LeetCode in Kotlin

3356. Zero Array Transformation II

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:

• Decrement the value at each index in the range [li, ri] in nums by at most vali.
• The amount by which each value is decremented can be chosen independently for each index.

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 i = 0 (l = 0, r = 2, val = 1):
  • Decrement values at indices [0, 1, 2] by [1, 0, 1] respectively.
  • The array will become [1, 0, 1].
• For i = 1 (l = 0, r = 2, val = 1):
  • Decrement values at indices [0, 1, 2] by [1, 0, 1] respectively.
  • 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:

• For i = 0 (l = 1, r = 3, val = 2):
  • Decrement values at indices [1, 2, 3] by [2, 2, 1] respectively.
  • The array will become [4, 1, 0, 0].
• For i = 1 (l = 0, r = 2, val = 1):
  • Decrement values at indices [0, 1, 2] by [1, 1, 0] respectively.
  • The array will become [3, 0, 0, 0], which is not a Zero Array.

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] <= 5 * 105
• 1 <= queries.length <= 105
• queries[i].length == 3
• 0 <= li <= ri < nums.length
• 1 <= vali <= 5

Solution

class Solution {
    fun minZeroArray(nums: IntArray, queries: Array<IntArray>): Int {
        val diff = IntArray(nums.size)
        var idx = 0
        var d = 0
        for (i in nums.indices) {
            d += diff[i]
            while (nums[i] + d > 0 && idx < queries.size) {
                val q = queries[idx]
                if (i >= q[0] && i <= q[1]) {
                    d -= q[2]
                }
                diff[q[0]] -= q[2]
                if (q[1] + 1 < nums.size) {
                    diff[q[1] + 1] += q[2]
                }
                idx++
            }
            if (nums[i] + d > 0) {
                return -1
            }
        }
        return idx
    }
}

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