Floor and Ceil in Sorted Array in DSA Go - Execution Trace

Concept Flow - Floor and Ceil in Sorted Array

Start with sorted array and target

↓

Initialize low=0, high=n-1

↓

While low <= high

↓

Calculate mid = (low+high)/2

↓

Compare arr[mid

↓

Update floor

↓

Move low=mid+1

↓

Repeat loop

↓

Return floor and ceil values

We use binary search to find floor and ceil by narrowing the search range and updating floor or ceil pointers based on comparisons.

Execution Sample

DSA Go

arr := []int{1, 3, 5, 7, 9}
target := 6
floor, ceil := -1, -1
low, high := 0, len(arr)-1
for low <= high {
  mid := (low + high) / 2
  if arr[mid] == target {
    floor, ceil = arr[mid], arr[mid]
    break
  } else if arr[mid] < target {
    floor = arr[mid]
    low = mid + 1
  } else {
    ceil = arr[mid]
    high = mid - 1
  }
}

This code finds floor and ceil of 6 in a sorted array using binary search.

Execution Table

Step	Operation	mid	arr[mid]	Comparison	floor	ceil	low	high	Visual State
1	Initialize	-	-	-	-1	-1	0	4	[1, 3, 5, 7, 9]
2	Calculate mid	2	5	5 < 6	5	-1	3	4	[1, 3, 5, 7, 9]
3	Calculate mid	3	7	7 > 6	5	7	3	2	[1, 3, 5, 7, 9]
4	Exit loop	-	-	low(3) > high(2)	5	7	3	2	[1, 3, 5, 7, 9]

💡 Loop ends because low (3) > high (2), floor is 5, ceil is 7

Variable Tracker

Variable	Start	After Step 2	After Step 3	Final
floor	-1	5	5	5
ceil	-1	-1	7	7
low	0	3	3	3
high	4	4	2	2
mid	-	2	3	-

Key Moments - 3 Insights

Why do we update floor when arr[mid] < target?

Because arr[mid] is less than target and could be the closest smaller or equal value, so we save it as floor (see step 2 in execution_table).

Why do we update ceil when arr[mid] > target?

Because arr[mid] is greater than target and could be the closest greater or equal value, so we save it as ceil (see step 3 in execution_table).

Why does the loop stop when low > high?

Because the search space is empty, meaning we have found the closest floor and ceil values (see step 4 in execution_table).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of floor after step 2?

A 5

B -1

C 7

D 3

💡 Hint

Check the 'floor' column in row for step 2 in execution_table.

At which step does the condition low <= high become false?

A Step 2

B Step 3

C Step 4

D Never

💡 Hint

Look at the 'low' and 'high' values in execution_table rows and see when low > high.

If the target was 7, what would floor and ceil be after the loop?

A floor=5, ceil=7

B floor=7, ceil=7

C floor=7, ceil=9

D floor=3, ceil=5

💡 Hint

If arr[mid] == target, floor and ceil both equal target (see code in execution_sample).

Concept Snapshot

Floor and Ceil in Sorted Array:
- Use binary search with low=0, high=n-1
- Calculate mid = (low+high)/2
- If arr[mid] == target, floor=ceil=arr[mid]
- If arr[mid] < target, floor=arr[mid], move low=mid+1
- If arr[mid] > target, ceil=arr[mid], move high=mid-1
- Loop ends when low > high, return floor and ceil

Full Transcript

We find floor and ceil in a sorted array by binary searching for the target. We keep track of floor as the largest value less than or equal to target, and ceil as the smallest value greater than or equal to target. At each step, we calculate mid and compare arr[mid] with target. If equal, floor and ceil are the same. If arr[mid] is less, update floor and move low up. If arr[mid] is greater, update ceil and move high down. The loop ends when low passes high, returning the closest floor and ceil values.

Practice

1. What does the floor of a number mean in a sorted array?

easy

A. The greatest element less than or equal to the number

B. The smallest element greater than the number

C. The average of all elements less than the number

D. The smallest element less than the number

Solution

1. Step 1: Understand the floor definition

   The floor of a number in a sorted array is the greatest element that is less than or equal to the given number.

2. Step 2: Compare options with definition

   The greatest element less than or equal to the number matches this definition exactly, while others describe different concepts.

3. Final Answer:

   The greatest element less than or equal to the number -> Option A

4. Quick Check:

   Floor = greatest ≤ number [OK]

Hint: Floor is always ≤ number, pick greatest such element [OK]

Common Mistakes:

• Confusing floor with ceil
• Thinking floor is strictly less than number
• Mixing floor with average or median

2. Which Go code snippet correctly declares a function to find the floor in a sorted array?

easy

A. func floor(arr int[], x int) int { /* code */ }

B. func floor(arr []int, x int) int { /* code */ }

C. function floor(arr []int, x int) int { /* code */ }

D. func floor(arr []int, x int) void { /* code */ }

Solution

1. Step 1: Check Go function syntax

   In Go, functions are declared with func, parameter types follow names, and return type is after parameters.

2. Step 2: Validate each option

   func floor(arr []int, x int) int { /* code */ } uses correct Go syntax: func floor(arr []int, x int) int. func floor(arr int[], x int) int { /* code */ } uses incorrect array syntax, C uses wrong keyword, D has wrong return type.

3. Final Answer:

   func floor(arr []int, x int) int { /* code */ } -> Option B

4. Quick Check:

   Go func syntax = func floor(arr []int, x int) int { /* code */ } [OK]

Hint: Go arrays use []type, func keyword, return type after params [OK]

Common Mistakes:

• Using 'function' keyword instead of 'func'
• Incorrect array type syntax
• Wrong return type declaration

3. What is the output of the following Go code snippet?

arr := []int{1, 3, 5, 7, 9}
x := 6
floor := -1
for _, val := range arr {
    if val <= x {
        floor = val
    } else {
        break
    }
}
fmt.Println(floor)

medium

A. 5

B. 6

C. 7

D. -1

Solution

1. Step 1: Trace the loop over array elements

   The loop checks each element: 1 ≤ 6 (floor=1), 3 ≤ 6 (floor=3), 5 ≤ 6 (floor=5), 7 ≤ 6? No, break loop.

2. Step 2: Determine final floor value

   Last floor assigned is 5, so printed output is 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Floor of 6 in arr = 5 [OK]

Hint: Floor is last element ≤ x before break [OK]

Common Mistakes:

• Assuming floor equals x if not present
• Not breaking loop early
• Confusing floor with ceil

4. Identify the bug in this Go code to find the ceil of x in a sorted array arr:

func ceil(arr []int, x int) int {
    ceil := -1
    for _, val := range arr {
        if val >= x {
            ceil = val
        }
    }
    return ceil
}

medium

A. It uses wrong comparison operator

B. It initializes ceil with -1 incorrectly

C. It returns ceil before loop ends

D. It does not break after finding first valid ceil

Solution

1. Step 1: Analyze loop behavior for ceil

   The loop updates ceil whenever val ≥ x, but does not stop after first valid ceil, so it ends with the last such value, which may not be the smallest ceil.

2. Step 2: Identify correct fix

   To find the smallest ceil, the loop should break immediately after first val ≥ x is found.

3. Final Answer:

   It does not break after finding first valid ceil -> Option D

4. Quick Check:

   Ceil needs first ≥ x, break loop [OK]

Hint: Break loop on first val ≥ x for correct ceil [OK]

Common Mistakes:

• Not breaking loop early
• Confusing ceil with floor logic
• Incorrect initialization of ceil

5. Given a sorted array arr := []int{2, 4, 6, 8, 10} and a number x = 5, write a Go function that returns both floor and ceil of x. What is the correct output of floor and ceil?

hard

A. Floor: -1, Ceil: -1

B. Floor: 6, Ceil: 4

C. Floor: 4, Ceil: 6

D. Floor: 5, Ceil: 5

Solution

1. Step 1: Find floor of 5 in arr

   Floor is greatest element ≤ 5. Elements ≤ 5 are 2 and 4, greatest is 4.

2. Step 2: Find ceil of 5 in arr

   Ceil is smallest element ≥ 5. Elements ≥ 5 are 6, 8, 10, smallest is 6.

3. Final Answer:

   Floor: 4, Ceil: 6 -> Option C

4. Quick Check:

   Floor ≤ x ≤ Ceil = 4 ≤ 5 ≤ 6 [OK]

Hint: Floor ≤ x ≤ Ceil in sorted array [OK]

Common Mistakes:

• Swapping floor and ceil values
• Assuming floor or ceil equals x if not present
• Returning -1 when valid floor or ceil exists