LeetCode in Kotlin

3489. Zero Array Transformation IV

Medium

You are given an integer array nums of length n and a 2D array queries, where queries[i] = [li, ri, vali].

Each queries[i] represents the following action on nums:

• Select a subset of indices in the range [li, ri] from nums.
• Decrement the value at each selected index by exactly vali.

A Zero Array is an array with all its elements equal to 0.

Return the minimum possible non-negative value of k, such that after processing the first k queries in sequence, nums becomes a Zero Array. If no such k exists, return -1.

Example 1:

Input: nums = [2,0,2], queries = [[0,2,1],[0,2,1],[1,1,3]]

Output: 2

Explanation:

• For query 0 (l = 0, r = 2, val = 1):
  • Decrement the values at indices [0, 2] by 1.
  • The array will become [1, 0, 1].
• For query 1 (l = 0, r = 2, val = 1):
  • Decrement the values at indices [0, 2] by 1.
  • The array will become [0, 0, 0], which is a Zero Array. Therefore, the minimum value of k is 2.

Example 2:

Input: nums = [4,3,2,1], queries = [[1,3,2],[0,2,1]]

Output: -1

Explanation:

It is impossible to make nums a Zero Array even after all the queries.

Example 3:

Input: nums = [1,2,3,2,1], queries = [[0,1,1],[1,2,1],[2,3,2],[3,4,1],[4,4,1]]

Output: 4

Explanation:

• For query 0 (l = 0, r = 1, val = 1):
  • Decrement the values at indices [0, 1] by 1.
  • The array will become [0, 1, 3, 2, 1].
• For query 1 (l = 1, r = 2, val = 1):
  • Decrement the values at indices [1, 2] by 1.
  • The array will become [0, 0, 2, 2, 1].
• For query 2 (l = 2, r = 3, val = 2):
  • Decrement the values at indices [2, 3] by 2.
  • The array will become [0, 0, 0, 0, 1].
• For query 3 (l = 3, r = 4, val = 1):
  • Decrement the value at index 4 by 1.
  • The array will become [0, 0, 0, 0, 0]. Therefore, the minimum value of k is 4.

Example 4:

Input: nums = [1,2,3,2,6], queries = [[0,1,1],[0,2,1],[1,4,2],[4,4,4],[3,4,1],[4,4,5]]

Output: 4

Constraints:

• 1 <= nums.length <= 10
• 0 <= nums[i] <= 1000
• 1 <= queries.length <= 1000
• queries[i] = [li, ri, vali]
• 0 <= li <= ri < nums.length
• 1 <= vali <= 10

Solution

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

class Solution {
    private fun solve(q: Array<IntArray>, i: Int, target: Int, k: Int, dp: Array<IntArray>): Int {
        // we found  a valid sum equal to target so return current index of query.
        if (target == 0) {
            return k
        }
        // return a larger number to invalidate this flow
        if (k >= q.size || target < 0) {
            return q.size + 1
        }
        if (dp[target][k] != -1) {
            return dp[target][k]
        }
        // skip current query val
        var res = solve(q, i, target, k + 1, dp)
        // pick the val if its range is in the range of target index
        if (q[k][0] <= i && i <= q[k][1]) {
            res = min(res, solve(q, i, target - q[k][2], k + 1, dp))
        }
        dp[target][k] = res
        return res
    }

    fun minZeroArray(nums: IntArray, queries: Array<IntArray>): Int {
        var ans = -1
        for (i in nums.indices) {
            val dp = Array<IntArray>(nums[i] + 1) { IntArray(queries.size) }
            dp.forEach { row: IntArray -> row.fill(-1) }
            ans = max(ans, solve(queries, i, nums[i], 0, dp))
        }
        return if (ans > queries.size) -1 else ans
    }
}

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