Trapping Rain Water Problem in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the total trapped water to zero.

DSA C++

int trap(vector<int>& height) {
    int water = [1];
    return water;
}

Drag options to blanks, or click blank then click option'

A0

B1

C-1

Dheight.size()

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing water with 1 or negative values.

We start counting trapped water from zero because initially no water is trapped.

2fill in blank

medium

Complete the code to find the maximum height to the left of each bar.

DSA C++

vector<int> leftMax(height.size());
leftMax[0] = height[0];
for (int i = 1; i < height.size(); i++) {
    leftMax[i] = [1](leftMax[i - 1], height[i]);
}

Drag options to blanks, or click blank then click option'

Amin

Babs

Cmax

Dpow

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max.

We use max to keep track of the highest bar on the left side.

3fill in blank

hard

Fix the error in the loop that calculates trapped water by choosing the correct function to find the minimum of leftMax and rightMax.

DSA C++

for (int i = 0; i < height.size(); i++) {
    water += [1](leftMax[i], rightMax[i]) - height[i];
}

Drag options to blanks, or click blank then click option'

Aabs

Bmin

Cmax

Dpow

Attempts:

3 left

💡 Hint

Common Mistakes

Using max instead of min causes incorrect water calculation.

We use min because water trapped depends on the shorter boundary between left and right.

4fill in blank

hard

Fill both blanks to correctly compute the maximum height to the right of each bar.

DSA C++

vector<int> rightMax(height.size());
rightMax[height.size() - 1] = height[height.size() - 1];
for (int i = height.size() - 2; i >= 0; i--) {
    rightMax[i] = [1](rightMax[i [2] 1], height[i]);
}

Drag options to blanks, or click blank then click option'

Amax

B-

C+

Dmin

Attempts:

3 left

💡 Hint

Common Mistakes

Using - instead of + in the index causes incorrect logic and potential out-of-bounds access.

We use max to find the tallest bar on the right, and i + 1 to refer to the rightMax already computed for the right side.

5fill in blank

hard

Fill all three blanks to complete the function that calculates total trapped water.

DSA C++

int trap(vector<int>& height) {
    int water = 0;
    vector<int> leftMax(height.size()), rightMax(height.size());
    leftMax[0] = height[0];
    for (int i = 1; i < height.size(); i++) {
        leftMax[i] = [1](leftMax[i - 1], height[i]);
    }
    rightMax[height.size() - 1] = height[height.size() - 1];
    for (int i = height.size() - 2; i >= 0; i--) {
        rightMax[i] = [2](rightMax[i + 1], height[i]);
    }
    for (int i = 0; i < height.size(); i++) {
        water += [3](leftMax[i], rightMax[i]) - height[i];
    }
    return water;
}

Drag options to blanks, or click blank then click option'

Amax

Bmin

Attempts:

3 left

💡 Hint

Common Mistakes

Using min for leftMax or rightMax causes wrong boundaries.

Use max to find leftMax and rightMax, and min to find the limiting height for trapped water.

Practice

1. What does the Trapping Rain Water problem ask you to find given an array of heights?

easy

A. The total amount of water that can be trapped between the bars after raining

B. The highest bar in the array

C. The number of bars taller than a given height

D. The sum of all bar heights

Solution

1. Step 1: Understand the problem context

   The problem gives an array representing heights of bars and asks how much water can be trapped between them after it rains.

2. Step 2: Identify the goal

   The goal is to calculate the total trapped water volume, not just heights or counts.

3. Final Answer:

   The total amount of water that can be trapped between the bars after raining -> Option A

4. Quick Check:

   Trapping rain water = total trapped water [OK]

Hint: Think of bars as walls holding water in gaps [OK]

Common Mistakes:

• Confusing trapped water with highest bar height
• Summing bar heights instead of trapped water
• Counting bars taller than a certain height

2. Which of the following C++ code snippets correctly initializes a vector named heights with values {0,1,0,2,1,0,1,3,2,1,2,1}?

easy

A. std::vector<int> heights = (0,1,0,2,1,0,1,3,2,1,2,1);

B. std::vector<int> heights = {0,1,0,2,1,0,1,3,2,1,2,1};

C. vector<int> heights(0,1,0,2,1,0,1,3,2,1,2,1);

D. std::vector heights = {0,1,0,2,1,0,1,3,2,1,2,1};

Solution

1. Step 1: Check vector initialization syntax

   In C++, std::vector<int> can be initialized with curly braces and values inside, like {0,1,...}.

2. Step 2: Identify correct syntax

   std::vector<int> heights = {0,1,0,2,1,0,1,3,2,1,2,1}; uses correct syntax with std::vector<int> and brace initialization.

3. Final Answer:

   std::vector<int> heights = {0,1,0,2,1,0,1,3,2,1,2,1}; -> Option B

4. Quick Check:

   Correct vector init uses braces and type [OK]

Hint: Use curly braces {} for vector initialization [OK]

Common Mistakes:

• Omitting std:: before vector
• Using parentheses instead of braces
• Not specifying vector type

3. What is the output of the following C++ code snippet that calculates trapped rain water?

std::vector<int> height = {0,1,0,2,1,0,1,3,2,1,2,1};
int n = height.size();
int left_max = 0, right_max = 0, left = 0, right = n - 1, water = 0;
while (left < right) {
  if (height[left] < height[right]) {
    if (height[left] >= left_max) left_max = height[left];
    else water += left_max - height[left];
    left++;
  } else {
    if (height[right] >= right_max) right_max = height[right];
    else water += right_max - height[right];
    right--;
  }
}
std::cout << water << std::endl;

medium

A. 6

B. 5

C. 7

D. 8

Solution

1. Step 1: Understand two-pointer approach

   Two pointers move inward, tracking max heights on left and right, accumulating trapped water where current height is less than max.

2. Step 2: Trace the code logic on given input

   For input {0,1,0,2,1,0,1,3,2,1,2,1}, the total trapped water calculated is 6 units.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Two-pointer trapped water = 6 [OK]

Hint: Use two pointers and track max heights from both ends [OK]

Common Mistakes:

• Not updating left_max or right_max correctly
• Adding negative water amounts
• Confusing left and right pointer moves

4. Identify the bug in this C++ code snippet for trapping rain water and select the fix:

std::vector<int> height = {4,2,0,3,2,5};
int n = height.size();
int left_max = 0, right_max = 0, left = 0, right = n - 1, water = 0;
while (left <= right) {
  if (height[left] < height[right]) {
    if (height[left] > left_max) left_max = height[left];
    else water += left_max - height[left];
    left++;
  } else {
    if (height[right] > right_max) right_max = height[right];
    else water += right_max - height[right];
    right--;
  }
}
std::cout << water << std::endl;

medium

A. Initialize left_max and right_max to height[0] and height[n-1] respectively

B. Use '>=' instead of '>' when updating left_max and right_max

C. Change while condition to 'left < right' to avoid double counting

D. No bug, code is correct

Solution

1. Step 1: Analyze loop condition impact

   The loop uses 'left <= right', which causes the middle element to be processed twice, leading to incorrect water calculation.

2. Step 2: Fix loop condition

   Changing condition to 'left < right' ensures each element is processed once, preventing double counting.

3. Final Answer:

   Change while condition to 'left < right' to avoid double counting -> Option C

4. Quick Check:

   Loop condition 'left < right' fixes double count bug [OK]

Hint: Use strict less than in two-pointer loops [OK]

Common Mistakes:

• Using <= instead of < in while loop
• Not updating max heights correctly
• Assuming no bug without testing

5. Given an array representing heights of bars, how would you modify the standard trapping rain water algorithm to handle very large input efficiently without extra space?

hard

A. Use recursion to calculate trapped water for each bar

B. Precompute left_max and right_max arrays and sum trapped water using extra arrays

C. Sort the array and calculate trapped water based on sorted positions

D. Use two pointers from both ends, track max heights, and calculate trapped water on the fly

Solution

1. Step 1: Understand space constraints

   Precomputing arrays uses extra space O(n), which is costly for very large inputs.

2. Step 2: Use two-pointer approach for O(1) space

   Two pointers from both ends track max heights and calculate trapped water without extra arrays, making it efficient.

3. Final Answer:

   Use two pointers from both ends, track max heights, and calculate trapped water on the fly -> Option D

4. Quick Check:

   Two-pointer approach = efficient O(1) space [OK]

Hint: Two pointers save space by avoiding extra arrays [OK]

Common Mistakes:

• Using extra arrays causing high memory use
• Sorting changes bar order, invalid for trapping water
• Recursion adds overhead and complexity