Container With Most Water in DSA Go - Time & Space Complexity

Time Complexity: Container With Most Water

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the container with the most water changes as the number of lines grows.

How does the number of steps grow when we have more lines to check?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

func maxArea(height []int) int {
    left, right := 0, len(height)-1
    maxArea := 0
    for left < right {
        width := right - left
        var h int
        if height[left] < height[right] {
            h = height[left]
            left++
        } else {
            h = height[right]
            right--
        }
        area := width * h
        if area > maxArea {
            maxArea = area
        }
    }
    return maxArea
}

This code finds the largest area between two lines by moving pointers from both ends towards the center.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: A single loop moving two pointers inward.
• How many times: The loop runs at most once for each element, so about n times where n is the number of lines.

How Execution Grows With Input

As the number of lines increases, the steps grow roughly the same as the number of lines.

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

Pattern observation: The steps increase linearly with input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the largest container grows in direct proportion to the number of lines.

Common Mistake

[X] Wrong: "Checking every pair of lines is needed, so time is O(n²)."

[OK] Correct: The two-pointer method smartly skips unnecessary pairs, so it only needs to check each line once.

Interview Connect

Understanding this approach shows you can optimize by avoiding needless checks, a key skill in problem solving.

Self-Check

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

Practice

1. What does the problem Container With Most Water ask you to find given an array of heights?

easy

A. The sum of all heights

B. The minimum height in the array

C. The maximum area of water that can be contained between two lines

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 calculate

   We calculate the area between two lines as the width between them times the smaller height.

3. Final Answer:

   The maximum area of water that can be contained between two lines -> Option C

4. Quick Check:

   Container with most water = maximum area [OK]

Hint: Focus on max area between two heights, not sum or count [OK]

Common Mistakes:

• Confusing max area with sum of heights
• Ignoring the smaller height when calculating area
• Thinking about counting lines instead of area

2. Which of the following Go code snippets correctly declares a slice of heights for the Container With Most Water problem?

easy

A. heights := []int{1, 8, 6, 2, 5, 4, 8, 3, 7}

B. heights := [1, 8, 6, 2, 5, 4, 8, 3, 7]

C. heights := (1, 8, 6, 2, 5, 4, 8, 3, 7)

D. heights := {1, 8, 6, 2, 5, 4, 8, 3, 7}

Solution

1. Step 1: Recall Go slice declaration syntax

   In Go, slices are declared using []type{values}, for example []int{1,2,3}.

2. Step 2: Check each option

   heights := []int{1, 8, 6, 2, 5, 4, 8, 3, 7} uses []int{...} which is correct. Options A, B, C use invalid syntax for slices.

3. Final Answer:

   heights := []int{1, 8, 6, 2, 5, 4, 8, 3, 7} -> Option A

4. Quick Check:

   Slice declaration uses []type{values} [OK]

Hint: Use []int{...} to declare slices in Go [OK]

Common Mistakes:

• Using parentheses or braces instead of [] for slices
• Confusing arrays and slices syntax
• Missing type declaration in slice

3. What is the output of the following Go code snippet that implements the Container With Most Water logic?

package main
import "fmt"
func maxArea(height []int) int {
    left, right := 0, len(height)-1
    maxArea := 0
    for left < right {
        width := right - left
        var h int
        if height[left] < height[right] {
            h = height[left]
            left++
        } else {
            h = height[right]
            right--
        }
        area := width * h
        if area > maxArea {
            maxArea = area
        }
    }
    return maxArea
}
func main() {
    heights := []int{1,8,6,2,5,4,8,3,7}
    fmt.Println(maxArea(heights))
}

medium

A. 49

B. 56

C. 48

D. 64

Solution

1. Step 1: Understand the two-pointer approach

   Start with left=0 and right=8, calculate area and move the pointer with smaller height inward.

2. Step 2: Trace the iterations

   Initial area: min(1,7)*8=8 Move left to 1 Next area: min(8,7)*7=49 Move right to 7 Next area: min(8,3)*6=18 Move right to 6 Next area: min(8,8)*5=40 Move right to 5 Next area: min(8,4)*4=16 Move right to 4 Next area: min(8,5)*3=15 Move right to 3 Next area: min(8,2)*2=4 Move right to 2 Next area: min(8,6)*1=1 Move left to 2 Max area found is 49.

3. Final Answer:

   49 -> Option A

4. Quick Check:

   Max area calculated by two-pointer = 49 [OK]

Hint: Use two pointers moving inward, track max area [OK]

Common Mistakes:

• Calculating area with max height instead of min
• Not moving the pointer with smaller height
• Confusing width calculation

4. Identify the bug in this Go function for Container With Most Water and select the fix:

func maxArea(height []int) int {
    left, right := 0, len(height)-1
    maxArea := 0
    for left <= right {
        width := right - left
        var h int
        if height[left] < height[right] {
            h = height[left]
            left++
        } else {
            h = height[right]
            right--
        }
        area := width * h
        if area > maxArea {
            maxArea = area
        }
    }
    return maxArea
}

medium

A. Use maxArea = area without checking if area > maxArea

B. Initialize maxArea to -1 instead of 0

C. Increment both left and right pointers inside the loop

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

Solution

1. Step 1: Analyze the loop condition

   The loop runs while left <= right, which allows left and right to be equal, causing width to be zero or negative.

2. Step 2: Fix the loop condition

   Changing to left < right ensures width is always positive and valid for area calculation.

3. Final Answer:

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

4. Quick Check:

   Loop must run while left < right [OK]

Hint: Use strict less than for pointers to avoid zero width [OK]

Common Mistakes:

• Using <= instead of < causing invalid area
• Updating both pointers at once
• Not checking if area is greater before updating maxArea

5. Given the heights slice []int{4, 3, 2, 1, 4}, what is the maximum water container area? Use the two-pointer approach and explain why the maximum area is found at those positions.

hard

A. 12

B. 16

C. 8

D. 6

Solution

1. Step 1: Initialize pointers and calculate areas

   Start with left=0 (height=4), right=4 (height=4), width=4, area=4*4=16.

2. Step 2: Move pointers and check other areas

   Since heights[left] == heights[right], move right pointer left: Now left=0, right=3 (height=1), width=3, area=min(4,1)*3=3 Move right to 2 (height=2), area= min(4,2)*2=4 Move right to 1 (height=3), area= min(4,3)*1=3 Max area remains 16.

3. Final Answer:

   16 -> Option B

4. Quick Check:

   Max area between positions 0 and 4 = 16 [OK]

Hint: Max area often at widest distance with min height [OK]

Common Mistakes:

• Not considering both ends first
• Moving wrong pointer when heights are equal
• Calculating area incorrectly with wrong width