Trapping Rain Water Problem in DSA Java - Time & Space Complexity

Time Complexity: Trapping Rain Water Problem

O(n)

Understanding Time Complexity

We want to understand how the time needed to solve the Trapping Rain Water problem changes as the input size grows.

Specifically, how the number of operations depends on the number of bars in the elevation map.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int trap(int[] height) {
    int left = 0, right = height.length - 1;
    int leftMax = 0, rightMax = 0;
    int trappedWater = 0;
    while (left <= right) {
        if (height[left] < height[right]) {
            if (height[left] > leftMax) leftMax = height[left];
            else trappedWater += leftMax - height[left];
            left++;
        } else {
            if (height[right] > rightMax) rightMax = height[right];
            else trappedWater += rightMax - height[right];
            right--;
        }
    }
    return trappedWater;
}
    

This code calculates how much water can be trapped between bars represented by the array.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves two pointers from both ends towards the center.
• How many times: Each element is visited at most once by either the left or right pointer, so the loop runs roughly n times where n is the array length.

How Execution Grows With Input

As the number of bars increases, the number of steps grows roughly the same as the number of bars.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The operations grow linearly with input size; doubling the bars roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to solve the problem grows directly in proportion to the number of bars.

Common Mistake

[X] Wrong: "Because there are nested conditions, the time complexity must be quadratic O(n²)."

[OK] Correct: The loop moves pointers inward without repeating the same elements multiple times, so each element is processed once, keeping the time linear.

Interview Connect

Understanding this linear time solution shows you can optimize problems by clever pointer use instead of brute force, a skill valued in real coding challenges.

Self-Check

"What if we used a nested loop to check every pair of bars instead of two pointers? How would the time complexity change?"

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 maximum height of the bars in the array

C. The minimum height of the bars in the array

D. The number of bars that are taller than their neighbors

Solution

1. Step 1: Understand the problem statement

   The problem asks to calculate how much water can be trapped between bars of different heights after it rains.

2. Step 2: Identify the goal

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

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 volume [OK]

Hint: Focus on trapped water volume, not bar heights [OK]

Common Mistakes:

• Confusing trapped water with max bar height
• Counting bars taller than neighbors instead of water
• Calculating minimum height instead of trapped water

2. Which of the following Java code snippets correctly initializes an array of heights for the Trapping Rain Water problem?

easy

A. int[] heights = new int(6) {4, 2, 0, 3, 2, 5};

B. int heights = [4, 2, 0, 3, 2, 5];

C. int[] heights = (4, 2, 0, 3, 2, 5);

D. int[] heights = {4, 2, 0, 3, 2, 5};

Solution

1. Step 1: Check Java array initialization syntax

   In Java, arrays are initialized with curly braces {} and type declaration like int[].

2. Step 2: Validate each option

   int[] heights = {4, 2, 0, 3, 2, 5}; uses correct syntax: int[] heights = {4, 2, 0, 3, 2, 5}; Options B, C, and D use invalid syntax.

3. Final Answer:

   int[] heights = {4, 2, 0, 3, 2, 5}; -> Option D

4. Quick Check:

   Java array init uses {} with type[] [OK]

Hint: Use curly braces {} with type[] to initialize arrays [OK]

Common Mistakes:

• Using square brackets [] without type[]
• Using parentheses () instead of braces {}
• Trying to initialize array with new int(size) and values together

3. What is the output of the following Java code snippet for trapping rain water?

int[] height = {0,1,0,2,1,0,1,3,2,1,2,1};
System.out.println(trap(height));

public static int trap(int[] height) {
    int left = 0, right = height.length - 1;
    int leftMax = 0, rightMax = 0, water = 0;
    while (left < right) {
        if (height[left] < height[right]) {
            if (height[left] >= leftMax) leftMax = height[left];
            else water += leftMax - height[left];
            left++;
        } else {
            if (height[right] >= rightMax) rightMax = height[right];
            else water += rightMax - height[right];
            right--;
        }
    }
    return water;
}

medium

A. 5

B. 6

C. 7

D. 8

Solution

1. Step 1: Understand two-pointer approach

   We use two pointers left and right, track max heights on both sides, and add trapped water where possible.

2. Step 2: Trace the code with given input

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

3. Final Answer:

   6 -> Option B

4. Quick Check:

   Two-pointer trap calculation = 6 [OK]

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

Common Mistakes:

• Not updating leftMax and rightMax correctly
• Adding negative water amounts
• Confusing left and right pointer movements

4. Identify the bug in this Java method for trapping rain water:

public static int trap(int[] height) {
    int left = 0, right = height.length - 1;
    int leftMax = 0, rightMax = 0, water = 0;
    while (left <= right) {
        if (height[left] < height[right]) {
            if (height[left] > leftMax) leftMax = height[left];
            else water += leftMax - height[left];
            left++;
        } else {
            if (height[right] > rightMax) rightMax = height[right];
            else water += rightMax - height[right];
            right--;
        }
    }
    return water;
}

medium

A. The water calculation should add height[left] instead of subtracting

B. The comparison should be 'height[left] <= height[right]' instead of '<'

C. The loop condition should be 'left < right' instead of 'left <= right'

D. leftMax and rightMax should be initialized to 1 instead of 0

Solution

1. Step 1: Analyze loop condition impact

   The loop uses 'left <= right', which causes double counting when left == right.

2. Step 2: Correct loop condition

   Changing to 'left < right' prevents overlapping pointers and avoids incorrect water calculation.

3. Final Answer:

   The loop condition should be 'left < right' instead of 'left <= right' -> Option C

4. Quick Check:

   Loop condition 'left < right' avoids overlap [OK]

Hint: Use strict less than in loop to avoid pointer overlap [OK]

Common Mistakes:

• Using <= causes double counting at same index
• Changing comparison operators incorrectly
• Initializing max heights incorrectly

5. Given the height array {3, 0, 2, 0, 4}, what is the total trapped rain water? Explain how the trapped water is calculated step-by-step.

hard

A. 7

B. 6

C. 5

D. 8

Solution

1. Step 1: Calculate leftMax and rightMax arrays

   leftMax = [3, 3, 3, 3, 4], rightMax = [4, 4, 4, 4, 4]

2. Step 2: Calculate trapped water at each index

   Water trapped = min(leftMax[i], rightMax[i]) - height[i] = [0, 3, 1, 3, 0] Sum = 0 + 3 + 1 + 3 + 0 = 7

3. Step 3: Re-check calculation for correctness

   At index 1: min(3,4)-0=3 At index 2: min(3,4)-2=1 At index 3: min(3,4)-0=3 Sum = 7 units trapped water

4. Final Answer:

   7 -> Option A

5. Quick Check:

   Sum trapped water = 7 units [OK]

Hint: Use leftMax and rightMax arrays to find trapped water [OK]

Common Mistakes:

• Forgetting to take minimum of leftMax and rightMax
• Subtracting height incorrectly
• Not summing trapped water at all indices