container-with-most-water solution (ts, py)

Author: hi-rachel

container-with-most-water/hi-rachel.py

@@ -0,0 +1,36 @@
+"""
+너비: e - s
+높이: min(height[s], height[e])
+
+넓이 = e - s * min(height[s], height[e])
+"""
+
+# TC: O(N^2), SC: O(1)
+# 모든 경우의 수를 따져 가장 넓이가 크게 나오는 경우를 찾는 풀이 -> Time Limit Exceeded
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        max_area = 0
+
+        for s in range(len(height) - 1):
+            for e in range(s + 1, len(height)):
+                area = (e - s) * min(height[s], height[e])
+                max_area = max(area, max_area)
+        return max_area
+    
+
+# 투 포인터 풀이
+# TC: O(N), SC: O(1)
+class Solution:
+    def maxArea(self, height: List[int]) -> int:
+        max_area = 0
+        s, e = 0, len(height) - 1
+        while s < e:
+            area = (e - s) * min(height[s], height[e])
+            max_area = max(area, max_area)
+            if height[s] < height[e]:
+                s += 1
+            else:
+                e -= 1
+
+        return max_area
+    

container-with-most-water/hi-rachel.ts

@@ -0,0 +1,23 @@
+/**
+너비: (e - s)
+높이: Math.min(height[s], height[e])
+
+넓이: (e - s) * Math.min(height[s], height[e])
+
+# TC: O(N), SC: O(1)
+*/
+
+function maxArea(height: number[]): number {
+  let maxArea = 0;
+  let s = 0,
+    e = height.length - 1;
+  while (s < e) {
+    maxArea = Math.max((e - s) * Math.min(height[s], height[e]), maxArea);
+    if (height[s] > height[e]) {
+      e -= 1;
+    } else {
+      s += 1;
+    }
+  }
+  return maxArea;
+}