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You are given a list of positive integers nums
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and a positive integer k.
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Return the length of the shortest sublist (can be empty sublist )
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you can delete such that the resulting list's sum is divisible by k.
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 You cannot delete the entire list.
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If it's not possible, return -1.
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Constraints
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1 ≤ n ≤ 100,000 where n is the length of nums
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Example 1
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Input
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nums = [1, 8, 6, 4, 5]
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k = 7
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Output
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2
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Explanation
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We can remove the sublist [6, 4] to get [1, 8, 5]
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which sums to 14 and is divisible by 7.

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class Solution {
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public:
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    int solve(vector<int>& nums, int k) {
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        int sum = 0;
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        for(auto i : nums)
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        {
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            sum += i;
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        }
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        sum = sum % k;
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        map<int, int> up;
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        up[0] = -1;
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        int pre = 0, n = nums.size();
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        int ans = nums.size();
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        for (int i = 0; i < n; i++)
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        {
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            pre += nums[i];
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            int re = pre % k;
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            up[re] = i;
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            int x = pre - sum;
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            int m = ((x % k) + k) % k;
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            if (up.count(m)) ans = min(ans, i - up[m]);
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        }
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        return ans == nums.size() ? -1 : ans;
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    }
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};