1562. Find Latest Group of Size M

Description

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.length
• 1 <= m <= n <= 105
• 1 <= arr[i] <= n
• All integers in arr are distinct.

 

Solutions

Solution: Hash Table

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[]} arr
 * @param {number} m
 * @return {number}
 */
const findLatestStep = function (arr, m) {
  const size = arr.length;
  const sizes = new Array(size + 2).fill(0);
  const counts = new Array(size + 2).fill(0);

  return arr.reduce((result, position, index) => {
    const left = sizes[position - 1];
    const right = sizes[position + 1];
    const length = right + left + 1;

    sizes[position] = sizes[position - left] = sizes[position + right] = length;
    counts[left] -= 1;
    counts[right] -= 1;
    counts[length] += 1;
    return counts[m] ? index + 1 : result;
  }, -1);
};