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Java-based LeetCode algorithm problem solutions, regularly updated

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package g1201_1300.s1223_dice_roll_simulation;

// #Hard #Array #Dynamic_Programming #2022_03_12_Time_9_ms_(90.98%)_Space_41.1_MB_(86.47%)

/**
 * 1223 - Dice Roll Simulation\.
 *
 * Hard
 *
 * A die simulator generates a random number from `1` to `6` for each roll. You introduced a constraint to the generator such that it cannot roll the number `i` more than `rollMax[i]` ( **1-indexed** ) consecutive times.
 *
 * Given an array of integers `rollMax` and an integer `n`, return _the number of distinct sequences that can be obtained with exact_ `n` _rolls_. Since the answer may be too large, return it **modulo** 10⁹ + 7.
 *
 * Two sequences are considered different if at least one element differs from each other.
 *
 * **Example 1:**
 *
 * **Input:** n = 2, rollMax = [1,1,2,2,2,3]
 *
 * **Output:** 34
 *
 * **Explanation:** There will be 2 rolls of die, if there are no constraints on the die, there are 6 \* 6 = 36 possible combinations. In this case, looking at rollMax array, the numbers 1 and 2 appear at most once consecutively, therefore sequences (1,1) and (2,2) cannot occur, so the final answer is 36-2 = 34.
 *
 * **Example 2:**
 *
 * **Input:** n = 2, rollMax = [1,1,1,1,1,1]
 *
 * **Output:** 30
 *
 * **Example 3:**
 *
 * **Input:** n = 3, rollMax = [1,1,1,2,2,3]
 *
 * **Output:** 181
 *
 * **Constraints:**
 *
 * *   `1 <= n <= 5000`
 * *   `rollMax.length == 6`
 * *   `1 <= rollMax[i] <= 15`
**/
public class Solution {
    private static final long MOD = 1000000007;

    public int dieSimulator(int n, int[] rollMax) {
        long[][] all = new long[6][15 + 1];
        long[] countsBySide = new long[6];
        long total = 0;
        long newTotal;
        int max;
        for (int j = 0; j < all.length; j++) {
            all[j][1] = 1;
            countsBySide[j] = 1;

            total = 6;
        }
        for (int i = 1; i < n; i++) {
            newTotal = total;
            for (int j = 0; j < all.length; j++) {
                all[j][0] = (total - countsBySide[j]) % MOD;
                max = rollMax[j];
                newTotal = (newTotal - all[j][max] + all[j][0]);
                countsBySide[j] = (total - all[j][max]) % MOD;
                System.arraycopy(all[j], 0, all[j], 1, max);
            }
            total = newTotal;
        }
        return (int) (total % MOD);
    }
}