Welcome to Subscribe On Youtube
3699. Number of ZigZag Arrays I
Description
You are given three integers n, l, and r.
A ZigZag array of length n is defined as follows:
- Each element lies in the range
[l, r]. - No two adjacent elements are equal.
- No three consecutive elements form a strictly increasing or strictly decreasing sequence.
Return the total number of valid ZigZag arrays.
Since the answer may be large, return it modulo 109 + 7.
A sequence is said to be strictly increasing if each element is strictly greater than its previous one (if exists).
A sequence is said to be strictly decreasing if each element is strictly smaller than its previous one (if exists).
Example 1:
Input: n = 3, l = 4, r = 5
Output: 2
Explanation:
There are only 2 valid ZigZag arrays of length n = 3 using values in the range [4, 5]:
[4, 5, 4][5, 4, 5]
Example 2:
Input: n = 3, l = 1, r = 3
Output: 10
Explanation:
There are 10 valid ZigZag arrays of length n = 3 using values in the range [1, 3]:
[1, 2, 1],[1, 3, 1],[1, 3, 2][2, 1, 2],[2, 1, 3],[2, 3, 1],[2, 3, 2][3, 1, 2],[3, 1, 3],[3, 2, 3]
All arrays meet the ZigZag conditions.
Constraints:
3 <= n <= 20001 <= l < r <= 2000
Solutions
Solution 1
-
int zigZagArrays(int n, int low, int high) { int range = high - low; int mod = 1000000007, *dp, *ptr, *end, i = 1, goingUp = 1; long long ans = 0; if (range < 1 || !(dp = malloc(range * sizeof(int)))) return 0; ptr = dp; end = dp + range; while (ptr < end) *ptr++ = 1; ptr = dp + 1; while (ptr < end) *ptr += ptr[-1], ptr++; for (; i < n - 1; i++) { if (goingUp) { ptr = dp + range - 2; while (ptr >= dp) *ptr += ptr[1], *ptr -= *ptr >= mod ? mod : 0, ptr--; } else { ptr = dp + 1; while (ptr < end) *ptr += ptr[-1], *ptr -= *ptr >= mod ? mod : 0, ptr++; } goingUp ^= 1; } ptr = dp; while (ptr < end) ans += *ptr++; free(dp); return (int) (ans * 2 % mod); }