Container With Most Water in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the problem 'Container With Most Water' ask you to find?

It asks to find two lines in an array that, together with the x-axis, form a container that holds the most water (maximum area).

intermediate

What is the main idea behind the two-pointer approach in 'Container With Most Water'?

Use two pointers at the start and end of the array, calculate area, then move the pointer with the smaller height inward to try to find a bigger area.

intermediate

Why do we move the pointer with the smaller height in the two-pointer approach?

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

intermediate

What is the time complexity of the two-pointer solution for 'Container With Most Water'?

The time complexity is O(n), where n is the number of lines, because each pointer moves at most n times.

beginner

Show the formula to calculate the area between two lines at positions i and j with heights h[i] and h[j].

Area = min(h[i], h[j]) * (j - i), where min(h[i], h[j]) is the shorter height and (j - i) is the distance between lines.

In the 'Container With Most Water' problem, what limits the amount of water a container can hold?

A The sum of the two lines' heights

B The taller of the two lines

C The shorter of the two lines

D The average height of the two lines

The container's water level is limited by the shorter line, so the shorter height determines the maximum water.

What is the initial position of the two pointers in the optimal solution?

A One at the start and one at the end of the array

B Both at the end of the array

C Both at the start of the array

D Both in the middle of the array

One pointer starts at the beginning and the other at the end to consider the widest container first.

When moving pointers, which pointer do you move to try to find a larger area?

A The pointer at the taller line

B The pointer at the shorter line

C Both pointers simultaneously

D Neither pointer moves

Moving the pointer at the shorter line may find a taller line and increase the area.

What is the time complexity of the two-pointer approach for this problem?

A O(n)

B O(n^2)

C O(n log n)

D O(log n)

Each pointer moves inward at most n times, so the total time is linear, O(n).

If the heights array is [1,8,6,2,5,4,8,3,7], what is the maximum water container area?

A 36

B 56

C 48

D 49

The max area is between lines at index 1 (height 8) and index 8 (height 7): min(8,7) * (8-1) = 7 * 7 = 49.

Explain the two-pointer approach to solve the 'Container With Most Water' problem.

Think about how to find the biggest container by checking pairs from the edges inward.

Describe why moving the pointer at the shorter line can lead to finding a larger container.

Focus on what limits the container's height and how changing pointers affects area.

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