[LeetCode][JAVA] 11. Container With Most Water

https://leetcode.com/problems/container-with-most-water/?envType=study-plan-v2&envId=leetcode-75

My first solving was using brute-force. I new this won’t be the optimal answer but there was no other way.

private static int maxArea(int[] height) {

        int n = height.length;

        //if the size of array is 2
        if (n == 2) {
            return Math.min(height[0], height[1]);
        }

        int maxP = Integer.MIN_VALUE;
        int maxSize = Integer.MIN_VALUE;

        //loop until before the end
        for (int p = 0; p <= n - 2; p++) {
            //if the stick is taller than the maxP
            if (maxP < height[p]) {
                for (int j = n - 1; j >= p + 1; j--) {
                    int size = (j - p) * Math.min(height[p], height[j]);
                    maxSize = Math.max(size, maxSize);
                    maxP = height[p];
                }
            }
        }

        return maxSize;

    }

This is using optimal 2 pointers. I studied with most voted code at the discussion tab.

private static int maxArea(int[] height) {
        int left = 0;
        int right = height.length - 1;
        int maxSize = 0;

        while (left < right) {
            maxSize = Math.max(maxSize, (right - left) * Math.min(height[left], height[right]));
            if (height[left] < height[right]) left++;
            else if (height[left] > height[right]) {
                right--;
            } else {
                left++;
                right--;
            }
        }

        return maxSize;
    }