Team problems

distincDifferenceArray.py

@@ -0,0 +1,3 @@
+class Solution: 
+    def distinctDifferenceArray(self, nums: List[int]) -> List[int]:
+        return [len(set(nums[:i])) - len(set(nums[i:])) for i in range(1, len(nums) + 1)]

numOfUneqTriplets.py

@@ -0,0 +1,11 @@
+class Solution: 
+    def unequalTriplets(self, nums: List[int]) -> int:
+        count = 0 
+
+        for i in range(len(nums)-2): 
+            for j in range(i+1, len(nums)-1): 
+                for k in range(j+1, len(nums)):
+                    if nums[i] != nums[j] and nums[j] != nums[k] and nums[i] != nums[k]: 
+                        count += 1
+
+        return count

python/expressionMatching.py

@@ -0,0 +1,28 @@
+class Solution:
+    def matches(self, s: str, p: str, i: int, j: int, memory) -> bool:
+        # Base cases 
+        if i >= len(s) and j >= len(p): 
+            return True 
+        elif j >= len(p): 
+            return False 
+
+        # Check if we have already computed the result
+        if memory[i][j] is not None: 
+            return memory[i][j]
+
+        matches = i < len(s) and (s[i] == p[j] or p[j] == '.')
+        isAMatch = None
+
+        if j < len(p) - 1 and p[j+1] == "*": 
+            isAMatch = (matches and self.matches(s,p,i+1,j,memory) or self.matches(s,p,i,j+2,memory))
+        else: 
+            isAMatch = matches and self.matches(s,p,i+1,j+1,memory)
+
+        # We register the result in memory
+        memory[i][j] = isAMatch
+
+        return isAMatch 
+
+    def isMatch(self, s: str, p: str) -> bool:
+        memory = [[None] * (len(p)) for i in range(len(s)+1)]
+        return self.matches(s,p,0,0,memory)

python/longestSubstring.py

@@ -0,0 +1,19 @@
+class Solution: 
+    def lengthOfLongestSubstring(self, s: str) -> int: 
+        visited = {}
+        startOfSubstring = 0
+        largestSubstringLen = 0
+
+        for idx, c in enumerate(s):
+            j = visited.get(c,-1)
+
+            if j >= startOfSubstring: 
+                startOfSubstring = j + 1
+            else: 
+                newLargest = idx - startOfSubstring + 1
+                if newLargest > largestSubstringLen: 
+                    largestSubstringLen = newLargest
+            visited[c] = idx
+
+
+        return largestSubstringLen