LeetCode-in-Java
1562. Find Latest Group of Size M
Medium
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: “00100”, groups: [“1”]
Step 2: “00101”, groups: [“1”, “1”]
Step 3: “10101”, groups: [“1”, “1”, “1”]
Step 4: “11101”, groups: [“111”, “1”]
Step 5: “11111”, 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: “00100”, groups: [“1”]
Step 2: “10100”, groups: [“1”, “1”]
Step 3: “10101”, groups: [“1”, “1”, “1”]
Step 4: “10111”, groups: [“1”, “111”]
Step 5: “11111”, groups: [“11111”]
No group of size 2 exists during any step.
Constraints:
n == arr.length1 <= m <= n <= 1051 <= arr[i] <= n- All integers in
arrare distinct.
Solution
public class Solution {
public int findLatestStep(int[] arr, int m) {
int[] lengthAtIndex = new int[arr.length + 2];
int[] countOfLength = new int[arr.length + 1];
int res = -1;
int step = 1;
for (int i : arr) {
int leftLength = lengthAtIndex[i - 1];
int rightLength = lengthAtIndex[i + 1];
int newLength = leftLength + rightLength + 1;
lengthAtIndex[i] = newLength;
lengthAtIndex[i - leftLength] = newLength;
lengthAtIndex[i + rightLength] = newLength;
countOfLength[newLength] += 1;
countOfLength[leftLength] -= 1;
countOfLength[rightLength] -= 1;
if (countOfLength[m] > 0) {
res = step;
}
step++;
}
return res;
}
}