LeetCode in Kotlin

3468. Find the Number of Copy Arrays

Medium

You are given an array original of length n and a 2D array bounds of length n x 2, where bounds[i] = [ui, vi].

You need to find the number of possible arrays copy of length n such that:

1. (copy[i] - copy[i - 1]) == (original[i] - original[i - 1]) for 1 <= i <= n - 1.
2. ui <= copy[i] <= vi for 0 <= i <= n - 1.

Return the number of such arrays.

Example 1:

Input: original = [1,2,3,4], bounds = [[1,2],[2,3],[3,4],[4,5]]

Output: 2

Explanation:

The possible arrays are:

• [1, 2, 3, 4]
• [2, 3, 4, 5]

Example 2:

Input: original = [1,2,3,4], bounds = [[1,10],[2,9],[3,8],[4,7]]

Output: 4

Explanation:

The possible arrays are:

• [1, 2, 3, 4]
• [2, 3, 4, 5]
• [3, 4, 5, 6]
• [4, 5, 6, 7]

Example 3:

Input: original = [1,2,1,2], bounds = [[1,1],[2,3],[3,3],[2,3]]

Output: 0

Explanation:

No array is possible.

Constraints:

• 2 <= n == original.length <= 105
• 1 <= original[i] <= 109
• bounds.length == n
• bounds[i].length == 2
• 1 <= bounds[i][0] <= bounds[i][1] <= 109

Solution

import kotlin.math.max
import kotlin.math.min

class Solution {
    fun countArrays(original: IntArray, bounds: Array<IntArray>): Int {
        var low = bounds[0][0]
        var high = bounds[0][1]
        var ans = high - low + 1
        for (i in 1..<original.size) {
            val diff = original[i] - original[i - 1]
            low = max((low + diff), bounds[i][0])
            high = min((high + diff), bounds[i][1])
            ans = min(ans, high - low + 1)
        }
        return max(ans, 0)
    }
}

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