LeetCode //C - 11. Container With Most Water

11. Container With Most Water

You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).

Find two lines that together with the x-axis form a container, such that the container contains the most water.

Return the maximum amount of water a container can store.

Notice that you may not slant the container.
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Example 1:

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 (blue section) the container can contain is 49.

Example 2:

Input: height = [1,1]
Output: 1

Constraints:

• n == height.length
• $2<=n<=10^5$
• $0<=height[i]<=10^4$

From: LeetCode
Link: 11. Container With Most Water

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Solution:

Ideas：

• We start with two pointers, one at the beginning (left) and one at the end (right) of the array.
• We calculate the area formed between the two pointers and compare it to the maximum area found so far.
• We move the pointer that points to the shorter line inward, as this could potentially increase the area.
• We continue this process until the two pointers meet.
• We return the maximum area found.

Code:

int maxArea(int* height, int heightSize) {
    int max_area = 0;
    int left = 0;
    int right = heightSize - 1;
    
    while (left < right) {
        int width = right - left;
        int current_height = height[left] < height[right] ? height[left] : height[right];
        int current_area = width * current_height;
        max_area = current_area > max_area ? current_area : max_area;
        
        if (height[left] < height[right]) {
            left++;
        } else {
            right--;
        }
    }
    
    return max_area;
}