# Leetcode – Height Checker Solution

This article will dissect the Height Checker problem, shed light on its underlying principles, present diverse solution strategies, and offer Python implementations.

1. Problem Statement
2. Grasping the Essence of the Problem
3. Unveiling Solution Strategies
4. Python Implementations
5. Analysis of Time Complexity
6. Conclusion and Takeaways

## 1. Problem Statement:

Students are asked to stand in non-decreasing order of heights for an annual photo. Return the minimum number of students that must move in order to ensure all students are standing in non-decreasing order of height.

### Example:

Input: heights = [1,1,4,2,1,3]
Output: 3

## 2. Grasping the Essence of the Problem:

The problem, in essence, asks us to find out how many students are out of place if we were to rearrange the array in a sorted order. In the given example, the sorted order is [1,1,1,2,3,4]. The students at positions 3, 5, and 6 in the original order are out of place, hence the answer is 3.

## 3. Unveiling Solution Strategies:

1. Sorting Approach: Create a sorted version of the heights array and compare the two arrays to count the differences.
2. Counting Sort Approach: Use a variation of counting sort to track the frequency of each height. This works especially well if the range of heights is limited.

## 4. Python Implementations:

### 1. Sorting Approach:

This is the most intuitive approach. After sorting the array, a simple iteration comparing the original and sorted array can give us the number of students out of place.

def heightChecker_sorting(heights):
sorted_heights = sorted(heights)
count = 0
for i in range(len(heights)):
if heights[i] != sorted_heights[i]:
count += 1
return count

### 2. Counting Sort Approach:

If we know the range of the heights (e.g., between 1 and 100), we can use a counting array to store the frequency of each height. This approach avoids the O(nlog⁡n) time complexity of sorting and provides a faster solution.

def heightChecker_counting(heights):
height_to_freq =  * 101  # As per problem constraints, height is between 1 and 100
for height in heights:
height_to_freq[height] += 1

count = 0
i = 0

for height in range(1, 101):
while height_to_freq[height] > 0:
if heights[i] != height:
count += 1
i += 1
height_to_freq[height] -= 1

return count

## 5. Analysis of Time Complexity:

• Sorting Approach: The main overhead is the sorting operation, which has a time complexity of O(nlog⁡n). The subsequent iteration has a linear time complexity of O(n), but the dominant factor remains O(nlog⁡n).
• Counting Sort Approach: This method involves initializing the counting array (which has a constant size based on height constraints) and two linear traversals. Thus, its overall time complexity is O(n).

## 6. Conclusion and Takeaways:

The “Height Checker” problem offers two valuable lessons:

1. Multiple Avenues: For many algorithmic challenges, there is often more than one solution. The key lies in discerning which is the most efficient for the given constraints.
2. Know Your Data: Understanding the nature of the data (e.g., constraints on the range of values) can often guide you to more optimized solutions.

In conclusion, problems like “Height Checker” not only enhance our algorithmic skills but also make us appreciate the application of these techniques in real-world scenarios. Whether you are a novice or an experienced coder, tackling such challenges continuously hones your problem-solving abilities.