Container With Most Water in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Container With Most Water Calculation

What is the output of the following Java code that calculates the maximum water container area?

DSA Java

public class Solution {
    public static int maxArea(int[] height) {
        int maxArea = 0;
        int left = 0, right = height.length - 1;
        while (left < right) {
            int width = right - left;
            int currentArea = Math.min(height[left], height[right]) * width;
            maxArea = Math.max(maxArea, currentArea);
            if (height[left] < height[right]) {
                left++;
            } else {
                right--;
            }
        }
        return maxArea;
    }

    public static void main(String[] args) {
        int[] height = {1,8,6,2,5,4,8,3,7};
        System.out.println(maxArea(height));
    }
}

A 49

B 56

C 48

D 50

💡 Hint

Think about the two-pointer approach and how the area is calculated by the shorter line times the distance.

The maximum area is 49, formed between the lines at index 1 (height 8) and index 8 (height 7). The width is 7, so area = 7 * 7 = 49.

🧠 Conceptual

intermediate

Understanding the Two-Pointer Approach

Why does the two-pointer approach work efficiently for the Container With Most Water problem?

A Because moving the pointer with the smaller height can potentially find a taller line and increase area.

B Because moving both pointers inward at the same time always increases the area.

C Because the maximum area is always between the first and last lines.

D Because the area depends only on the tallest line in the array.

💡 Hint

Consider what limits the area between two lines.

The area is limited by the shorter line, so moving the pointer at the smaller height may find a taller line and increase the area.

🔧 Debug

advanced

Identify the Error in Container With Most Water Code

What error will this Java code produce when run?

DSA Java

public class Solution {
    public static int maxArea(int[] height) {
        int maxArea = 0;
        int left = 0, right = height.length;
        while (left < right) {
            int width = right - left;
            int currentArea = Math.min(height[left], height[right]) * width;
            maxArea = Math.max(maxArea, currentArea);
            if (height[left] < height[right]) {
                left++;
            } else {
                right--;
            }
        }
        return maxArea;
    }

    public static void main(String[] args) {
        int[] height = {1,8,6,2,5,4,8,3,7};
        System.out.println(maxArea(height));
    }
}

A No error, outputs 49

B NullPointerException

C Compilation error

D ArrayIndexOutOfBoundsException

💡 Hint

Check the initialization of the right pointer and array indexing.

The right pointer is initialized to height.length, which is out of bounds for array indexing (max index is length-1), causing ArrayIndexOutOfBoundsException.

🚀 Application

advanced

Maximum Water Container for Custom Input

Given the array heights = {4,3,2,1,4}, what is the maximum water container area?

A 8

B 12

C 16

D 6

💡 Hint

Calculate area between pairs and find the maximum.

The maximum area is between the first and last lines (index 0 and 4), min height 4, width 4, area = 16.

❓ Predict Output

expert

Output of Modified Container With Most Water Code with Equal Heights

What is the output of this Java code?

DSA Java

public class Solution {
    public static int maxArea(int[] height) {
        int maxArea = 0;
        int left = 0, right = height.length - 1;
        while (left < right) {
            int width = right - left;
            int currentArea = Math.min(height[left], height[right]) * width;
            maxArea = Math.max(maxArea, currentArea);
            if (height[left] <= height[right]) {
                left++;
            } else {
                right--;
            }
        }
        return maxArea;
    }

    public static void main(String[] args) {
        int[] height = {5,5,5,5,5};
        System.out.println(maxArea(height));
    }
}

A 25

B 20

C 15

D 10

💡 Hint

All heights are equal; area depends on width.

The max area is between index 0 and 4, height 5, width 4, area = 20.

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