Merge pull request #536 from Anupam-Panwar/containerWithMostWater

Added containerWithMostWater.cpp

Author: smv1999

Arrays/containerWithMostWater.cpp

@@ -0,0 +1,57 @@
+/* 
+Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n 
+vertical lines are drawn such that the two endpoints of the line i is at (i, ai) and (i, 0). 
+Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.
+
+Notice that you may not slant the container.
+
+
+Input: height = [1,8,6,2,5,4,8,3,7]
+Output: 49
+Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. 
+In this case, the max area of water the container can contain is 49.
+
+Constraints:
+ n == height.length
+ 2 <= n <= 105
+ 0 <= height[i] <= 104
+  
+*/
+
+/*
+Time Complexity: O(N), where N is the size of the array
+Space Complexity: O(1), Constant Space
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+int maxArea(vector<int> &height)
+{
+    int n = height.size(); 
+    int i = 0; // one pointer to the start of the array
+    int j = n - 1; // one poninter to the end of the array
+    int mx = min(height[i], height[j]) * (j - i); // contains the max value of maximum water that can be stored
+    while (i < j) 
+    {
+        if (height[i] < height[j]) // Checking condition if height pointed by i is less than that pointed by j if yes => incrementing i and updating value of mx
+        {
+            i++;
+            mx = max(mx, (min(height[i], height[j]) * (j - i)));
+        }
+        else // otherwise =>  decrementing j and updating value of mx
+        {
+            j--;
+            mx = max(mx, (min(height[i], height[j]) * (j - i)));
+        }
+    }
+    return mx;
+}
+
+int main() // driver function
+{
+    vector<int> heights = {1,8,6,2,5,4,8,3,7};
+    int ans = maxArea(heights);
+    cout<<ans;
+    return 0;
+}