LeetCode Entry
1051. Height Checker
Count unsorted elements in array
1051. Height Checker easy
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Problem TLDR
Count unsorted elements in array #easy
Intuition
We can use bucket sort to do this in O(n).
Approach
Let’s just use a simple sort to save the effort.
Complexity
-
Time complexity: \(O(nlog(n))\)
-
Space complexity: \(O(n)\)
Code
fun heightChecker(heights: IntArray) = heights
.toList().sorted().withIndex()
.count { (i, h) -> h != heights[i] }
pub fn height_checker(heights: Vec<i32>) -> i32 {
let mut s = heights.clone(); s.sort_unstable();
(0..s.len()).map(|i| (s[i] != heights[i]) as i32).sum()
}