LC 3254. Find the Power of K-Size Subarrays I (Rust)

Author: raul-sauco

README.md

@@ -790,6 +790,7 @@ to the solution in this repository.
 | [3110. Score of a String][lc3110]                                                             | 🟢 Easy    | [![rust](res/rs.png)][lc3110rs]                                                                       |
 | [3133. Minimum Array End][lc3133]                                                             | 🟠 Medium  | [![rust](res/rs.png)][lc3133rs]                                                                       |
 | [3217. Delete Nodes From Linked List Present in Array][lc3217]                                | 🟠 Medium  | [![python](res/py.png)][lc3217py]                                                                     |
+| [3254. Find the Power of K-Size Subarrays I][lc3254]                                          | 🟠 Medium  | [![rust](res/rs.png)][lc3254rs]                                                                       |
 
 [🔝 Back to Top 🔝](#coding-challenges)
 
@@ -2523,6 +2524,8 @@ to the solution in this repository.
 [lc3133rs]: leetcode/minimum-array-end.rs
 [lc3217]: https://leetcode.com/problems/delete-nodes-from-linked-list-present-in-array/
 [lc3217py]: leetcode/delete-nodes-from-linked-list-present-in-array.py
+[lc3254]: https://leetcode.com/problems/find-the-power-of-k-size-subarrays-i/
+[lc3254rs]: leetcode/find-the-power-of-k-size-subarrays-i.rs
 
 ## CodeWars problems
 

leetcode/find-the-power-of-k-size-subarrays-i.rs

@@ -0,0 +1,74 @@
+// 3254. Find the Power of K-Size Subarrays I
+// 🟠 Medium
+//
+// https://leetcode.com/problems/find-the-power-of-k-size-subarrays-i/
+//
+// Tags: Array - Sliding Window
+
+struct Solution;
+impl Solution {
+    /// Iterate over the input vector, keep track of the count of increasing values up to the
+    /// current point, if we have k or more, the k sized subarray is good, we can return its
+    /// greatest value, the last one as the result for this subarray, otherwise -1.
+    ///
+    /// Time complexity: O(n) - We iterate over the input and do constant time workfor each
+    /// element that we visit.
+    /// Space complexity: O(1) - We store the count of good elements.
+    ///
+    /// Runtime 0 ms Beats 100%
+    /// Memory 2.17 MB Beats 87%
+    pub fn results_array(nums: Vec<i32>, k: i32) -> Vec<i32> {
+        if k == 1 {
+            return nums;
+        }
+        let n = nums.len();
+        let k = k as usize;
+        let mut res = vec![-1; n - k + 1];
+        let mut good = 1;
+        for i in 0..n - 1 {
+            good = if nums[i] + 1 == nums[i + 1] {
+                good + 1
+            } else {
+                1
+            };
+            if good >= k {
+                res[2 + i - k] = nums[i + 1];
+            }
+        }
+        res
+    }
+}
+
+// Tests.
+fn main() {
+    let tests = [
+        (vec![1, 2, 3, 4, 3, 2, 5], 3, vec![3, 4, -1, -1, -1]),
+        (vec![2, 2, 2, 2, 2], 4, vec![-1, -1]),
+        (vec![3, 2, 3, 2, 3, 2], 2, vec![-1, 3, -1, 3, -1]),
+    ];
+    println!("\n\x1b[92m» Running {} tests...\x1b[0m", tests.len());
+    let mut success = 0;
+    for (i, t) in tests.iter().enumerate() {
+        let res = Solution::results_array(t.0.clone(), t.1);
+        if res == t.2 {
+            success += 1;
+            println!("\x1b[92m✔\x1b[95m Test {} passed!\x1b[0m", i);
+        } else {
+            println!(
+                "\x1b[31mx\x1b[95m Test {} failed expected: {:?} but got {:?}!!\x1b[0m",
+                i, t.2, res
+            );
+        }
+    }
+    println!();
+    if success == tests.len() {
+        println!("\x1b[30;42m✔ All tests passed!\x1b[0m")
+    } else if success == 0 {
+        println!("\x1b[31mx \x1b[41;37mAll tests failed!\x1b[0m")
+    } else {
+        println!(
+            "\x1b[31mx\x1b[95m {} tests failed!\x1b[0m",
+            tests.len() - success
+        )
+    }
+}