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Author: wuduhren

problems/python3/container-with-most-water.py

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+"""
+Time: O(N)
+Space: O(1)
+
+i and j starts from the leftest and rightest. Move the one that has less height.
+
+Why we won't miss any i and j?
+For example, currently i is heigher than j.
+If between i~j, there are no height that is larger or equal to i, than since the area is `min(height[i], height[j]) * (j-i)`, you cannot find any area that is larger than the current one.
+If between i~j, there is a height that is larger or equal to i, j will on it, and it will be tested.
+Thus, given any i and j, any other future i and j that have the potential of forming larger area will be tested.
+"""
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        i = 0
+        j = len(height)-1
+        ans = 0
+        
+        while i<j:
+            ans = max(ans, min(height[i], height[j]) * (j-i))
+            if height[i]>height[j]:
+                j -= 1
+            else:
+                i += 1
+        return ans