Day - 80 work

Author: mandliya

README.md

@@ -6,9 +6,9 @@
 
 | Current Status|     Stats     |
 | :------------: | :----------: |
-| Total Problems | 116 |
-| Current Streak | 79 days |
-| Longest Streak | 79 ( August 17, 2015 - November 03, 2015 ) |
+| Total Problems | 117 |
+| Current Streak | 80 days |
+| Longest Streak | 80 ( August 17, 2015 - November 04, 2015 ) |
 
 </center>
 
@@ -193,3 +193,6 @@ Include contains single header implementation of data structures and some algori
 |Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.For example, Given nums = [0, 1, 3] return 2.| [missing_number.cpp](leet_code_problems/missing_number.cpp)|
 |Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array.| [find_min_rotated.cpp](leet_code_problems/find_min_rotated.cpp)|
 |Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.| [threeSumClosest.cpp](leet_code_problems/threeSumClosest.cpp)|
+|Given n non-negative integers a<sub>1</sub>, a<sub>2</sub>, ..., a<sub>n</sub>, where each represents a point at coordinate (i, a<sub>i</sub>). n vertical lines are drawn such that the two endpoints of line i is at (i, a<sub>i</sub>) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water. Note: You may not slant the container.| [maxArea.cpp](leet_code_problems/maxArea.cpp)|
+
+

leet_code_problems/maxArea.cpp

@@ -0,0 +1,44 @@
+/**
+ * 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 line i is at (i, ai) and (i, 0).
+ * Find two lines, which together with x-axis forms a container, such that the container contains the most water.
+ * Note: You may not slant the container.
+ */
+
+#include <iostream>
+#include <vector>
+#include <limits>
+
+int maxArea(std::vector<int>& height) {
+	size_t l = 0;
+	size_t r = height.size() - 1;
+	int maxAr = std::numeric_limits<int>::min();
+	while( l < r ) {
+		int curAr = ( std::min(height[l], height[r]) * ( r - l ));
+		if ( curAr > maxAr ) {
+			maxAr = curAr;
+		}
+		if ( height[l] < height[r]) {
+			l++;
+		} else {
+			r--;
+		}
+	}
+	return maxAr;
+}
+
+void printVec( std::vector<int> & vec ) {
+	std::cout << "Heights:";
+	for( auto v : vec ) {
+		std::cout << v << " ";
+	}
+	std::cout << std::endl;
+}
+
+int main()
+{
+	std::vector<int> heights{ 4, 5, 1 };
+	printVec(heights);
+	std::cout << "Max possible area with above heights:" << maxArea(heights) << std::endl;
+	return 0;
+}