Trapping Rain Water Problem in DSA Go - 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 Go code that calculates trapped rain water.

func trap(height []int) int {
    n := len(height)
    left, right := 0, n-1
    leftMax, rightMax := 0, 0
    water := 0
    for 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
}
    

This code uses two pointers moving inward to calculate how much water is trapped between bars.

Identify Repeating Operations

Look for loops or repeated steps in the code.

• Primary operation: The single for-loop that moves two pointers from both ends towards the center.
• How many times: The loop runs at most n times, where n is the number of bars.

How Execution Grows With Input

As the number of bars increases, the loop runs once per bar.

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

Pattern observation: The number of steps grows linearly with input size.

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 two pointers, the time is doubled or quadratic."

[OK] Correct: Both pointers move inward without repeating positions, so total steps are still proportional to n, not n².

Interview Connect

Understanding this linear time approach shows you can optimize problems by clever pointer use instead of nested loops, a valuable skill in interviews.

Self-Check

"What if we used a nested loop to check water trapped at each bar? 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 minimum height of the bars

B. The maximum height of the bars

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

D. The number of bars in the array

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 rain.

2. Step 2: Identify the goal

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

3. Final Answer:

   The total amount of water trapped between bars -> Option C

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 bar heights
• Counting bars instead of water volume
• Ignoring gaps between bars

2. Which of the following Go code snippets correctly initializes an integer slice to represent bar heights for the Trapping Rain Water problem?

easy

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

B. heights := int[]{4, 2, 0, 3, 2, 5}

C. heights := [6]int{4, 2, 0, 3, 2, 5}

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

Solution

1. Step 1: Recall Go slice initialization syntax

   In Go, slices are initialized with []int{values}, arrays with [size]int{values}.

2. Step 2: Identify correct slice syntax

   heights := []int{4, 2, 0, 3, 2, 5} uses []int{...} which is correct for a slice. heights := [6]int{4, 2, 0, 3, 2, 5} is an array, also valid but less flexible.

3. Final Answer:

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

4. Quick Check:

   Slice syntax = []int{...} [OK]

Hint: Use []int{} for slices, [n]int{} for arrays [OK]

Common Mistakes:

• Using int[] like in other languages
• Trying to use new keyword for slices
• Confusing arrays and slices syntax

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

package main
import "fmt"
func trap(height []int) int {
    left, right := 0, len(height)-1
    leftMax, rightMax := 0, 0
    water := 0
    for 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
}
func main() {
    fmt.Println(trap([]int{0,1,0,2,1,0,1,3,2,1,2,1}))
}

medium

A. 6

B. 5

C. 7

D. 8

Solution

1. Step 1: Understand two-pointer approach

   The code uses two pointers moving inward, tracking max heights on left and right to calculate trapped water.

2. Step 2: Trace the example input

   For input [0,1,0,2,1,0,1,3,2,1,2,1], the trapped water calculated by this method is 6 units.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Two-pointer trap water = 6 [OK]

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

Common Mistakes:

• Not updating max heights correctly
• Adding negative water amounts
• Confusing left and right pointer moves

4. Identify the bug in the following Go code for trapping rain water and select the correct fix:

func trap(height []int) int {
    left, right := 0, len(height)-1
    leftMax, rightMax := 0, 0
    water := 0
    for 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. Change '>' to '>=' in max height comparisons

B. Change loop condition to 'left < right' instead of 'left <= right'

C. Initialize water to 1 instead of 0

D. Swap left and right pointer initial values

Solution

1. Step 1: Analyze loop condition impact

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

2. Step 2: Fix loop condition

   Changing to 'left < right' prevents overlap and corrects water calculation.

3. Final Answer:

   Change loop condition to 'left < right' instead of 'left <= right' -> Option B

4. Quick Check:

   Loop condition fix = 'left < right' [OK]

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

Common Mistakes:

• Using <= causes double counting
• Changing comparison operators incorrectly
• Incorrect pointer initialization

5. Given the heights array [5, 2, 1, 2, 1, 5], what is the total trapped rain water? Use the two-pointer approach to solve.

hard

A. 10

B. 12

C. 8

D. 14

Solution

1. Step 1: Initialize pointers and max heights

   left=0, right=5, leftMax=0, rightMax=0, water=0.

2. Step 2: Move pointers inward calculating trapped water

   - left=0 (height=5) >= leftMax(0), leftMax=5, left++ - left=1 (2) < right=5 (5), height[left]

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Two-pointer sum water = 10 [OK]

Hint: Sum min(leftMax, rightMax) - height at each step [OK]

Common Mistakes:

• Adding water twice for same positions
• Not updating max heights correctly
• Stopping loop too early