Why Container With Most Water in DSA C++?

The Big Idea

What if you could find the biggest water container without checking every pair of lines?

The Scenario

Imagine you have a row of vertical lines on a graph, each with different heights. You want to find two lines that, together with the x-axis, can hold the most water. Doing this by checking every possible pair manually is like trying to find the biggest bucket by testing every pair of sticks one by one.

The Problem

Checking every pair of lines manually means a lot of work. If you have 100 lines, you would check 4,950 pairs! This takes too long and is easy to make mistakes. It's slow and frustrating, especially when the number of lines grows.

The Solution

The "Container With Most Water" approach uses two pointers starting at the ends of the line array. It moves the pointers inward smartly, skipping unnecessary checks. This way, it finds the biggest container quickly without testing every pair.

Before vs After

✗ Before

int maxArea = 0;
for (int i = 0; i < n; i++) {
  for (int j = i + 1; j < n; j++) {
    int area = min(height[i], height[j]) * (j - i);
    maxArea = max(maxArea, area);
  }
}

✓ After

int left = 0, right = n - 1, maxArea = 0;
while (left < right) {
  int area = min(height[left], height[right]) * (right - left);
  maxArea = max(maxArea, area);
  if (height[left] < height[right]) left++;
  else right--;
}

What It Enables

This method lets you find the largest water container fast and efficiently, even with thousands of lines.

Real Life Example

Think of trying to find the best place between two walls to store the most rainwater. Instead of measuring every pair of walls, you quickly narrow down the best spot by comparing the shortest walls first.

Key Takeaways

Manual checking of all pairs is slow and inefficient.

Two-pointer technique smartly skips unnecessary checks.

Efficiently finds the maximum water container in linear time.

Practice

1. What is the main goal of the Container With Most Water problem?

easy

A. Find two lines that together with the x-axis form a container holding the most water

B. Sort the array of heights in ascending order

C. Find the maximum height in the array

D. Calculate the sum of all heights

Solution

1. Step 1: Understand the problem statement

   The problem asks to find two vertical lines which, together with the x-axis, form a container that holds the maximum amount of water.

2. Step 2: Identify what is being maximized

   The area between two lines is calculated by the minimum height of the two lines multiplied by the distance between them. We want to maximize this area.

3. Final Answer:

   Find two lines that together with the x-axis form a container holding the most water -> Option A

4. Quick Check:

   Goal = Max water container [OK]

Hint: Max area = min height * width between lines [OK]

Common Mistakes:

• Thinking it's about sorting heights
• Trying to sum heights instead of area
• Looking for max single height only

2. Which of the following C++ code snippets correctly initializes two pointers for the two-pointer approach in the Container With Most Water problem?

easy

A. int left = 1, right = heights.size();

B. int left = 0, right = heights.size() - 1;

C. int left = 0, right = heights.size();

D. int left = 1, right = heights.size() - 1;

Solution

1. Step 1: Understand pointer initialization

   We start with two pointers: one at the beginning (index 0) and one at the end (index size-1) of the array.

2. Step 2: Check each option

   int left = 0, right = heights.size() - 1; correctly sets left = 0 and right = size - 1, which are valid indices. Other options use invalid indices or off-by-one errors.

3. Final Answer:

   int left = 0, right = heights.size() - 1; -> Option B

4. Quick Check:

   Pointers start at ends = int left = 0, right = heights.size() - 1; [OK]

Hint: Start pointers at array ends: 0 and size-1 [OK]

Common Mistakes:

• Using size() as index (out of bounds)
• Starting pointers at 1 instead of 0
• Swapping left and right initialization

3. What is the output of the following C++ code snippet?

std::vector<int> heights = {1, 8, 6, 2, 5, 4, 8, 3, 7};
int left = 0, right = heights.size() - 1;
int maxArea = 0;
while (left < right) {
    int height = std::min(heights[left], heights[right]);
    int width = right - left;
    maxArea = std::max(maxArea, height * width);
    if (heights[left] < heights[right]) {
        left++;
    } else {
        right--;
    }
}
std::cout << maxArea << std::endl;

medium

A. 56

B. 50

C. 48

D. 49

Solution

1. Step 1: Trace the two-pointer approach

   Start with left=0 (height=1), right=8 (height=7), area= min(1,7)*8=8. Move left pointer since height[left]

2. Step 2: Continue updating maxArea

   Next left=1 (8), right=8 (7), area= min(8,7)*7=49 (maxArea=49). Move right pointer since height[left]>height[right]. Then left=1, right=7 (3), area= min(8,3)*6=18 (maxArea=49). Move right pointer. Continue similarly until pointers meet. The maximum area found is 49.

3. Final Answer:

   49 -> Option D

4. Quick Check:

   Max area = 49 [OK]

Hint: Move pointer at smaller height to find max area [OK]

Common Mistakes:

• Not updating maxArea correctly
• Moving wrong pointer after comparison
• Calculating area with max height instead of min

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

int maxArea(std::vector<int>& heights) {
    int left = 0, right = heights.size();
    int max_area = 0;
    while (left < right) {
        int height = std::min(heights[left], heights[right]);
        int width = right - left;
        max_area = std::max(max_area, height * width);
        if (heights[left] < heights[right]) {
            left++;
        } else {
            right--;
        }
    }
    return max_area;
}

medium

A. The loop condition should be left <= right

B. Use max instead of min to calculate height

C. right should be initialized to heights.size() - 1

D. Increment right instead of decrementing

Solution

1. Step 1: Check pointer initialization

   right is initialized to heights.size(), which is out of bounds since valid indices go up to size()-1.

2. Step 2: Correct initialization and loop condition

   right should be heights.size() - 1 to point to the last valid element. The loop condition left < right is correct.

3. Final Answer:

   right should be initialized to heights.size() - 1 -> Option C

4. Quick Check:

   Index out of bounds fixed by right should be initialized to heights.size() - 1 [OK]

Hint: Array indices go from 0 to size-1 [OK]

Common Mistakes:

• Using size() as index causing runtime error
• Changing loop condition incorrectly
• Using max instead of min for height

5. Given an array of heights representing vertical lines, how would you modify the two-pointer approach to find the maximum water container if the array can contain zero heights (lines with zero height)?

hard

A. Skip zero height lines by moving pointers inward until non-zero heights are found

B. Treat zero heights as normal and proceed with the standard two-pointer approach

C. Remove zero height lines from the array before processing

D. Use a nested loop to check all pairs instead of two-pointer

Solution

1. Step 1: Understand zero height impact

   Lines with zero height cannot hold water, so including them wastes time and area calculations.

2. Step 2: Modify two-pointer approach

   Skip zero height lines by moving pointers inward until both point to non-zero heights. Then apply the standard two-pointer logic to maximize area.

3. Final Answer:

   Skip zero height lines by moving pointers inward until non-zero heights are found -> Option A

4. Quick Check:

   Skip zeros for max area [OK]

Hint: Ignore zero heights by moving pointers inward [OK]

Common Mistakes:

• Processing zero height lines as normal
• Removing elements changes indices and complicates logic
• Switching to nested loops loses efficiency