1051. Height Checker

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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

Solutions

Solution 1: Sorting

We can first sort the heights of the students, then compare the sorted heights with the original heights, and count the positions that are different.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the number of students.

Python3

class Solution:
    def heightChecker(self, heights: List[int]) -> int:
        expected = sorted(heights)
        return sum(a != b for a, b in zip(heights, expected))

Java

class Solution {
    public int heightChecker(int[] heights) {
        int[] expected = heights.clone();
        Arrays.sort(expected);
        int ans = 0;
        for (int i = 0; i < heights.length; ++i) {
            if (heights[i] != expected[i]) {
                ++ans;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int heightChecker(vector<int>& heights) {
        vector<int> expected = heights;
        sort(expected.begin(), expected.end());
        int ans = 0;
        for (int i = 0; i < heights.size(); ++i) {
            ans += heights[i] != expected[i];
        }
        return ans;
    }
};

Go

func heightChecker(heights []int) (ans int) {
	expected := slices.Clone(heights)
	sort.Ints(expected)
	for i, v := range heights {
		if v != expected[i] {
			ans++
		}
	}
	return
}

TypeScript

function heightChecker(heights: number[]): number {
    const expected = [...heights].sort((a, b) => a - b);
    return heights.reduce((acc, cur, i) => acc + (cur !== expected[i] ? 1 : 0), 0);
}

Solution 2: Counting Sort

Since the height of the students in the problem does not exceed $100$, we can use counting sort. Here we use an array $cnt$ of length $101$ to count the number of times each height $h_i$ appears.

The time complexity is $O(n + M)$, and the space complexity is $O(M)$. Where $n$ is the number of students, and $M$ is the maximum height of the students. In this problem, $M = 101$.

Python3

class Solution:
    def heightChecker(self, heights: List[int]) -> int:
        cnt = [0] * 101
        for h in heights:
            cnt[h] += 1
        ans = i = 0
        for j in range(1, 101):
            while cnt[j]:
                cnt[j] -= 1
                if heights[i] != j:
                    ans += 1
                i += 1
        return ans

Java

class Solution {
    public int heightChecker(int[] heights) {
        int[] cnt = new int[101];
        for (int h : heights) {
            ++cnt[h];
        }
        int ans = 0;
        for (int i = 0, j = 0; i < 101; ++i) {
            while (cnt[i] > 0) {
                --cnt[i];
                if (heights[j++] != i) {
                    ++ans;
                }
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int heightChecker(vector<int>& heights) {
        vector<int> cnt(101);
        for (int& h : heights) ++cnt[h];
        int ans = 0;
        for (int i = 0, j = 0; i < 101; ++i) {
            while (cnt[i]) {
                --cnt[i];
                if (heights[j++] != i) ++ans;
            }
        }
        return ans;
    }
};

Go

func heightChecker(heights []int) int {
	cnt := make([]int, 101)
	for _, h := range heights {
		cnt[h]++
	}
	ans := 0
	for i, j := 0, 0; i < 101; i++ {
		for cnt[i] > 0 {
			cnt[i]--
			if heights[j] != i {
				ans++
			}
			j++
		}
	}
	return ans
}

TypeScript

function heightChecker(heights: number[]): number {
    const cnt = Array(101).fill(0);
    for (const i of heights) {
        cnt[i]++;
    }
    let ans = 0;
    for (let j = 1, i = 0; j < 101; j++) {
        while (cnt[j]--) {
            if (heights[i++] !== j) {
                ans++;
            }
        }
    }
    return ans;
}