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| 1 | +/** |
| 2 | + * 2318. Number of Distinct Roll Sequences |
| 3 | + * https://leetcode.com/problems/number-of-distinct-roll-sequences/ |
| 4 | + * Difficulty: Hard |
| 5 | + * |
| 6 | + * You are given an integer n. You roll a fair 6-sided dice n times. Determine the total number |
| 7 | + * of distinct sequences of rolls possible such that the following conditions are satisfied: |
| 8 | + * 1. The greatest common divisor of any adjacent values in the sequence is equal to 1. |
| 9 | + * 2. There is at least a gap of 2 rolls between equal valued rolls. More formally, if the |
| 10 | + * value of the ith roll is equal to the value of the jth roll, then abs(i - j) > 2. |
| 11 | + * |
| 12 | + * Return the total number of distinct sequences possible. Since the answer may be very large, |
| 13 | + * return it modulo 109 + 7. |
| 14 | + * |
| 15 | + * Two sequences are considered distinct if at least one element is different. |
| 16 | + */ |
| 17 | + |
| 18 | +/** |
| 19 | + * @param {number} n |
| 20 | + * @return {number} |
| 21 | + */ |
| 22 | +var distinctSequences = function(n) { |
| 23 | + const MOD = 1e9 + 7; |
| 24 | + const gcd = (a, b) => b === 0 ? a : gcd(b, a % b); |
| 25 | + |
| 26 | + const dp = new Array(n + 1) |
| 27 | + .fill() |
| 28 | + .map(() => new Array(7).fill().map(() => new Array(7).fill(0))); |
| 29 | + |
| 30 | + for (let i = 1; i <= 6; i++) { |
| 31 | + dp[1][i][0] = 1; |
| 32 | + } |
| 33 | + |
| 34 | + for (let len = 2; len <= n; len++) { |
| 35 | + for (let curr = 1; curr <= 6; curr++) { |
| 36 | + for (let prev = 0; prev <= 6; prev++) { |
| 37 | + for (let prevPrev = 0; prevPrev <= 6; prevPrev++) { |
| 38 | + if (dp[len - 1][prev][prevPrev] === 0) continue; |
| 39 | + if (curr === prev || curr === prevPrev) continue; |
| 40 | + if (prev !== 0 && gcd(curr, prev) !== 1) continue; |
| 41 | + dp[len][curr][prev] = (dp[len][curr][prev] + dp[len - 1][prev][prevPrev]) % MOD; |
| 42 | + } |
| 43 | + } |
| 44 | + } |
| 45 | + } |
| 46 | + |
| 47 | + let result = 0; |
| 48 | + for (let curr = 1; curr <= 6; curr++) { |
| 49 | + for (let prev = 0; prev <= 6; prev++) { |
| 50 | + result = (result + dp[n][curr][prev]) % MOD; |
| 51 | + } |
| 52 | + } |
| 53 | + |
| 54 | + return result; |
| 55 | +}; |
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