How do you solve Height Checker on LeetCode?

Height Checker (LeetCode #1051) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Sorting helps us find the expected order quickly.

What is the time complexity of Height Checker?

The optimal solution for Height Checker runs in O(n) time and uses O(1) space. The counting step runs in O(n) time, and since the count array size is constant (101), the space complexity is O(1).

What algorithm or data structure is used to solve Height Checker?

Height Checker uses the following concepts: Array, Sorting, Counting Sort. The recommended patterns to study are: Counting Sort, Array.

What are the interview tips for Height Checker?

Explain your thought process clearly while coding. Consider edge cases, such as all students being of the same height. Practice explaining your solution to someone else to solidify your understanding.

What are common mistakes when solving Height Checker?

Not considering the range of heights when using arrays. Overlooking the need to compare the original and expected arrays correctly.

Height Checker — LeetCode #1051 (Easy)

Tags: Array, Sorting, Counting Sort

Related patterns: Counting Sort, Array

Brute Force approach

Time complexity: O(n²). Space complexity: O(n).

We can simply sort the heights to find the expected order and then compare each student's height with the expected height. This is straightforward but inefficient.

The sorting step takes O(n log n) time, and the comparison takes O(n), making the overall complexity O(n log n). However, if we consider the comparison as part of the sorting, we can say it's O(n²) in a naive implementation.

Step 1: Create a copy of the heights array and sort it to get the expected order.
Step 2: Initialize a counter to zero to keep track of mismatches.
Step 3: Iterate through the original heights array and the sorted array simultaneously, incrementing the counter whenever the heights do not match.

1. heights = [1, 1, 4, 2, 1, 3], expected = sorted(heights) = [1, 1, 1, 2, 3, 4] 2. Initialize count = 0. 3. Compare heights[0] (1) with expected[0] (1) -> match, count = 0. 4. Compare heights[1] (1) with expected[1] (1) -> match, count = 0. 5. Compare heights[2] (4) with expected[2] (1) -> mismatch, count = 1. 6. Compare heights[3] (2) with expected[3] (2) -> match, count = 1. 7. Compare heights[4] (1) with expected[4] (3) -> mismatch, count = 2. 8. Compare heights[5] (3) with expected[5] (4) -> mismatch, count = 3. Final count = 3.

Optimal Solution approach

Time complexity: O(n). Space complexity: O(1).

Instead of sorting the array, we can count the occurrences of each height and then determine how many heights are out of place based on their expected positions. This is more efficient.

The counting step runs in O(n) time, and since the count array size is constant (101), the space complexity is O(1).

Step 1: Create a count array of size 101 (since heights range from 1 to 100) to count occurrences of each height.
Step 2: Populate the count array by iterating through the heights array.
Step 3: Initialize a counter and iterate through the count array to determine the expected heights and compare with the original heights, counting mismatches.

1. heights = [1, 1, 4, 2, 1, 3], count = [0, 3, 1, 1, 1, 0, ..., 0]. 2. Initialize mismatches = 0, index = 0. 3. For height = 1: count[1] = 3, check heights[0] (1) -> match, index = 1; check heights[1] (1) -> match, index = 2; check heights[2] (4) -> mismatch, mismatches = 1, index = 3. 4. For height = 2: count[2] = 1, check heights[3] (2) -> match, index = 4. 5. For height = 3: count[3] = 1, check heights[4] (1) -> mismatch, mismatches = 2, index = 5. 6. For height = 4: count[4] = 1, check heights[5] (3) -> mismatch, mismatches = 3. Final mismatches = 3.

Key Insights

Sorting helps us find the expected order quickly.
Counting occurrences can help us avoid sorting and improve efficiency.

Common Mistakes

Not considering the range of heights when using arrays.
Overlooking the need to compare the original and expected arrays correctly.

Interview Tips

Explain your thought process clearly while coding.
Consider edge cases, such as all students being of the same height.
Practice explaining your solution to someone else to solidify your understanding.

#1051

Height Checker

Easy

Array↗Sorting↗Counting Sort↗Counting Sort↗Array↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of sorting the array, we can count the occurrences of each height and then determine how many heights are out of place based on their expected positions. This is more efficient.

⚙️

Algorithm

3 steps

1Step 1: Create a count array of size 101 (since heights range from 1 to 100) to count occurrences of each height.
2Step 2: Populate the count array by iterating through the heights array.
3Step 3: Initialize a counter and iterate through the count array to determine the expected heights and compare with the original heights, counting mismatches.

solution.py13 lines

1def heightChecker(heights):
2    count = [0] * 101
3    for h in heights:
4        count[h] += 1
5    mismatches = 0
6    index = 0
7    for height in range(1, 101):
8        while count[height] > 0:
9            if heights[index] != height:
10                mismatches += 1
11            index += 1
12            count[height] -= 1
13    return mismatches

ℹ

Complexity note: The counting step runs in O(n) time, and since the count array size is constant (101), the space complexity is O(1).

1Sorting helps us find the expected order quickly.
2Counting occurrences can help us avoid sorting and improve efficiency.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.