{leetcode}/problems/minimum-time-to-make-array-sum-at-most-x/[LeetCode - 2809. Minimum Time to Make Array Sum At Most x ^]
You are given two 0-indexed integer arrays nums1 and nums2 of equal length. Every second, for all indices 0 ⇐ i < nums1.length, value of nums1[i] is incremented by nums2[i]. After this is done, you can do the following operation:
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Choose an index
0 ⇐ i < nums1.lengthand makenums1[i] = 0.
You are also given an integer x.
Return the minimum time in which you can make the sum of all elements of _`nums1` to be* less than or equal* to `x`, _or _`-1` if this is not possible._
Example 1:
Input: nums1 = [1,2,3], nums2 = [1,2,3], x = 4 Output: 3 Explanation: For the 1st second, we apply the operation on i = 0. Therefore nums1 = [0,2+2,3+3] = [0,4,6]. For the 2nd second, we apply the operation on i = 1. Therefore nums1 = [0+1,0,6+3] = [1,0,9]. For the 3rd second, we apply the operation on i = 2. Therefore nums1 = [1+1,0+2,0] = [2,2,0]. Now sum of nums1 = 4. It can be shown that these operations are optimal, so we return 3.
Example 2:
Input: nums1 = [1,2,3], nums2 = [3,3,3], x = 4 Output: -1 Explanation: It can be shown that the sum of nums1 will always be greater than x, no matter which operations are performed.
Constraints:
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<font face="monospace">1 ⇐ nums1.length ⇐ 103 -
1 ⇐ nums1[i] ⇐ 103 -
0 ⇐ nums2[i] ⇐ 103 -
nums1.length == nums2.length -
0 ⇐ x ⇐ 106