l33t

Author: stevestar888

1512-num_of_good_pairs.py

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+"""
+https://leetcode.com/problems/number-of-good-pairs/submissions/
+
+Strat:
+    Since we only have to calculate the # of good pairs (instead of find all
+    the good pairs themselves), we can employ a trick: given [1, 1, 1, 1], the 
+    number of good pairs for the first 1 is 3 (all the 1s preceding it). The 
+    number of good pairs for the second 1 is 2 (all the 1s preceding it).
+    Apply this trick for the rest of the 1s, and the pattern of 
+    1 + 2 + 3 + ... + n = (n)(n-1)/2 emerges.
+    
+    So use a dictionary to count the number of occurances for each #. Iterate through
+    all the counts, and apply the above series.
+    
+Stats:
+    Runtime: 16 ms, faster than 98.00% of Python online submissions for Number of Good Pairs.
+    Memory Usage: 12.7 MB, less than 100.00% of Python online submissions for Number of Good Pairs.
+"""
+class Solution(object):
+    def numIdenticalPairs(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        nums_counts = {}
+        
+        for num in nums:
+            nums_counts[num] = nums_counts.get(num, 0) + 1 
+        
+        result = 0
+        for count in nums_counts.values():
+            result += self.count_good_pairs(count)
+        
+        return result
+    
+    
+    def count_good_pairs(self, count):
+        return count * (count - 1) / 2
+        

344-reverse_string.py

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+"""
+https://leetcode.com/problems/reverse-string/submissions/
+
+Strat:
+    Use two pointers, and narrow in on results.
+    
+Stats:
+    Runtime: 240 ms, faster than 16.60% of Python online submissions for Reverse String.
+    Memory Usage: 19.6 MB, less than 66.41% of Python online submissions for Reverse String.
+
+Cool resource linked: http://pythontutor.com/live.html#mode=edit
+"""
+class Solution(object):
+    def reverseString(self, s):
+        """
+        :type s: List[str]
+        :rtype: None Do not return anything, modify s in-place instead.
+        """    
+        l, r = 0, len(s) - 1
+        while l < r:
+            s[l], s[r] = s[r], s[l]
+            l += 1
+            r -= 1
+        return s