LeetCode 1051: Height Checker

Problem Description

A school is trying to take an annual photo of all the students. The students are asked to stand in a single file line in non-decreasing order by height. Let this ordering be represented by the integer array expected where expected[i] is the expected height of the i^th student in line.

You are given an integer array heights representing the current order that the students are standing in. Each heights[i] is the height of the i^th student in line (0-indexed).

Return the number of indices where heights[i] != expected[i].

Example 1:

Input: heights = [1,1,4,2,1,3]
Output: 3
Explanation: 
heights:  [1,1,4,2,1,3]
expected: [1,1,1,2,3,4]
Indices 2, 4, and 5 do not match.

Example 2:

Input: heights = [5,1,2,3,4]
Output: 5
Explanation:
heights:  [5,1,2,3,4]
expected: [1,2,3,4,5]
All indices do not match.

Example 3:

Input: heights = [1,2,3,4,5]
Output: 0
Explanation:
heights:  [1,2,3,4,5]
expected: [1,2,3,4,5]
All indices match.

Constraints:

1 <= heights.length <= 100
1 <= heights[i] <= 100

Difficulty: Easy

Tags: array, sorting, counting sort

Rating: 93.46%

Complexity Analysis

The solution has the following complexity characteristics:

Time Complexity: O(n+r) where n is the number of elements in the input array and r is the range of the elements
Space Complexity: O(r)

Solution

Here’s my Python solution to this problem:

class Solution:
    def heightChecker(self, heights: List[int]) -> int:
        minv, maxv = min(heights), max(heights)
        r = maxv-minv+1

        count = [0] * r
        for h in heights:
            count[h-minv] += 1

        s = 0
        h = 0
        for i in range(r):
            while count[i] > 0:
                if heights[h] != i + minv:
                    s += 1
                h += 1
                count[i] -= 1

        return s