1562. Find Latest Group of Size M: AC

Author: tan-wei

src/solution/mod.rs

@@ -1178,3 +1178,4 @@ mod s1558_minimum_numbers_of_function_calls_to_make_target_array;
 mod s1559_detect_cycles_in_2d_grid;
 mod s1560_most_visited_sector_in_a_circular_track;
 mod s1561_maximum_number_of_coins_you_can_get;
+mod s1562_find_latest_group_of_size_m;

src/solution/s1562_find_latest_group_of_size_m.rs

@@ -0,0 +1,101 @@
+/**
+ * [1562] Find Latest Group of Size M
+ *
+ * Given an array arr that represents a permutation of numbers from 1 to n.
+ * You have a binary string of size n that initially has all its bits set to zero. At each step i (assuming both the binary string and arr are 1-indexed) from 1 to n, the bit at position arr[i] is set to 1.
+ * You are also given an integer m. Find the latest step at which there exists a group of ones of length m. A group of ones is a contiguous substring of 1's such that it cannot be extended in either direction.
+ * Return the latest step at which there exists a group of ones of length exactly m. If no such group exists, return -1.
+ *  
+ * Example 1:
+ *
+ * Input: arr = [3,5,1,2,4], m = 1
+ * Output: 4
+ * Explanation:
+ * Step 1: "00<u>1</u>00", groups: ["1"]
+ * Step 2: "0010<u>1</u>", groups: ["1", "1"]
+ * Step 3: "<u>1</u>0101", groups: ["1", "1", "1"]
+ * Step 4: "1<u>1</u>101", groups: ["111", "1"]
+ * Step 5: "111<u>1</u>1", groups: ["11111"]
+ * The latest step at which there exists a group of size 1 is step 4.
+ *
+ * Example 2:
+ *
+ * Input: arr = [3,1,5,4,2], m = 2
+ * Output: -1
+ * Explanation:
+ * Step 1: "00<u>1</u>00", groups: ["1"]
+ * Step 2: "<u>1</u>0100", groups: ["1", "1"]
+ * Step 3: "1010<u>1</u>", groups: ["1", "1", "1"]
+ * Step 4: "101<u>1</u>1", groups: ["1", "111"]
+ * Step 5: "1<u>1</u>111", groups: ["11111"]
+ * No group of size 2 exists during any step.
+ *
+ *  
+ * Constraints:
+ *
+ * 	n == arr.length
+ * 	1 <= m <= n <= 10^5
+ * 	1 <= arr[i] <= n
+ * 	All integers in arr are distinct.
+ *
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.com/problems/find-latest-group-of-size-m/
+// discuss: https://leetcode.com/problems/find-latest-group-of-size-m/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn find_latest_step(arr: Vec<i32>, m: i32) -> i32 {
+        let (n, m) = (arr.len(), m as usize);
+        if n == m {
+            return n as i32;
+        }
+        let mut length = vec![0; n + 2];
+        let mut count = vec![0; n + 1];
+        let mut result = -1;
+
+        for (i, &a) in arr.iter().enumerate() {
+            let (a, low, high) = (a as usize, length[a as usize - 1], length[a as usize + 1]);
+            let sum = low + high + 1;
+            length[a - low] = sum;
+            length[a + high] = sum;
+            count[low] -= 1;
+            count[high] -= 1;
+            count[sum] += 1;
+            if count[m] > 0 {
+                result = i as i32 + 1;
+            }
+        }
+
+        result
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_1562_example_1() {
+        let arr = vec![3, 5, 1, 2, 4];
+        let m = 1;
+
+        let result = 4;
+
+        assert_eq!(Solution::find_latest_step(arr, m), result);
+    }
+
+    #[test]
+    fn test_1562_example_2() {
+        let arr = vec![3, 1, 5, 4, 2];
+        let m = 2;
+
+        let result = -1;
+
+        assert_eq!(Solution::find_latest_step(arr, m), result);
+    }
+}