Hex.*

LC1051 - Height Checker

Posted on In
Leetcode: 127/2658 (53+60+14)

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 ith 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 ith 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

Solution

  • Sort the array heights and store it in the variable expected
  • Count the number of heights[i] ≠ expected[i]
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#O(nlogn) time | O(1) space
class Solution:
    def heightChecker(self, heights: List[int]) -> int:
        expected = sorted(heights)
        cnt =0
        for i in range(len(heights)):
            if expected[i] != heights[i]:
                cnt+=1
        return cnt