Initialize the value of the dp array
        dp[0] = 0
        
        # Let's compute dp array for all values from 1 to amount
        for i in range(1, amount+1):

amarjitdhillon

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        num_rows,num_cols,num_islands = len(grid), len(grid[0]),0
        visited = set()
        
        def findIsland(row,col):
            # add this cell to the visited set
            visited.add((row,col))
            
            # add left, right, top and down neighbours of this cell respectively. Here the first element is row id and second one is the column id

amarjitdhillon

class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        
        # sort the intervals by the start value, in case they are not sorted by default
        intervals.sort(key = lambda x: x[0])
        
        # edge case. If num elements is one, then return the same list
        if len(intervals) == 1:
            return intervals
        

amarjitdhillon

class Solution:
    def kthFactor(self, n: int, k: int) -> int:
        
        # counter represent the i^th factor of the number found so far
        counter = 0
        
        # go over the number from 1 to n as we need to divide the n by this number to find if that number is a factor
        for i in range(1, n+1):
                        
            if n%i == 0:

amarjitdhillon

class Solution:
    def mostVisitedPattern(self, username: List[str], timestamp: List[int], website: List[str]) -> List[str]:

        # zip the time, user and website to one list
        tuw = list(zip(timestamp,username,website))

        # we need to sort them by time --> username --> website
        sorted_tuw = sorted(tuw)

        # We will polulate a user history hashmap of various pages visited. The hashmap is of type {user: [pages]}

amarjitdhillon

class ParkingSystem:
    # initialize the parking object
    def __init__(self, big: int, medium: int, small: int):
        # to hold various types of parking we are using a dict in constructor
        self.lot = {}  
        
        # add the parking counters in this lot dict
        self.lot[1] = big
        self.lot[2] = medium
        self.lot[3] = small

amarjitdhillon

class Solution:
    def wordPattern(self, pattern: str, s: str) -> bool:
        hm = {}
        
        # we need to split the s over the space(" ")
        s = s.split(" ")   
        
        # if the lengths of pattern and s are not same then return False
        if len(s) != len(pattern):
            return False