3388. Count Beautiful Splits in an Array

Posted on Edited on In

3388. Count Beautiful Splits in an Array

Description

You are given an array nums.

A split of an array nums is beautiful if:

  • The array nums is split into three subarrays: nums1, nums2, and nums3, such that nums can be formed by concatenating nums1, nums2, and nums3 in that order.
  • The subarray nums1 is a prefix of nums2 OR nums2 is a prefix of nums3.

Return the number of ways you can make this split.

Example 1:

1
2
3
Input: nums = [1,1,2,1]

Output: 2

Explanation:

The beautiful splits are:

  • A split with nums1 = [1], nums2 = [1,2], nums3 = [1].
  • A split with nums1 = [1], nums2 = [1], nums3 = [2,1].

Example 2:

1
2
3
Input: nums = [1,2,3,4]

Output: 0

Explanation:

There are 0 beautiful splits.

Constraints:

  • 1 <= nums.length <= 5000
  • 0 <= nums[i] <= 50

Hints/Notes

Solution

Language: C++

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
class Solution {
public:
    vector<int> calc_z(vector<int> nums) {
        int n = nums.size();
        vector<int> z(n, 0);
        int box_l = 0, box_r = 0;
        for (int i = 1; i < n; i++) {
            if (i <= box_r) {
                z[i] = min(z[i - box_l], box_r - i + 1);
            }
            while (i + z[i] < n && nums[z[i]] == nums[i + z[i]]) {
                box_l = i;
                box_r = i + z[i];
                z[i]++;
            }
        }
        return z;
    }

    int beautifulSplits(vector<int>& nums) {
        int n = nums.size(), res = 0;
        vector<int> all = calc_z(nums);
        for (int i = 1; i < n; i++) {
            // this following code is wrong, because it skips the subarray
            // that nums1 is a prefix, but nums2 is short than nums1
            // if (all[i] >= i) {
            //     res += n - 2 * i;
            //     continue;
            // }
            vector<int> cur = calc_z(vector<int>(nums.begin() + i, nums.end()));
            for (int j = i + 1; j < n; j++) {
                // the 1st part: [0, i)
                // the 2nd part: [i, j)
                // the 3rd part: [j, n)
                if (i <= j - i && all[i] >= i || cur[j - i] >= j - i) {
                    res++;
                }
            }
        }
        return res;
    }
};