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 <= 1001 <= 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[101] = {}, 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;
}
}