LeetCode #1223 — HARD

Dice Roll Simulation

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

2
[1,1,2,2,2,3]

Example 2

2
[1,1,1,1,1,1]

Example 3

3
[1,1,1,2,2,3]

Core Insight

What unlocks the optimal approach

Think on Dynamic Programming.
DP(pos, last) which means we are at the position pos having as last the last character seen.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1223: Dice Roll Simulation
class Solution {
    private Integer[][][] f;
    private int[] rollMax;

    public int dieSimulator(int n, int[] rollMax) {
        f = new Integer[n][7][16];
        this.rollMax = rollMax;
        return dfs(0, 0, 0);
    }

    private int dfs(int i, int j, int x) {
        if (i >= f.length) {
            return 1;
        }
        if (f[i][j][x] != null) {
            return f[i][j][x];
        }
        long ans = 0;
        for (int k = 1; k <= 6; ++k) {
            if (k != j) {
                ans += dfs(i + 1, k, 1);
            } else if (x < rollMax[j - 1]) {
                ans += dfs(i + 1, j, x + 1);
            }
        }
        ans %= 1000000007;
        return f[i][j][x] = (int) ans;
    }
}

// Accepted solution for LeetCode #1223: Dice Roll Simulation
func dieSimulator(n int, rollMax []int) int {
	f := make([][7][16]int, n)
	const mod = 1e9 + 7
	var dfs func(i, j, x int) int
	dfs = func(i, j, x int) int {
		if i >= n {
			return 1
		}
		if f[i][j][x] != 0 {
			return f[i][j][x]
		}
		ans := 0
		for k := 1; k <= 6; k++ {
			if k != j {
				ans += dfs(i+1, k, 1)
			} else if x < rollMax[j-1] {
				ans += dfs(i+1, j, x+1)
			}
		}
		f[i][j][x] = ans % mod
		return f[i][j][x]
	}
	return dfs(0, 0, 0)
}

# Accepted solution for LeetCode #1223: Dice Roll Simulation
class Solution:
    def dieSimulator(self, n: int, rollMax: List[int]) -> int:
        @cache
        def dfs(i, j, x):
            if i >= n:
                return 1
            ans = 0
            for k in range(1, 7):
                if k != j:
                    ans += dfs(i + 1, k, 1)
                elif x < rollMax[j - 1]:
                    ans += dfs(i + 1, j, x + 1)
            return ans % (10**9 + 7)

        return dfs(0, 0, 0)

// Accepted solution for LeetCode #1223: Dice Roll Simulation
struct Solution;

const M: i32 = 1_000_000_007;

impl Solution {
    fn die_simulator(n: i32, roll_max: Vec<i32>) -> i32 {
        let n = n as usize;
        let mut dp: Vec<Vec<Vec<i32>>> = vec![vec![vec![0; 5001]; 16]; 6];
        Self::dfs(n, -1, 1, &mut dp, &roll_max)
    }

    fn dfs(
        n: usize,
        prev: i32,
        repeat: usize,
        dp: &mut Vec<Vec<Vec<i32>>>,
        roll_max: &[i32],
    ) -> i32 {
        if n == 0 {
            1
        } else {
            if prev >= 0 && dp[prev as usize][repeat][n] > 0 {
                return dp[prev as usize][repeat][n];
            }
            let mut sum = 0;
            for i in 0..6 {
                if i == prev {
                    if repeat < roll_max[i as usize] as usize {
                        sum += Self::dfs(n - 1, i, repeat + 1, dp, roll_max);
                        sum %= M;
                    }
                } else {
                    sum += Self::dfs(n - 1, i, 1, dp, roll_max);
                    sum %= M;
                }
            }
            if prev >= 0 {
                dp[prev as usize][repeat][n] = sum;
            }
            sum
        }
    }
}

#[test]
fn test() {
    let n = 2;
    let roll_max = vec![1, 1, 2, 2, 2, 3];
    let res = 34;
    assert_eq!(Solution::die_simulator(n, roll_max), res);
    let n = 2;
    let roll_max = vec![1, 1, 1, 1, 1, 1];
    let res = 30;
    assert_eq!(Solution::die_simulator(n, roll_max), res);
    let n = 3;
    let roll_max = vec![1, 1, 1, 2, 2, 3];
    let res = 181;
    assert_eq!(Solution::die_simulator(n, roll_max), res);
    let n = 4;
    let roll_max = vec![2, 1, 1, 3, 3, 2];
    let res = 1082;
    assert_eq!(Solution::die_simulator(n, roll_max), res);
}

// Accepted solution for LeetCode #1223: Dice Roll Simulation
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1223: Dice Roll Simulation
// class Solution {
//     private Integer[][][] f;
//     private int[] rollMax;
// 
//     public int dieSimulator(int n, int[] rollMax) {
//         f = new Integer[n][7][16];
//         this.rollMax = rollMax;
//         return dfs(0, 0, 0);
//     }
// 
//     private int dfs(int i, int j, int x) {
//         if (i >= f.length) {
//             return 1;
//         }
//         if (f[i][j][x] != null) {
//             return f[i][j][x];
//         }
//         long ans = 0;
//         for (int k = 1; k <= 6; ++k) {
//             if (k != j) {
//                 ans += dfs(i + 1, k, 1);
//             } else if (x < rollMax[j - 1]) {
//                 ans += dfs(i + 1, j, x + 1);
//             }
//         }
//         ans %= 1000000007;
//         return f[i][j][x] = (int) ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × k^2 × M)

Space

O(n × k × M)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.