Container With Most Water medium

Description

You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the i-th 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.

Examples

> Case 1:
    height = [1, 8, 6, 2, 5, 4, 8, 3, 7]
    Output: 49
    Explanation: lines at indices 1 and 8 (heights 8 and 7) form a container
                 of width 7 and bottleneck height 7 → area = 49.
 
> Case 2:
    height = [1, 1]
    Output: 1

Constraints

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

State design

1. Flavor? Opposite-ends.
2. Pointers? l = 0, r = n - 1. Width is r - l, height is min(height[l], height[r]).
3. Movement? Move the shorter side inward. Moving the taller side can only decrease the bottleneck (the shorter line stays the bottleneck) while shrinking the width — strictly worse. The shorter line is the only direction with hope.
4. Stop? When l >= r.
5. Record? Max area at every step.

Try it yourself

def max_area(height):
  # two pointers from the ends; always move the shorter side inward
  pass

Code

int maxArea(vector<int>& height) {
    int l = 0, r = height.size() - 1, best = 0;
    while (l < r) {
        int area = min(height[l], height[r]) * (r - l);
        best = max(best, area);
        if (height[l] < height[r]) l++;
        else                       r--;
    }
    return best;
}

def max_area(height):
    l, r = 0, len(height) - 1
    best = 0
    while l < r:
        area = min(height[l], height[r]) * (r - l)
        best = max(best, area)
        if height[l] < height[r]: l += 1
        else:                     r -= 1
    return best

public int maxArea(int[] height) {
    int l = 0, r = height.length - 1, best = 0;
    while (l < r) {
        int area = Math.min(height[l], height[r]) * (r - l);
        best = Math.max(best, area);
        if (height[l] < height[r]) l++;
        else                       r--;
    }
    return best;
}

A tie-breaking note

When height[l] == height[r], it doesn’t matter which side we move — both produce identical optimal values. Many solutions move l by default; some move both. Either is correct.

Analysis

• Time: O(n). Each pointer moves at most n - 1 times.
• Space: O(1).

Same skin

• Trapping Rain Water — similar opposite-ends idea, but tracking running max heights on each side.
• Maximize Distance to Closest Person — opposite-ends-style scan for the largest gap.
• Boats to Save People — opposite-ends greedy pairing for “lightest + heaviest fit in one boat?”