Container With Most Water - Practice Coding Problems

Container With Most Water - Problem

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: You may not slant the container.

Input & Output

Example 1 — Basic Case

$ Input: height = [1,8,6,2,5,4,8,3,7]

› Output: 49

💡 Note: The vertical lines are at positions with heights [1,8,6,2,5,4,8,3,7]. The maximum area is between lines at indices 1 and 8: area = (8-1) × min(8,7) = 7 × 7 = 49

Example 2 — Minimum Case

$ Input: height = [1,1]

› Output: 1

💡 Note: Only two lines available: area = (1-0) × min(1,1) = 1 × 1 = 1

Example 3 — Equal Heights

$ Input: height = [4,3,2,1,4]

› Output: 16

💡 Note: Maximum area is between first and last lines: area = (4-0) × min(4,4) = 4 × 4 = 16

Constraints

n == height.length
2 ≤ n ≤ 10⁵
0 ≤ height[i] ≤ 10⁴

Visualization

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Understanding the Visualization

Input

Array of heights representing vertical lines

Process

Find two lines that form the largest water container

Output

Maximum water area possible

Key Takeaway

🎯 Key Insight: Water level is always limited by the shorter line, so use two pointers and always move from the shorter side

Asked in

A Amazon 42 G Google 38 M Microsoft 35 🍎 Apple 28 F Facebook 24

The key insight is that water level is limited by the shorter line, so when using two pointers, always move the pointer at the shorter line inward. This greedy approach ensures we don't miss the optimal solution. Best approach is Two Pointers with Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(n²)	O(1)	Check every pair of lines to find maximum water area
Two Pointers	O(n)	O(1)	Use two pointers starting from both ends, moving inward optimally

Brute Force — Algorithm Steps

Iterate through each line as left boundary
For each left line, try all possible right boundaries
Calculate area = min(height[left], height[right]) × (right - left)
Keep track of maximum area found

Visualization

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Step-by-Step Walkthrough

Try All Pairs

Check every combination of two lines

Calculate Area

Area = width × min(height1, height2)

Track Maximum

Keep the largest area found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* height, int n) {
    int maxArea = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            int width = j - i;
            int minHeight = (height[i] < height[j]) ? height[i] : height[j];
            int area = width * minHeight;
            if (area > maxArea) {
                maxArea = area;
            }
        }
    }
    
    return maxArea;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    // Parse JSON array manually
    int height[1000];
    int n = 0;
    char* token = strtok(line + 1, ",]"); // Skip opening bracket
    
    while (token != NULL) {
        height[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(height, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops check all n×(n-1)/2 pairs of lines

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant variables for tracking maximum area

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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