Container With Most Water - Practice Coding Problems

Container With Most Water - Problem

Imagine you're designing a water collection system using vertical walls! You're given an array of n positive integers representing the height of vertical lines positioned at equal distances along the x-axis.

Each line i extends from point (i, 0) to (i, height[i]). Your goal is to find two lines that, together with the x-axis, form a container that can hold the maximum amount of water.

Key Rules:

The container cannot be tilted (must be upright)
Water level is limited by the shorter of the two walls
The wider the container, the more water it can potentially hold
Return the maximum area of water that can be contained

Formula: Area = min(height[i], height[j]) × (j - i)

Input & Output

example_1.py — Standard Case

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

› Output: 49

💡 Note: The maximum area is between lines at indices 1 and 8 (heights 8 and 7). Area = min(8,7) × (8-1) = 7 × 7 = 49.

example_2.py — Minimal Case

$ Input: [1,1]

› Output: 1

💡 Note: Only two lines available, both with height 1. Area = min(1,1) × (1-0) = 1 × 1 = 1.

example_3.py — Ascending Heights

$ Input: [2,3,4,5,18,17,6]

› Output: 17

💡 Note: Maximum area is between indices 4 and 5 (heights 18 and 17). Area = min(18,17) × (5-4) = 17 × 1 = 17.

Visualization

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Understanding the Visualization

Set Up Walls

Visualize vertical lines as walls that can contain water

Start Wide

Begin with leftmost and rightmost walls (maximum width)

Move Strategically

Always move the pointer at the shorter wall inward

Find Maximum

Continue until pointers meet, tracking maximum area

Key Takeaway

🎯 Key Insight: Always move the pointer at the shorter wall because moving the taller wall would only decrease the total area while reducing width.

Time & Space Complexity

Time Complexity

⏱️

O(n²)

We check all possible pairs of lines, which gives us n×(n-1)/2 combinations

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track maximum area and current calculations

✓ Linear Space

Constraints

n == height.length
2 ≤ n ≤ 10⁵
0 ≤ height[i] ≤ 10⁴
At least two lines are always provided

Asked in

G Google 42 a Amazon 38 f Meta 31 ⊞ Microsoft 28 Apple 22

The optimal solution uses the two pointers technique with O(n) time complexity. Start with the widest container and greedily move the pointer at the shorter wall inward, since moving the taller wall would only decrease the area. This greedy approach guarantees finding the maximum water area efficiently.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n²)	O(1)	Check all possible pairs of lines to find maximum water area
Two Pointers (Optimal)	O(n)	O(1)	Use two pointers from both ends, always moving the shorter wall inward

Brute Force (Nested Loops) — Algorithm Steps

Initialize maxArea to 0
Use nested loops to try all pairs (i, j) where i < j
For each pair, calculate area = min(height[i], height[j]) × (j - i)
Update maxArea if current area is larger
Return maxArea after checking all pairs

Visualization

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Step-by-Step Walkthrough

Initialize

Start with first pair (0,1) and maxArea = 0

Calculate Area

Area = min(height[i], height[j]) × (j - i)

Update Maximum

If current area > maxArea, update maxArea

Next Pair

Move to next pair and repeat until all tested

Code -

solution.c — C

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

int min(int a, int b) {
    return (a < b) ? a : b;
}

int max(int a, int b) {
    return (a > b) ? a : b;
}

int maxArea(int* height, int heightSize) {
    int maxArea = 0;
    
    // Try all possible pairs
    for (int i = 0; i < heightSize; i++) {
        for (int j = i + 1; j < heightSize; j++) {
            // Calculate area between lines i and j
            int width = j - i;
            int minHeight = min(height[i], height[j]);
            int area = width * minHeight;
            maxArea = max(maxArea, area);
        }
    }
    
    return maxArea;
}

int main() {
    char input[1000];
    fgets(input, sizeof(input), stdin);
    
    int height[1000];
    int count = 0;
    
    // Simple parsing
    char* token = strtok(input, "[],");
    while (token != NULL) {
        if (strlen(token) > 0) {
            height[count++] = atoi(token);
        }
        token = strtok(NULL, "[],");
    }
    
    printf("%d\n", maxArea(height, count));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

We check all possible pairs of lines, which gives us n×(n-1)/2 combinations

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track maximum area and current calculations

✓ Linear Space

Constraints

n == height.length
2 ≤ n ≤ 10⁵
0 ≤ height[i] ≤ 10⁴
At least two lines are always provided

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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