[SunaDu] Week 4

coin-change/dusunax.py

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+'''
+# 322. Coin Change
+
+use a queue for BFS & iterate through the coins and check the amount is down to 0.
+use a set to the visited check.
+
+## Time and Space Complexity
+
+```
+TC: O(n * Amount)
+SC: O(Amount)
+```
+
+#### TC is O(n * Amount):
+- sorting the coins = O(n log n)
+- reversing the coins = O(n)
+- iterating through the queue = O(Amount)
+- iterating through the coins and check the remaining amount is down to 0 = O(n)
+
+#### SC is O(Amount):
+- using a queue to store (the remaining amount, the count of coins) tuple = O(Amount)
+- using a set to store the visited check = O(Amount)
+'''
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        if amount == 0:
+            return 0
+        if len(coins) == 1 and coins[0] == amount:
+            return 1
+
+        coins.sort() # TC: O(n log n)
+        coins.reverse() # TC: O(n)
+
+        queue = deque([(amount, 0)]) # SC: O(Amount)
+        visited = set() # SC: O(Amount)
+
+        while queue: # TC: O(Amount)
+            remain, count = queue.popleft()
+
+            for coin in coins: # TC: O(n)
+                next_remain = remain - coin
+
+                if next_remain == 0:
+                    return count + 1
+                if next_remain > 0 and next_remain not in visited:
+                    queue.append((next_remain, count + 1))
+                    visited.add(next_remain)
+
+        return -1

merge-two-sorted-lists/dusunax.py

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+'''
+# 21. Merge Two Sorted Lists
+
+A. iterative approach: use a two pointers to merge the two lists.
+B. recursive approach: use recursion to merge the two lists.
+
+
+## Time and Space Complexity
+
+### A. Iterative Approach
+
+```
+TC: O(n + m)
+SC: O(1)
+```
+
+#### TC is O(n + m):
+- iterating through the two lists just once for merge two sorted lists. = O(n + m)
+
+#### SC is O(1):
+- temp node is used to store the result. = O(1)
+
+### B. Recursive Approach
+
+```
+TC: O(n + m)
+SC: O(n + m)
+```
+
+#### TC is O(n + m):
+- iterating through the two lists just once for merge two sorted lists. = O(n + m)
+
+#### SC is O(n + m):
+- because of the recursive call stack. = O(n + m)
+'''
+class Solution:
+    '''
+    A. Iterative Approach
+    - use a temp node to store the result.
+    - use a current node to iterate through the two lists.
+    '''
+    def mergeTwoListsIterative(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
+        temp = ListNode(-1)
+        current = temp
+
+        while list1 is not None and list2 is not None:
+            if list1.val < list2.val:
+                current.next = list1
+                list1 = list1.next
+            else:
+                current.next = list2
+                list2 = list2.next
+            current = current.next
+
+        if list1 is not None:
+            current.next = list1
+        elif list2 is not None:
+            current.next = list2
+
+        return temp.next
+
+    '''
+    B. Recursive Approach
+    - use recursion to merge the two lists.
+    '''
+    def mergeTwoListsRecursive(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
+        if list1 is None:
+            return list2
+        elif list2 is None:
+            return list1
+
+        if list1.val < list2.val:
+            list1.next = self.mergeTwoLists(list1.next, list2)
+            return list1
+        else:
+            list2.next = self.mergeTwoLists(list1, list2.next)
+            return list2

missing-number/dusunax.py

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+'''
+# 268. Missing Number
+
+A. iterative approach: sort the array and find the missing number.
+B. XOR approach: use XOR to find the missing number.
+    - a ^ a = 0, a ^ 0 = a
+
+## Time and Space Complexity
+
+### A. Iterative Approach
+
+```
+TC: O(n log n)
+SC: O(1)
+```
+
+#### TC is O(n):
+- sorting the array. = O(n log n)
+- iterating through the array just once to find the missing number. = O(n)
+
+#### SC is O(1):
+- no extra space is used. = O(1)
+
+### B. XOR Approach
+
+```
+TC: O(n)
+SC: O(1)
+```
+
+#### TC is O(n):
+- iterating through the array just once to find the missing number. = O(n)
+
+#### SC is O(1):
+- no extra space is used. = O(1)
+
+'''
+class Solution:
+    '''
+    A. Iterative Approach
+    '''
+    def missingNumberIterative(self, nums: List[int]) -> int:
+        nums.sort()
+        n = len(nums)
+
+        for i in range(n):
+            if nums[i] != i:
+                return i
+        return n
+
+    '''
+    B. XOR Approach
+    '''
+    def missingNumberXOR(self, nums: List[int]) -> int:
+        n = len(nums)
+        xor_nums = 0
+
+        for i in range(n + 1):
+            if i < n:
+                xor_nums ^= nums[i]
+            xor_nums ^= i
+
+        return xor_nums

palindromic-substrings/dusunax.py

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+'''
+# 647. Palindromic Substrings
+
+A. use dynamic programming table to store the palindrome status.
+B. use two pointers to expand around the center.
+
+## Time and Space Complexity
+
+### A. Dynamic Programming Table
+```
+TC: O(n^2)
+SC: O(n^2)
+```
+
+#### TC is O(n^2):
+- filling DP table by iterating through all substrings. 
+- each cell (i, j) checks if a substring is a palindrome & counting the cases = O(n^2)
+
+#### SC is O(n^2):
+- storing the n x n dp table. = O(n^2)
+
+### B. Expand Around Center
+```
+TC: O(n^2)
+SC: O(1)
+```
+
+#### TC is O(n^2):
+- for each char, expand outwards to check for palindromes. = O(n^2)
+
+#### SC is O(1):
+- no additional data structures are used. `count` is a single integer. = O(1)
+'''
+class Solution:
+    def countSubstringsDPTable(self, s: str) -> int:
+        '''
+        A. Dynamic Programming Table
+        '''
+        n = len(s)
+        dp = [[False] * n for _ in range(n)] # List comprehension. = SC: O(n^2)
+        count = 0
+
+        for i in range(n): # TC: O(n)
+            dp[i][i] = True
+            count += 1
+
+        for i in range(n - 1):
+            if s[i] == s[i + 1]:
+                dp[i][i + 1] = True
+                count += 1
+
+        for s_len in range(3, n + 1): # TC: O(n)
+            for i in range(n - s_len + 1): # TC: O(n) 
+                j = i + s_len - 1
+
+                if s[i] == s[j] and dp[i + 1][j - 1]:
+                    dp[i][j] = True
+                    count += 1
+
+        return count
+    def countSubstrings(self, s: str) -> int:
+        '''
+        B. Expand Around Center
+        '''
+        count = 0
+
+        def expand(left, right):
+            nonlocal count
+            while left >= 0 and right < len(s) and s[left] == s[right]: # TC: O(n)
+                count += 1
+                left -= 1
+                right += 1
+
+        for i in range(len(s)): # TC: O(n)
+            expand(i, i)
+            expand(i, i + 1)
+
+        return count

word-search/dusunax.py

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+'''
+# 79. Word Search
+
+use backtracking(DFS) to search for the word in the board.
+
+## Time and Space Complexity
+
+```
+TC: O(n * m * 4^L)
+SC: O(L)
+```
+
+#### TC is O(n * m * 4^L):
+- n is the number of rows in the board.
+- m is the number of columns in the board.
+- L is the length of the word.
+  - 4^L is the number of directions we can go at each step. (explores 4 branches recursively)
+
+#### SC is O(L):
+- modifying the board in-place to mark visited cells. = O(L)
+'''
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        rows = len(board)
+        cols = len(board[0])
+
+        def backtracking(i, j, word_index): # TC: O(4^L), SC: O(L)
+            if word_index == len(word):
+                return True
+
+            if i < 0 or i >= rows or j < 0 or j >= cols or board[i][j] != word[word_index]:
+                return False
+
+            temp = board[i][j]
+            board[i][j] = "."
+
+            found = (
+                backtracking(i + 1, j, word_index + 1) or
+                backtracking(i - 1, j, word_index + 1) or
+                backtracking(i, j + 1, word_index + 1) or
+                backtracking(i, j - 1, word_index + 1)
+            )
+            board[i][j] = temp
+
+            return found
+
+        for row in range(rows): # TC: O(n * m)
+            for col in range(cols):
+                if backtracking(row, col, 0):
+                    return True
+
+        return False