practice-algos-python.txt

# Coin change ============================================================================

    # Summary: 
    #       - initialize a list of size amount + 1 to store the minimum qty amounts
    #       - for every coin, for every amount (represented by each index of list), store the minimum in that index between what's already there and list[that index value - coin] + 1
    #       - each index in the list will represent the least number of coins necessary to reach that index value.
    #       - return list[amount]
    def coinChange(self, coins: [int], amount: int) -> int:
        # function header describes variables + their type. (var : type)
        # -> int indicates return type
        
        # Bottom-up approach, recurrence relation: finding the minimum coins
        # needed for every value up till the final amount.
        
        # key points:
        # - the minimum qty of coins needed for a given (amount value) is equal to the minimum qty of coins needed for (amount value - C) + 1, where C is the last coin added to equal the amount value. This makes sense since you are taking off one coin to the amount value, this means that if you add this coin, it will take 1 more total qty of coins.       
        # - thus, we need to find the minimum qty of coins for each incremental amount leading up to the amount value we are looking for by comparing the minimum qty of coins that could be obtained from each C in the list.
        
        # more intuition: say your amount = 11, and coins = [1,2,5]. Then if your "last coin" is 1, then the remaining amount occupied by the other X number of coins is 10. You can then check what the minimum amount of coins is needed to obtain 10, etc. until the smallest coin value, and then repeat this process for coin '2' and '5'. (In the code, we do this the other way around, starting from the smallest coin value)
        
###################################### CODE BELOW ######################################

        # creating a list to store the number of coins needed for each incremental amount
        # float('inf') is a python code to indicate big value. The list will be of size amount+1 filled with 'inf'. The 'inf' will be replaced with smallest qty needed for each amount value. The amount value is indicated by the index of the elements in the list. It goes up to amount + 1 since we must account for the 0 index (amount+1's index nb = amount)
        dp = [float('inf')] * (amount + 1)
        dp[0] = 0
        
        # for each coin as last coin picked, find the minimum qty of coins needed for each amount value (which is indicated by index)
        # for loop over coins first so that we don't have to deal with negatives since x>=coin is guaranteed in the second loop.
        for coin in coins: 
            for x in range(coin, amount + 1):
                dp[x] = min(dp[x], dp[x - coin] + 1)
                
        return dp[amount] if dp[amount] != float('inf') else -1


# Linked List Cycle: determining if there is a cycle

# slower: change the value in each node, if re-encounter it, return true
def hasCycle(self, head: ListNode) -> bool:
    if head is None:
        return False
    temp = header           
    while temp.next is not None:
        if temp.val == float('inf'):
            return True
        temp.val = float('inf')
        temp = temp.next
    return False

# faster: use two pointers to race each other, one jumps faster so will loop around if cycle exists.
def hasCycle(self, head: ListNode) -> bool:
    if head is None:
        return False
    
    # two pointers
    ptr1 = head
    ptr2 = head.next
    
    while ptr2.next is not None and ptr2.next.next is not None:
        if ptr1 == ptr2:
            return True
        ptr1 = ptr1.next
        ptr2 = ptr2.next.next
    return False

# REVERSE LINKED LIST =====================================
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

# version 1: changing values not actual nodes, using stack O(n)
def reverseList(self, head: ListNode) -> ListNode:
    if head is None:
        return head
    
    stack = []
    temp = head
    temp2 = head
    
    while temp is not None:
        stack.append(temp.val)
        temp = temp.next
    
    while head is not None:
        head.val = stack.pop()
        head = head.next
    
    return temp2

# version 2 iterative: changing node positions. O(n) time and O(1) space
def reverseList(self, head: ListNode) -> ListNode:
    # Summary: retrieve each element in the list one by one from left to right and link them to each other in a 'separate list' to invert 
    #           the list nodes order.
    
    # -keep track of 'prev' element which always points to the head of the reversed list
    # -almost as if we are making a separate list. the 'prev' starts at null since it's the last element of the inverted list. 
    # -we update 'prev' each time by linking to it the next element in the list, then setting 'prev' to this new element as it's now 
    #  the head of inverted list.
    # -while doing this, we use 'temp' to store the reference to the rest of the list, and curr = temp moves the pointers forward.
    
    curr = head
    prev = None
    while curr is not None:
        temp = curr.next
        curr.next = prev
        prev = curr
        curr = temp # -> curr = curr.next
    return prev

# version 2 recursive: start from the end and link each node to it's previous one. think of it as the nodes at the end are already
#                      reversed so now how do you reverse the n-1 linkage.
def reverseList(self, head: ListNode) -> ListNode:
    if head is None or head.next is None:
        return head
    
    newHead = self.reverseList(head.next) # don't forget "self" since func not defined yet
    head.next.next = head # linking next node to previous! pointer manipulation
    head.next = None
    return newHead


# MERGE TWO SORTED LIST ============================================================

# similar to merge sort: compare the elements from the two list and link the smallest to the new linked list node.
#  -> must advance the pointers in the first list, second list and also resulting list.
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
head = ListNode() # new linked list (head will be placeholder, actually starts at head.next)
temp = head 

# iterate until one of the two list is done
while l1 is not None and l2 is not None:
    if l1.val < l2.val:
        temp.next = l1 # ok to set directly because l1 already moved from previous iteration
        l1 = l1.next # increment 1st list pointer
    else:
        temp.next = l2
        l2 = l2.next # increment 2nd list pointer
    temp = temp.next #increment result list pointer

# attach the remainder list
temp.next = l1 if l1 is not None else l2
return head.next

# BUY AND SELL STOCK ===================================================================

# brute force: O(n^2)
# iterate over array, for each element, check the other ones
# if the next element is larger than element AND larger than previous, store this value
# at the end of each iteration, compute maximum profit for that var, (0 if negative)
# return max profit
def maxProfit(self, prices: List[int]) -> int:
    maxProfit = 0
    max = 0
    # for each value, update max 'sell' value, then compare profit with previous
    for i in range(0, len(prices)):
        for j in range(i+1, len(prices)):
            if prices[j] > prices[i] and prices[j] > max:
                max = prices[j]
        # if profit greater than previous, update
        if maxProfit < max - prices[i]:
            maxProfit = max - prices[i]
        max = 0
        
    return maxProfit if maxProfit > 0 else 0

# One-Pass: Keep track of minimum buy value & max profit, equivalent
#           to the smallest valley followed by the max peak. O(n)

# key: sell must be after buy, so previous elements don't matter if they didn't work

# Since only the sell must be after buy, we can directly iterate over and store the 
# minimum buy value. If the value after it is smaller, than we can directly update.
# If the value after is greater, then we can directly compute new maxProfit and see if
# it's greater than the previous one. (If a given value Y has not provided us with
# a max profit up until the value X, and X is smaller than Y, then we can simply give
# up on Y and check for max profit with X for the rest of the array.)
def maxProfit(self, prices: List[int]) -> int:
    maxProfit = 0
    minValue = float('inf')
    for price in prices:
        if price < minValue:
            minValue = price
        elif maxProfit < price - minValue:
            maxProfit = price - minValue
    return maxProfit

# RANGE SUM QUERY - IMMUTABLE ==================================================================
# function to get sum of values in given range repetitively.
class NumArray:
    
    # If the sumRange function is used repetitively on similar ranges, we would have to
    # re-compute same values multiple times. Instead, we can use prefix sum to cache the
    # sums up till a given index, and simply subtract to return the sum in a given range.
    def __init__(self, nums: List[int]):
        self.numArray = nums
        # compute prefix sum
        for i in range(1, len(nums)):
            self.numArray[i] = nums[i] + nums[i-1]

    def sumRange(self, i: int, j: int) -> int:
        if i == 0:
            return self.numArray[j]
        else:
            return self.numArray[j] - self.numArray[i-1]

# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(i,j)

# PALINDROME ===========================================

# fastest
def isPalindrome(self, s: str) -> bool:
    if s == "":
        return True
    st = "".join(char for char in s if char.isalnum()).lower()
    # indexes the string, -1 indicates reverse [start:end:step]
    # st[::-1] returns the reversed string. fastest because no external 
    # function calls, directly uses slicing which is highly optimized.
    return st == st[::-1]

# slower
def isPalindrome(self, s: str) -> bool:
    if s == "":
        return True
    
    # strip the string of non-alpha chars
    strg = re.sub("[^a-z0-9]", "", s.lower())
        if strg[i] != strg[len(strg)-1-i]:
            return False
    return True

# MERGE INTERVALS ================================================

# need to create a separate 'merged' list to return.

# sort the intervals by their first element.
# once sorted, the intervals which need to be merged will be contiguous.
# then we can simply check if the next interval's first element is bigger or not.
# if bigger, simply add to list, if smaller or equal, merge!
def merge(self, intervals: List[List[int]]) -> List[List[int]]:
    # sort the intervals by 1st element:
    intervals.sort(key=lambda x: x[0])
    
    # merged intervals result list
    merged = [intervals[0]]
    
    # check if next interval's starting element is larger or not
    for interval in intervals:
        # if larger, no need to merge, so simply add to resulting list
        if interval[0] > merged[-1][1]:
            merged.append(interval)
        # if smaller or equal, then merge with the one already in resulting list
        else:
            merged[-1][1] = max(merged[-1][1], interval[1])
    return merged

# MAXIMUM SUBARRAY ==================================================

def maxSubArray(self, nums: List[int]) -> int:
# sliding window (find negative nb)
# if positive = keep adding
# 3 scenarios:
#   1. max subarray sum is the value currently stored in maxSum
#   2. max subarray includes the new found negative nb (counter += nb)
#   3. max subarray is the one which comes after the new found negative nb (counter = 0)

# if negative = reset a counter to 0, since there may be a bigger subarray but only
# after the negative nb OR with the negative number
# keep track of the max sum so far as well as the sum from after the negative nb.
maxSum, counter = nums[0], 0

for nb in nums:
    if counter < 0:
        counter = 0
    counter += nb
    maxSum = max(maxSum, counter)
return maxSum

# CISCO ASSESSMENT =====================================================

# Q1:
matrix = []

size = int(raw_input())
for i in range(0, size):
    # this splits a string into list of nbs, then converts all nbs into integers.
    nbs = map(int, str(raw_input()).split())
    matrix.append(nbs)

# this rotates matrix by 90 degrees clockwise
matrix = list(list(x)[::-1] for x in zip(*matrix))

for i in range(0, size):
    for j in range(0, size):
        print matrix[i][j],
    print 

# Q2: 3D MATRIX IN PYTHON
size = int(input())
# row starting with white and black
wRow = []
bRow = []
for i in range(0,size):
    wRow.append("W" if i % 2 == 0 else "B")
    bRow.append("W" if (i+1) % 2 == 0 else "B")

# board
board = []
count = 0
for i in range(0, size):
    if count % 2 == 0:
        board.append(wRow)
    else:
        board.append(bRow)
    count += 1

# Print
for i in range(0, len(board)):
    for j in range(0, len(board[i])):
        print(board[i][j], end=" ")
    print()

# FLIPPING MATRIX

def rotate_90_degree_anticlckwise(matrix):
    new_matrix = []
    for i in range(len(matrix[0]), 0, -1):
        new_matrix.append(list(map(lambda x: x[i-1], matrix)))

    return new_matrix


def rotate_90_degree_clckwise(matrix):
    new_matrix = []
    for i in range(len(matrix[0])):
        li = list(map(lambda x: x[i], matrix))
        li.reverse()
        new_matrix.append(li)

    return new_matrix

def rotate_90_degree_clckwise_by_zip_method(matrix):
    return list(list(x)[::-1] for x in zip(*matrix))

# https://treyhunner.com/2018/10/asterisks-in-python-what-they-are-and-how-to-use-them/

# Check if nb is odd or even BITWISE OPERATOR

if nb & 1:
    return 'odd'
else:
    return 'even'


# email regex
if re.search("^[a-zA-Z0-9_.]+@[a-z-]+\.com|net|org$", word):
    print("true")
else:
    print("false")


def numDuplicates(name, price, weight):
    # Write your code here
    
    dict = {}
    count = 0
    for (n, p, w) in zip(name, price, weight):
        if (n, p, w) in dict:
            count += 1
        dict[(n, p, w)] = 1
    return count

# TRAPPING RAIN WATER =======================================================
 """
- Squarepoint Capital

[Brute force O(n^2)]
For each element in list:
- find the max left and right pillar height
- take the minimum of the two
- from that minimum, minus the height of the current spot. 
This will give you the nb of raindrops in the current spot.
Don't forget to discard negatives when incrementing the total output.
"""
def trap(self, height: List[int]) -> int:
    if not height:
        return 0
    counter = 0
    max_left = height[0]
    max_right = height[len(height)-1]
    
    # iterate over each spot
    for i in range(1, len(height)-1):
        # check max left pillar height
        for l in range(1, i):
            if height[l] > max_left:
                max_left = height[l]
        # check max right pillar height
        for r in range(i+1, len(height)-1):
            if height[r] > max_right:
                max_right = height[r]
        # get current spot raindrop nb
        add = min(max_left, max_right) - height[i]
        if add > 0:
            counter += add
        max_left = height[0]
        max_right = height[len(height)-1]
    return counter

# Make The String Great ==================================
""" 
- also got brute force method with recursion 2 pass on leetcode.
- use stack instead to check last index element.
"""

def makeGood(self, s: str) -> str:
    stack = []

    for ch in s:
        if stack and stack[-1] == ch.swapcase():
            stack.pop()
        else:
            stack.append(ch)
    return "".join(stack)

# MAX DEPTH OF BINARY TREE

def maxDepth(self, root: TreeNode) -> int:
    if root is None:
        return 0
    
    left = self.maxDepth(root.left) + 1
    right = self.maxDepth(root.right) + 1
    return max(left,right)

# MISSING NUMBER IN ARRAY

def missingNumber(self, nums: List[int]) -> int:
    sum = ((len(nums)-1)*((len(nums)-1)+1))/2
    count = 0
    for i in nums:
        count += i
    return int(count - sum + 1)

# VALIDATE BINARY SEARCH TREE

def isValidBST(self, root: TreeNode) -> bool:
    if root is None:
        return True
    bst = []
    self.inorder(root, bst)
    return all(bst[i] > bst[i-1] for i in range(1, len(bst)))
    
def inorder(self, root, bst):
    if root is None:
        return
    self.inorder(root.left, bst)
    bst.append(root.val)
    self.inorder(root.right, bst)

# LOWEST COMMON ANCESTOR BINARY SEARCH TREE

def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
    if root is None:
        return root
    if p.val < root.val and q.val < root.val:
        return self.lowestCommonAncestor(root.left, p, q)
    if p.val > root.val and q.val > root.val:
        return self.lowestCommonAncestor(root.right, p, q)
    return root

# 211 DESIGN ADD AND SEARCH WORDS DATA STRUCTURE
"""
You used a dictionary and a set. The optimal solution requires a Trie.

Trie: tree with each node containing a hashmap<Character, Node> and a boolean. Each node links to another node until word is finished.
"""
class WordDictionary:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        # Used dictionary to append: (length of word, [list of word of this length])
        self.data = {}
        # Used set to checked if word already added.
        self._set = set()
        
    # Time complexity: O(1)
    def addWord(self, word: str) -> None:
        if word in self._set or not word.isalpha() or any(x.isupper() for x in word):
            return
        if len(word) not in self.data: 
            self.data[len(word)] = [word]
        else:
            self.data[len(word)].append(word)
        self._set.add(word)
        
    # Time complexity: O(d*w) -> where d is number of dots in word and w number of words of same length.
    # -> Looping through word of same length
    def search(self, word: str) -> bool:
        if any(x.isupper() for x in word) or len(word) not in self.data:
            return False
        if "." in word:
            word_list = self.data[len(word)]
            for w in word_list:
                if re.search("^" + word + "$", w):
                    return True
        else:
            return True if word in self._set else False
        return False

# SYMMETRIC TREE

def isSymmetric(self, root: TreeNode) -> bool:
    if root is None:
        return True
    
    return self.isMirror(root.left, root.right)

def isMirror(self, left, right):
    if left is None and right is None:
        return True
    if left is None or right is None:
        return False
    if left.val != right.val:
        return False
    
    return self.isMirror(left.left, right.right) and self.isMirror(left.right, right.left) 

# kTH SMALLEST ELEMENT IN A BST

def kthSmallest(self, root: TreeNode, k: int) -> int:
    if root is None:
        return -1

    def inorder(root, arr):
        if root is None:
            return
        inorder(root.left, arr)
        arr.append(root.val)
        inorder(root.right, arr)
        
    arr = []
    inorder(root, arr)
    return arr[k-1]    

# MINIMUM DEPTH OF A BINARY TREE

# O(n), faster if use if statements and prevent calls for Nones
# must use both max and min, in case of linked list type trees (if
# only one sided.)
def minDepth(self, root: TreeNode) -> int:
    if root is None:
        return 0
    
    if None in [root.left, root.right]:
        return max(self.minDepth(root.left), self.minDepth(root.right)) + 1
    else:
        return min(self.minDepth(root.left), self.minDepth(root.right)) + 1

# BINARY TREE RIGHT SIDE VIEW

"""
Getting the right most value of each level of the tree.

- Use a dictionary to store the depth as the key and the
value as the value in each node.
- Use pre-order traversal, so that the value for each depth updates
towards the right, and the most right nodes are the ones staying at the end.
The nodes which have higher depth will not be overwritten and will stay when
converting dictionary to list.
"""
def rightSideView(self, root: TreeNode) -> List[int]:
    right = {}
    def see_right(root, d):
        if root is None:
            return
        right[d] = root.val
        
        see_right(root.left, d + 1)
        see_right(root.right, d + 1)
    
    see_right(root, 0)
    return right.values()

# BFS implementation
"""
Remember the dequeue stack.. pop from the start (left) so that it's
BFS! Pop from the top of stack so that it's DFS.
"""
def rightSideView(self, root):
    if root is None:
        return []
    
    # create dequeue to allow O(1) from both ends 
    deq = collections.deque()
    deq.append(root)
    result = []
    
    while deq:
        size, val = len(deq), 0
        
        for i in range(size):
            # pop left since BFS
            node = deq.popleft()
            
            if node.left:
                deq.append(node.left)
            if node.right:
                deq.append(node.right)
                
            val = node.val
        
        result.append(val)
    return result

# RANGE SUM QUERY

class NumArray:
    """
    Initialize an additional array to hold blocks of size sqrt(n), where n is size of original list.
    These blocks would be the sums of those sqrt(n) nb of elements. We would then use these blocks
    to perform range sum queries to avoid having to iterate over each element to sum. O(sqrt(n))
    
    - Keep in mind we still need original array to perform sums on the two blocks on the end, since
    the picked index range could be a number in the middle of the block.
    """
    def __init__(self, nums: List[int]):
        # nums: original array
        # blocks_len: total length of blocks array
        # blocks: blocks array
        self.nums = nums
        block_size = math.sqrt(len(nums))
        self.blocks_len = int(math.ceil(len(nums) / block_size))
        self.blocks = [0] * self.blocks_len
        for i in range(len(nums)):
            self.blocks[i // self.blocks_len] += nums[i]
    
    """
    - Update block array value
    - Update original array value
    """
    def update(self, index: int, val: int) -> None:
        block_index = index // self.blocks_len
        self.blocks[block_index] = self.blocks[block_index] - self.nums[index] + val
        self.nums[index] = val
        
    
    """
    2 scenarios:
    - If in same block
    - If in different blocks
    """
    def sumRange(self, left: int, right: int) -> int:
        start_block = left // self.blocks_len
        end_block = right // self.blocks_len
        sum = 0
        
        if start_block == end_block:
            # means the indices are contained inside one block, just iterate and add.
            for i in range(left, right +1):
                sum += self.nums[i]
        else:
            # for left-most block. Convert the block index to the actual index!
            for i in range(left, (start_block + 1) * self.blocks_len):
                sum += self.nums[i]
            # for blocks in between
            for i in range(start_block + 1, end_block):
                sum += self.blocks[i]
            # for right-most block
            for i in range(end_block * self.blocks_len, right +1):
                sum += self.nums[i]
        return sum

# AVERAGE LEVEL BINARY TREE

def averageOfLevels(self, root: TreeNode) -> List[float]:
    if root is None:
        return
    
    yeet = []
    que = collections.deque()
    que.append(root)
    sum = 0
    
    while que:
        size = len(que)
        for i in range(size):
            node = que.popleft()
            sum += node.val
            if node.left:
                que.append(node.left)
            if node.right:
                que.append(node.right)
            
        yeet.append(sum / size)
        sum = 0
    
    return yeet

# MEDIUM: ROTTING ORANGES

    """
    BFS
    - The minimum number of minutes needed to rot all oranges is essentially the number of
      iterations needed to obtain the shortest path between the starting point and the 
      furthest away fresh orange.
    - It's why this can be computed with both DFS and BFS.
    - With BFS, we can find the number of iterations for the shortest path simply by incrementing.
    - With DFS, we can find this minimum number by finding the max depth of recursion of the search.
    """
    def orangesRotting(self, grid: List[List[int]]) -> int:
        """
        - Find the rotten oranges and place their indices in a queue.
        - Find the number of fresh oranges.
            - If there are none: finish.
        - Else, start BFS with the queue of rotten orange indices.
            - For each rotten orange, rot their neighbours.
            - Can check if neighbouring indices are out of bounds with a coordinates array.
            - Increment a count during each iteration.
            - Finish if either all oranges are rotted or if nb of fresh oranges reaches 0.
        """
        if not grid:
            return -1
        
        fresh_count = 0
        # queue of rotten orange indices
        que = collections.deque()
        for i in range(len(grid)):
            for k in range(len(grid[i])):
                if grid[i][k] == 2:
                    que.append((i,k))
                elif grid[i][k] == 1:
                    fresh_count += 1
                
        if fresh_count == 0:
            return 0
        
        minutes = 0
        # coordinates
        check = [(-1, 0), (1, 0), (0, 1), (0, -1)]
        # BFS
        while que and fresh_count:
            size = len(que)
            minutes += 1
            for i in range(size):
                orange = que.popleft()
                for coord in check:
                    row = orange[0] + coord[0]
                    col = orange[1] + coord[1]
                    # if new index exists and is a fresh orange, rot it and put in queue
                    if row >= 0 and row < len(grid) and col >= 0 and col < len(grid[row]) and grid[row][col] == 1:
                        grid[row][col] = 2
                        que.append((row, col))
                        fresh_count -= 1
        
        return minutes if fresh_count == 0 else -1
        
# DFS
"""
Iterate through the matrix and dfs on each cell.
Increment the minutes integer each time you dfs.
The max depth of the dfs minus two is the minimum number of minutes needed.

This works too because the depth of the DFS to an end is a "shortest path". This
value will be the same as when doing BFS since their results both reflect the "shortest path".

Time: O(M*N) -> Row * Columns. Each cell is traversed max two times.
Space: O(M*N): if only one case is rotten at a corner and all else are fresh. Big recursion!
"""
def orangesRotting(self, grid: List[List[int]]) -> int:
    for i in range(len(grid)):
        for k in range(len(grid[i])):
            if grid[i][k] == 2:
                self.dfs(grid, i, k, 2)
    
    min = 2
    for i in range(len(grid)):
        for k in range(len(grid[i])):
            if grid[i][k] == 1:
                return -1
            min = max(min, grid[i][k])
    
    return min-2
    
def dfs(self, grid, i, k, min):
    # check if out of bounds indices, empty, or rotten by anohter
    if i < 0 or k < 0 or i >= len(grid) or k >= len(grid[i]) or grid[i][k] == 0 or (grid[i][k] < min and grid[i][k] >= 2): 
        return

    grid[i][k] = min
    self.dfs(grid, i+1, k, min+1)
    self.dfs(grid, i-1, k, min+1)
    self.dfs(grid, i, k+1, min+1)
    self.dfs(grid, i, k-1, min+1)