Height Checker LeetCode Solution

Problem – Height Checker LeetCode Solution

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

Height Checker LeetCode Solution in Java

class Solution {
    public int heightChecker(int[] heights) {
        int[] heightToFreq = new int[101];
        
        for (int height : heights) {
            heightToFreq[height]++;
        }
        
        int result = 0;
        int curHeight = 0;
        
        for (int i = 0; i < heights.length; i++) {
            while (heightToFreq[curHeight] == 0) {
                curHeight++;
            }
            
            if (curHeight != heights[i]) {
                result++;
            }
            heightToFreq[curHeight]--;
        }
        
        return result;
    }
}

Height Checker LeetCode Solution in Python

class Solution:
    def heightChecker(self, heights: List[int]) -> int:
        max_nr = max(heights)
        # initialize frequency array with 0's
        count = [0] * (max_nr + 1)
        # get frequencies
        for number in heights:
            count[number] += 1
        # create a sumcount array
        sumcount = [0] * (max_nr + 1)
        for index, number in enumerate(count[1:],start=1):
            sumcount[index] = number + sumcount[index-1]
        # sumcount determines the index in sorted array
        # create output array
        output = [0] * len(heights)
        # loop backwards starting with last element for stable sort
        for p in range(len(heights)-1,-1,-1):
            output[sumcount[heights[p]]-1] = heights[p]
            sumcount[heights[p]] -= 1
		# return the difference compared to original array
        result = 0
        for index, number in enumerate(heights):
            if number != output[index]:
                result += 1
        return result

Height Checker LeetCode Solution in C++

    int heightChecker(vector<int>& heights) {
        int mismatch = 0; //count the number of mismatches
        vector<int> count(101, 0); // store the count of heights

        for(int i = 0; i < heights.size(); i++) count[heights[i]]++;

        int i = 1, j = 0; 
        while(i < 101){
			// check value of count
            if(count[i] == 0){
                i++;
            }
            else{
				// compare the index of count with value of height
                if(i != heights[j]) mismatch++;
                j++; count[i]--;
            }
        }

        return mismatch;
    }
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