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3700. Number of ZigZag Arrays II

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 <= 109
  • 1 <= l < r <= 75​​​​​​​

Solutions

Solution 1

Thinking

\(n\) can reach \(10^9\), so a length-by-length DP is impossible; the value range has length \(m=r-l+1\le 75\), which keeps the state space small. A zigzag array forbids equal neighbors and any strictly monotone triple, which means the comparison direction must flip at every step. We therefore encode a state by the last value and last direction (\(2m\) states). The transition is independent of the remaining length, so we build a \(2m\times 2m\) matrix, raise it to the \((n-1)\)-st power, and multiply by the length-\(1\) initial vector.

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class Solution {
    private static final long MOD = 1_000_000_007L;

    public int zigZagArrays(int n, int l, int r) {
        int m = r - l + 1;
        int size = 2 * m;

        long[][] trans = new long[size][size];
        for (int x = 0; x < m; x++) {
            for (int y = 0; y < x; y++) {
                trans[y][m + x] = 1;
            }
        }

        // down[x] -> up[y] where y > x
        for (int x = 0; x < m; x++) {
            for (int y = x + 1; y < m; y++) {
                trans[m + y][x] = 1;
            }
        }

        long[][] power = matrixPow(trans, n - 1);

        long[] init = new long[size];
        for (int i = 0; i < m; i++) {
            init[i] = 1;
            init[m + i] = 1;
        }

        long[] result = multiply(power, init);

        long ans = 0;
        for (long v : result) {
            ans = (ans + v) % MOD;
        }

        return (int) ans;
    }

    private long[] multiply(long[][] mat, long[] vec) {
        int n = mat.length;
        long[] res = new long[n];

        for (int i = 0; i < n; i++) {
            long sum = 0;
            for (int j = 0; j < n; j++) {
                sum = (sum + mat[i][j] * vec[j]) % MOD;
            }
            res[i] = sum;
        }

        return res;
    }

    private long[][] matrixPow(long[][] mat, long exp) {
        int n = mat.length;

        long[][] res = new long[n][n];
        for (int i = 0; i < n; i++) {
            res[i][i] = 1;
        }

        while (exp > 0) {
            if ((exp & 1) == 1) {
                res = multiply(res, mat);
            }

            mat = multiply(mat, mat);
            exp >>= 1;
        }

        return res;
    }

    private long[][] multiply(long[][] a, long[][] b) {
        int n = a.length;
        long[][] res = new long[n][n];

        for (int i = 0; i < n; i++) {
            for (int k = 0; k < n; k++) {
                if (a[i][k] == 0) continue;

                long aik = a[i][k];

                for (int j = 0; j < n; j++) {
                    if (b[k][j] == 0) continue;

                    res[i][j] = (res[i][j] + aik * b[k][j]) % MOD;
                }
            }
        }

        return res;
    }
}

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