Add solution for Valid Triangle Number problem

Implement two-pointer approach to count valid triangle numbers.

Author: 724thomas

leetcode3/최원준/611. Valid Triangle Number.py

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+#
+
+'''
+1. 아이디어 :
+투포인터를 사용.
+i에 위치한 숫자가 왼쪽+오른쪽 값보다 작으면 삼각형이 만들어진다.
+오른쪽 포인터를 이동하여 유요한지 확인
+else
+왼쪽 포인터를 이동.
+
+2. 시간복잡도 :
+    O(n*n)
+
+3. 자료구조/알고리즘 :
+투포인터
+
+'''
+class Solution:
+    def triangleNumber(self, nums: List[int]) -> int:
+        nums.sort()
+        n = len(nums)
+        ans = 0
+
+        for i in range(n-1, 1, -1):
+            left = 0
+            right = i - 1
+
+            while left < right:
+                if nums[left] + nums[right] > nums[i]:
+                    ans += right - left
+                    right -= 1
+                else:
+                    left += 1
+        return ans
+
+        # nums.sort()
+        # def binary_search(side1, side2, mid_index):
+        #     left = mid_index
+        #     right = n
+
+        #     while left < right:
+        #         mid = (left+right) // 2
+        #         longest = nums[mid]
+        #         if longest < side1 + side2:
+        #             left = mid+1
+        #         else:
+        #             right = mid
+        #     return left
+
+        # n = len(nums)
+        # ans = 0
+        # for i in range(n):
+        #     if nums[i] == 0:
+        #         continue
+        #     for j in range(i+1, n):
+        #         end = binary_search(nums[i], nums[j], j+1)
+        #         ans += end - (j+1)
+        # return ans