[Helena] Week 8

clone-graph/yolophg.py

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+# Time Complexity: O(N + E) - visit each node once and for each node, we iterate through its neighbors O(E).
+# Space Complexity: O(N) - store a copy of each node in the hashmap O(N).
+
+class Solution:
+    def cloneGraph(self, node):
+        if node is None:
+            return None
+
+        # dictionary to keep track of cloned nodes (original -> clone)
+        mp = {} 
+
+        def clone(node):
+            if node in mp:
+                # if the node has already been cloned, return the copy
+                return mp[node]  
+
+            # create a new node with the same value
+            cpy = Node(node.val)  
+            # store it in the map so don't clone it again
+            mp[node] = cpy  
+
+            # clone all neighbors and add them to the new node's neighbors list
+            for n in node.neighbors:
+                cpy.neighbors.append(clone(n))
+
+            return cpy
+
+        return clone(node)

longest-common-subsequence/yolophg.py

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+# Time Complexity: O(m * n) - each (i, j) pair is computed once and stored, reducing redundant calls.
+# Space Complexity: O(m * n) - memoization dictionary stores O(m * n) states.
+# - Recursion stack depth is O(m + n) in the worst case.
+
+class Solution:
+    def longestCommonSubsequence(self, t1: str, t2: str) -> int:
+        # to memoize results so we don't recompute the same subproblems
+        m = dict()
+
+        # recursive function to compute LCS
+        def s(i, j):
+            # base case: if we reach the end of either string, there's nothing left to compare
+            if i == len(t1) or j == len(t2):
+                return 0
+
+            # if already computed this state, just return the cached value
+            if (i, j) in m:
+                return m[(i, j)]
+
+            # if the characters match, we take this character and move diagonally
+            if t1[i] == t2[j]:
+                m[i, j] = 1 + s(i + 1, j + 1)
+            else:
+                # if they don't match, we either move forward in t1 or t2 and take the max
+                m[i, j] = max(s(i + 1, j), s(i, j + 1))
+
+            return m[i, j]
+
+        return s(0, 0)

longest-repeating-character-replacement/yolophg.py

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+# Time Complexity: O(n) - loop through the string once, and operations like `max(count.values())` are constant time because there are at most 26 characters.
+# Space Complexity: O(1) - `count` only stores counts for up to 26 characters.
+
+class Solution:
+    def characterReplacement(self, s: str, k: int) -> int:
+        # keep track of counts in the current window
+        count = {}
+
+        left, right = 0, 0
+        res = 0
+
+        # move the right pointer across the string
+        for right in range(len(s)):
+            # update the count for the character at the right pointer
+            count[s[right]] = count.get(s[right], 0) + 1
+
+            # if the window size minus the most frequent char count is bigger than k,
+            # need to shrink the window from the left
+            while (right - left + 1) - max(count.values()) > k:
+                # reduce the count of the char at the left pointer and move the left pointer
+                count[s[left]] -= 1
+                left += 1
+
+            # update the max length of the valid window
+            res = max(res, right - left + 1)
+
+        return res

number-of-1-bits/yolophg.py

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+# Time Complexity: O(k) - check each bit of n once. In the worst case, this is about 32 iterations.
+# Space Complexity: O(1) - only use a constant amount of extra space.
+
+class Solution:
+    def hammingWeight(self, n: int) -> int:
+        count = 0
+
+        # keep going till n becomes 0 (no more bits left to check)
+        while n:
+            # check if the last bit is 1 (n % 2 tells us this) and add it to the count
+            count += (n % 2)
+            # shift n to the right by 1 to move to the next bit
+            n = n >> 1
+
+        return count