[数据结构与算法]LeetCode 1590 Make Sum Divisible by P (前缀和 hash)

Given an array of positive integers?nums, remove the?smallest?subarray (possibly?empty) such that the?sum?of the remaining elements is divisible by?p. It is?not?allowed to remove the whole array.

Return?the length of the smallest subarray that you need to remove, or?-1?if it's impossible.

A?subarray?is defined as a contiguous block of elements in the array.

Example 1:

Input: nums = [3,1,4,2], p = 6
Output: 1
Explanation: The sum of the elements in nums is 10, which is not divisible by 6. We can remove the subarray [4], and the sum of the remaining elements is 6, which is divisible by 6.

Example 2:

Input: nums = [6,3,5,2], p = 9
Output: 2
Explanation: We cannot remove a single element to get a sum divisible by 9. The best way is to remove the subarray [5,2], leaving us with [6,3] with sum 9.

Example 3:

Input: nums = [1,2,3], p = 3
Output: 0
Explanation: Here the sum is 6. which is already divisible by 3. Thus we do not need to remove anything.

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109
• 1 <= p <= 109

题目链接：https://leetcode.com/problems/make-sum-divisible-by-p/

题目大意：删除一个长度最短的子数组使得剩余数字和是p的倍数

题目分析：预处理后缀和模值，并用map记录，对每个前缀和模值找其后最近的与其相加为p的后缀模值

58ms，时间击败59.31%

class Solution {
    public int minSubarray(int[] nums, int p) {
        int ans = Integer.MAX_VALUE, n = nums.length, sufmod = 0;
        Map<Integer, List<Integer>> pos = new HashMap<>();
        for (int i = n - 1; i >= 0; i--) {
            sufmod = (sufmod + nums[i]) % p;
            if (sufmod == 0) {
                ans = i;
            }
            if (!pos.containsKey(sufmod)) {
                pos.put(sufmod, new ArrayList<>());
            }
            pos.get(sufmod).add(i);
        }

        int premod = 0;
        for (int i = 0; i < n; i++) {
            premod = (premod + nums[i]) % p;
            if (premod == 0) {
                ans = Math.min(ans, n - i - 1);
            }
            if (pos.containsKey(p - premod)) {
                List<Integer> rpos = pos.get(p - premod);
                int sz = rpos.size();
                for (int j = sz - 1; j >= 0; j--) {
                    if (rpos.get(j) > i) {
                        ans = Math.min(ans, rpos.get(j) - i - 1);
                        break;
                    }
                }
            }
        }
    
        if (ans == Integer.MAX_VALUE) {
            return -1;
        }
        return ans;
    }
}