You are given an integer array nums and an integer k.
You may repeatedly choose any contiguous subarray of nums whose sum is divisible by k and delete it; after each deletion, the remaining elements close the gap.
Return the minimum possible sum of nums after performing any number of such deletions.
\n
Example 1:
\n\nInput: nums = [1,1,1], k = 2
\n\nOutput: 1
\n\nExplanation:
\n\nnums[0..1] = [1, 1], whose sum is 2 (divisible by 2), leaving [1].Example 2:
\n\nInput: nums = [3,1,4,1,5], k = 3
\n\nOutput: 5
\n\nExplanation:
\n\nnums[1..3] = [1, 4, 1], whose sum is 6 (divisible by 3), leaving [3, 5].nums[0..0] = [3], whose sum is 3 (divisible by 3), leaving [5].\n
Constraints:
\n\n1 <= nums.length <= 1051 <= nums[i] <= 1061 <= k <= 105给你一个整数数组 nums 和一个整数 k。
你可以 多次 选择 连续 子数组 nums，其元素和可以被 k 整除，并将其删除；每次删除后，剩余元素会填补空缺。
返回在执行任意次数此类删除操作后，nums 的最小可能 和。
\n\n
示例 1：
\n\n输入： nums = [1,1,1], k = 2
\n\n输出： 1
\n\n解释：
\n\nnums[0..1] = [1, 1]，其和为 2（可以被 2 整除），剩余 [1]。示例 2：
\n\n输入： nums = [3,1,4,1,5], k = 3
\n\n输出： 5
\n\n解释：
\n\nnums[1..3] = [1, 4, 1]，其和为 6（可以被 3 整除），剩余数组为 [3, 5]。nums[0..0] = [3]，其和为 3（可以被 3 整除），剩余数组为 [5]。\n\n
提示：
\n\n1 <= nums.length <= 1051 <= nums[i] <= 1061 <= k <= 105k precisely when the prefix sums at its two endpoints have the same remainder mod k.",
"Define dp[i] as the minimum total remaining sum after processing the first i elements.",
"Keep a map from each remainder to the best (smallest) value of dp[j] - prefixSum[j] you've seen for that remainder to update dp[i] in O(1).",
"Maintain a running prefix sum so you never recompute subarray sums from scratch."
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