Formatted question description: https://leetcode.ca/all/1051.html

# 1051. Height Checker (Easy)

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 for all students to be standing in non-decreasing order of height.

Notice that when a group of students is selected they can reorder in any possible way between themselves and the non selected students remain on their seats.

Example 1:

Input: heights = [1,1,4,2,1,3]
Output: 3
Explanation:
Current array : [1,1,4,2,1,3]
Target array  : [1,1,1,2,3,4]
On index 2 (0-based) we have 4 vs 1 so we have to move this student.
On index 4 (0-based) we have 1 vs 3 so we have to move this student.
On index 5 (0-based) we have 3 vs 4 so we have to move this student.


Example 2:

Input: heights = [5,1,2,3,4]
Output: 5


Example 3:

Input: heights = [1,2,3,4,5]
Output: 0


Constraints:

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

Related Topics:
Array

## Solution 1.

// OJ: https://leetcode.com/problems/height-checker/

// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
int heightChecker(vector<int>& A) {
auto B = A;
sort(begin(B), end(B));
int ans = 0;
for (int i = 0; i < A.size(); ++i) ans += A[i] != B[i];
return ans;
}
};


## Solution 2. Count Sort

// OJ: https://leetcode.com/problems/height-checker/

// Time: O(N)
// Space: O(1)
class Solution {
public:
int heightChecker(vector<int>& A) {
int cnt = {}, ans = 0;
for (int n : A) cnt[n]++;
for (int i = 1, j = 0; i <= 100; ++i) {
for (int k = 0; k < cnt[i]; ++k, ++j) {
if (A[j] != i) ++ans;
}
}
return ans;
}
};


Java

class Solution {
public int heightChecker(int[] heights) {
int length = heights.length;
int[] sorted = new int[length];
System.arraycopy(heights, 0, sorted, 0, length);
Arrays.sort(sorted);
int count = 0;
for (int i = 0; i < length; i++) {
if (heights[i] != sorted[i])
count++;
}
return count;
}
}