feat : container-with-most-water

Author: ekgns33

container-with-most-water/ekgns33.java

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+/*
+input : array of integer height
+output : maximum amount of water a container can store.
+
+we can build container with two different line (height[i], height[j] when i < j)
+
+example
+1 8 6 2 5 4 8 3 7
+choose i = 1.
+choose j = 4
+   8 ______5 
+    amount of water == (j - i) * min(height[i], height[j])
+   = 3 * 5 = 15
+
+we have to maximize amount of water.
+
+constraints:
+1) is the input array valid?
+length of array is in range [2, 10^5]
+2) positive integers?
+no. range of height is [0, 10 ^4]
+>> check amount can be overflow. 10^4 * 10 ^ 5 = 10^9 < Integer range
+edge:
+1) if length is 2
+return min(height[0], height[1]).
+...
+
+solution 1) brute force;
+iterate through the array from index i = 0 to n-1
+     when n is the length of input array
+    iterate through the array from index j = i + 1 to n;
+        calculate the amount of water and update max amount
+ds : array
+algo : x
+tc : O(n^2) ~= 10^10 TLE
+space : O(1)
+
+solution 2) better 
+ds : array
+algo: two pointer?
+
+two variant for calculating amount of water
+1. height 2. width
+
+set width maximum at first, check heights
+    decrease width one by one
+
+    - at each step width is maximum. 
+        so we have to maximize
+        the minimum between left and right pointer
+
+use left and right pointer 
+while left < right
+
+    calculate amount 
+    compare
+    if height[left] < height[right]
+        move left by one
+    else 
+        vice versa
+
+return max
+tc : O(n)
+sc : O(1)
+
+ */
+class Solution {
+  public int maxArea(int[] height) {
+    int left = 0;
+    int right = height.length - 1;
+    int maxAmount = 0;
+    while(left < right) {
+      int curAmount = (right - left) * Math.min(height[left], height[right]);
+      maxAmount = Math.max(maxAmount, curAmount);
+      if(height[left] < height[right]) {
+        left++;
+      } else {
+        right--;
+      }
+    }
+    return maxAmount;
+  }
+}