LeetCode in Kotlin

1764. Form Array by Concatenating Subarrays of Another Array

Medium

You are given a 2D integer array groups of length n. You are also given an integer array nums.

You are asked if you can choose n disjoint subarrays from the array nums such that the ith subarray is equal to groups[i] (0-indexed), and if i > 0, the (i-1)th subarray appears before the ith subarray in nums (i.e. the subarrays must be in the same order as groups).

Return true if you can do this task, and false otherwise.

Note that the subarrays are disjoint if and only if there is no index k such that nums[k] belongs to more than one subarray. A subarray is a contiguous sequence of elements within an array.

Example 1:

Input: groups = [[1,-1,-1],[3,-2,0]], nums = [1,-1,0,1,-1,-1,3,-2,0]

Output: true

Explanation: You can choose the 0^th subarray as [1,-1,0,1,-1,-1,3,-2,0] and the 1^st one as [1,-1,0,1,-1,-1,3,-2,0]. These subarrays are disjoint as they share no common nums[k] element.

Example 2:

Input: groups = [[10,-2],[1,2,3,4]], nums = [1,2,3,4,10,-2]

Output: false

Explanation: Note that choosing the subarrays [1,2,3,4,10,-2] and [1,2,3,4,10,-2] is incorrect because they are not in the same order as in groups. [10,-2] must come before [1,2,3,4].

Example 3:

Input: groups = [[1,2,3],[3,4]], nums = [7,7,1,2,3,4,7,7]

Output: false

Explanation: Note that choosing the subarrays [7,7,1,2,3,4,7,7] and [7,7,1,2,3,4,7,7] is invalid because they are not disjoint. They share a common elements nums[4] (0-indexed).

Constraints:

• groups.length == n
• 1 <= n <= 103
• 1 <= groups[i].length, sum(groups[i].length) <= 103
• 1 <= nums.length <= 103
• -107 <= groups[i][j], nums[k] <= 107

Solution

class Solution {
    fun canChoose(groups: Array<IntArray>, nums: IntArray): Boolean {
        var prev = 0
        for (i in groups.indices) {
            val temp = IntArray(groups[i].size)
            if (prev + groups[i].size > nums.size) {
                return false
            }
            var index = 0
            var j: Int
            j = prev
            while (j < prev + groups[i].size) {
                temp[index++] = nums[j]
                j++
            }
            if (temp.contentEquals(groups[i])) {
                prev = j
                continue
            }
            var k: Int = j
            while (k < nums.size) {
                var l: Int
                l = 0
                while (l < temp.size - 1) {
                    temp[l] = temp[l + 1]
                    l++
                }
                temp[l] = nums[k]
                if (temp.contentEquals(groups[i])) {
                    prev = k + 1
                    break
                }
                k++
            }
            if (k == nums.size) {
                return false
            }
        }
        return true
    }
}

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