CointainsWithMostWater(Array)

// This function is with time complexity of O(n^2) which use 2 for loops
    public static int containsMaxWater(int[] a) {
        int maxArea = 0;
        for (int i = 0; i < a.length; i++) {
            for (int j = i + 1; j < a.length; j++ ) {
                int distance = j - i;
                int height = Math.min(a[i], a[j]);
                int area = height * distance;
                maxArea = Math.max(area, maxArea);
            }
        }
        return maxArea;
    }


    // This function is with time complexity of O(n) which use 2 pointer approach
    public static int containsWithMaxWater(int[] a) {
        int l = 0, r = a.length - 1;
        int maxArea = 0;
        while (l < r) {
            int height = Math.min(a[l],a[r]);
            int distance = r - l;
            int area = distance * height;
            maxArea = Math.max(area, maxArea);
            if (a[l] < a[r])
                l++;
            else r--;
        }
        return maxArea;
    }