Container With Most Water in DSA Java - Build from Scratch

Container With Most Water

📖 Scenario: Imagine you have a row of vertical lines drawn on a wall. Each line has a different height. You want to find two lines that, together with the wall, can hold the most water between them.

🎯 Goal: You will write a Java program that finds the maximum amount of water that can be contained between two lines from a given list of heights.

📋 What You'll Learn

Create an integer array called heights with the exact values: 1, 8, 6, 2, 5, 4, 8, 3, 7

Create an integer variable called maxArea and set it to 0

Use a while loop with variables left and right to find the maximum water container area

Print the value of maxArea

💡 Why This Matters

🌍 Real World

This problem models situations like finding the best place to build a water container between walls or barriers to hold the most water.

💼 Career

Understanding this problem helps in learning efficient algorithms and two-pointer techniques, which are common in coding interviews and real-world software development.

1

Create the heights array

Create an integer array called heights with these exact values: 1, 8, 6, 2, 5, 4, 8, 3, 7

DSA Java

// Create the integer array heights with the given values
// Your code here

Hint

Use int[] heights = { ... }; to create the array with the exact numbers.

2

Initialize maxArea variable

Create an integer variable called maxArea and set it to 0 inside the main method after the heights array.

DSA Java

public class ContainerWithMostWater {
    public static void main(String[] args) {
        int[] heights = {1, 8, 6, 2, 5, 4, 8, 3, 7};
        // Create maxArea variable and set to 0
        // Your code here
    }
}

Hint

Use int maxArea = 0; to create the variable.

3

Find the maximum water container area

Use a while loop with two integer variables left set to 0 and right set to heights.length - 1. Inside the loop, calculate the area between heights[left] and heights[right]. Update maxArea if the current area is larger. Move left forward if heights[left] is less than heights[right], otherwise move right backward.

DSA Java

public class ContainerWithMostWater {
    public static void main(String[] args) {
        int[] heights = {1, 8, 6, 2, 5, 4, 8, 3, 7};
        int maxArea = 0;
        // Use two pointers left and right to find max area
        // Your code here
    }
}

Hint

Use two pointers starting at the ends of the array and move them towards each other while calculating areas.

4

Print the maximum area

Print the value of maxArea using System.out.println(maxArea); inside the main method after the loop.

DSA Java

public class ContainerWithMostWater {
    public static void main(String[] args) {
        int[] heights = {1, 8, 6, 2, 5, 4, 8, 3, 7};
        int maxArea = 0;
        int left = 0;
        int right = heights.length - 1;
        while (left  maxArea) {
                maxArea = currentArea;
            }
            if (heights[left] < heights[right]) {
                left++;
            } else {
                right--;
            }
        }
        // Print maxArea
        // Your code here
    }
}

Hint

Use System.out.println(maxArea); to show the result.

Practice

1. What does the Container With Most Water problem ask you to find given an array of heights?

easy

A. The minimum height of the lines in the array.

B. The maximum area of water that can be contained between two lines.

C. The total sum of all heights in the array.

D. The number of lines taller than a given height.

Solution

1. Step 1: Understand the problem statement

   The problem asks to find two lines which together with the x-axis form a container that holds the most water.

2. Step 2: Identify what to maximize

   We want to maximize the area between two lines, which is the width between them times the smaller height.

3. Final Answer:

   The maximum area of water that can be contained between two lines. -> Option B

4. Quick Check:

   Max area = max water container [OK]

Hint: Look for two lines maximizing min height * distance [OK]

Common Mistakes:

• Confusing maximum area with sum of heights
• Ignoring the smaller height between two lines
• Thinking it's about counting lines

2. Which of the following is the correct way to declare the two-pointer variables for solving the Container With Most Water problem in Java?

easy

A. int left = 0, right = heights.length;

B. int left = 1, right = heights.length;

C. int left = 1, right = heights.length - 1;

D. int left = 0, right = heights.length - 1;

Solution

1. Step 1: Understand array indexing in Java

   Arrays start at index 0, so the left pointer should start at 0.

2. Step 2: Set right pointer to last index

   The right pointer should be at the last element, which is length - 1.

3. Final Answer:

   int left = 0, right = heights.length - 1; -> Option D

4. Quick Check:

   Array indices start at 0 [OK]

Hint: Start pointers at 0 and length-1 for arrays [OK]

Common Mistakes:

• Starting pointers at 1 instead of 0
• Using length instead of length - 1 for last index
• Confusing array length with last index

3. What is the output of the following Java code snippet for the Container With Most Water problem?

int[] heights = {1, 8, 6, 2, 5, 4, 8, 3, 7};
int left = 0, right = heights.length - 1;
int maxArea = 0;
while (left < right) {
    int height = Math.min(heights[left], heights[right]);
    int width = right - left;
    maxArea = Math.max(maxArea, height * width);
    if (heights[left] < heights[right]) {
        left++;
    } else {
        right--;
    }
}
System.out.println(maxArea);

medium

A. 49

B. 56

C. 48

D. 50

Solution

1. Step 1: Initialize pointers and maxArea

   left=0, right=8, maxArea=0

2. Step 2: Iterate and calculate areas

   Calculate areas step-by-step: - left=0, right=8: min(1,7)=1, width=8, area=8, maxArea=8 - left=1, right=8: min(8,7)=7, width=7, area=49, maxArea=49 - left=1, right=7: min(8,3)=3, width=6, area=18, maxArea=49 - left=1, right=6: min(8,8)=8, width=5, area=40, maxArea=49 - left=1, right=5: min(8,4)=4, width=4, area=16, maxArea=49 - left=1, right=4: min(8,5)=5, width=3, area=15, maxArea=49 - left=1, right=3: min(8,2)=2, width=2, area=4, maxArea=49 - left=1, right=2: min(8,6)=6, width=1, area=6, maxArea=49 Loop ends when left >= right. This misses some pointer moves. Let's trace fully: - Start: left=0, right=8 area = min(1,7)*8=8 heights[left] left++ => left=1 - left=1, right=8 area = min(8,7)*7=49 heights[left]>heights[right] => right-- => right=7 - left=1, right=7 area = min(8,3)*6=18 heights[left]>heights[right] => right-- => right=6 - left=1, right=6 area = min(8,8)*5=40 heights[left]==heights[right] => right-- => right=5 - left=1, right=5 area = min(8,4)*4=16 heights[left]>heights[right] => right-- => right=4 - left=1, right=4 area = min(8,5)*3=15 heights[left]>heights[right] => right-- => right=3 - left=1, right=3 area = min(8,2)*2=4 heights[left]>heights[right] => right-- => right=2 - left=1, right=2 area = min(8,6)*1=6 heights[left]>heights[right] => right-- => right=1 - left=1, right=1 loop ends Max area found is 49. But the maximum area for this input is actually 49. However, 56 is 56, which is incorrect. Let's check if 56 is possible: - Between indices 1 and 6: min(8,8)*5=40 - Between indices 1 and 8: min(8,7)*7=49 - Between indices 2 and 8: min(6,7)*6=36 - Between indices 1 and 7: min(8,3)*6=18 - Between indices 0 and 8: min(1,7)*8=8 No 56 found. Therefore, correct answer is 49 -> 49.

3. Final Answer:

   49 -> Option A

4. Quick Check:

   Max area found = 49 [OK]

Hint: Move pointer with smaller height inward [OK]

Common Mistakes:

• Calculating area with max height instead of min
• Not updating maxArea correctly
• Moving wrong pointer in the loop

4. Identify the bug in the following Java code snippet for the Container With Most Water problem:

int[] heights = {1, 2, 4, 3};
int left = 0, right = heights.length - 1;
int maxArea = 0;
while (left <= right) {
    int height = Math.min(heights[left], heights[right]);
    int width = right - left;
    maxArea = Math.max(maxArea, height * width);
    if (heights[left] < heights[right]) {
        left++;
    } else {
        right--;
    }
}
System.out.println(maxArea);

medium

A. The pointers should both start at 0.

B. The width calculation should be right + left.

C. The loop condition should be left < right, not left <= right.

D. The maxArea should be initialized to Integer.MAX_VALUE.

Solution

1. Step 1: Analyze loop condition

   The loop runs while left <= right, which allows left == right, meaning width = 0.

2. Step 2: Understand impact on area calculation

   When left == right, width = 0, so area is zero and unnecessary iteration occurs. The correct condition is left < right to avoid zero width.

3. Final Answer:

   The loop condition should be left < right, not left <= right. -> Option C

4. Quick Check:

   Loop must stop before pointers meet [OK]

Hint: Use left < right to avoid zero width area [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Incorrect width calculation
• Wrong initialization of maxArea

5. Given an array of heights representing vertical lines, how would you modify the two-pointer approach to find the container with the most water if the lines can only be chosen at even indices?

hard

A. Initialize pointers at the first and last even indices, then move inward skipping odd indices.

B. Use a nested loop to check all pairs of even indices only.

C. Ignore the even index condition and apply the standard two-pointer approach.

D. Sort the array and then apply the two-pointer approach.

Solution

1. Step 1: Adjust pointers to even indices

   Start left pointer at index 0 (first even index) and right pointer at the last even index (length-1 or length-2 depending on parity).

2. Step 2: Move pointers inward skipping odd indices

   When moving pointers, increment or decrement by 2 to stay on even indices only.

3. Step 3: Calculate area as usual

   Calculate area using min height and width between pointers, update max area accordingly.

4. Final Answer:

   Initialize pointers at the first and last even indices, then move inward skipping odd indices. -> Option A

5. Quick Check:

   Restrict pointers to even indices [OK]

Hint: Move pointers by 2 to stay on even indices [OK]

Common Mistakes:

• Using standard two-pointer without skipping odd indices
• Sorting array which changes line positions
• Using nested loops causing higher complexity