Description
You have n dice, and each dice has k faces numbered from 1 to k.
Given three integers n, k, and target, return the number of possible ways (out of the kn total ways) to roll the dice, so the sum of the face-up numbers equals target. Since the answer may be too large, return it modulo 109 + 7.
Example 1:
Input: n = 1, k = 6, target = 3
Output: 1
Explanation: You throw one die with 6 faces.
There is only one way to get a sum of 3.
Example 2:
Input: n = 2, k = 6, target = 7
Output: 6
Explanation: You throw two dice, each with 6 faces.
There are 6 ways to get a sum of 7: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1.
Example 3:
Input: n = 30, k = 30, target = 500
Output: 222616187
Explanation: The answer must be returned modulo 109 + 7.
Constraints:
1 <= n, k <= 301 <= target <= 1000
Solutions
Solution 1: Dynamic Programming
We define f[i][j] as the number of ways to get a sum of j using i dice. Then, we can obtain the following state transition equation:
f[i][j] = ∑_{h=1}min(j, k) f[i-1][j-h]
where h represents the number of points on the i-th die.
Initially, we have f[0][0] = 1, and the final answer is f[n][target].
The time complexity is O(n × k × target), and the space complexity is O(n × target).
We notice that the state f[i][j] only depends on f[i-1][], so we can use a rolling array to optimize the space complexity to O(target).
PythonJavaC++GoTypeScriptRust
class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
f = [[0] * (target + 1) for _ in range(n + 1)]
f[0][0] = 1
mod = 10**9 + 7
for i in range(1, n + 1):
for j in range(1, min(i * k, target) + 1):
for h in range(1, min(j, k) + 1):
f[i][j] = (f[i][j] + f[i - 1][j - h]) % mod
return f[n][target](code-box)
class Solution {
public int numRollsToTarget(int n, int k, int target) {
final int mod = (int) 1e9 + 7;
int[][] f = new int[n + 1][target + 1];
f[0][0] = 1;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= Math.min(target, i * k); ++j) {
for (int h = 1; h <= Math.min(j, k); ++h) {
f[i][j] = (f[i][j] + f[i - 1][j - h]) % mod;
}
}
}
return f[n][target];
}
}(code-box)
class Solution {
public:
int numRollsToTarget(int n, int k, int target) {
const int mod = 1e9 + 7;
int f[n + 1][target + 1];
memset(f, 0, sizeof f);
f[0][0] = 1;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= min(target, i * k); ++j) {
for (int h = 1; h <= min(j, k); ++h) {
f[i][j] = (f[i][j] + f[i - 1][j - h]) % mod;
}
}
}
return f[n][target];
}
};(code-box)
func numRollsToTarget(n int, k int, target int) int {
const mod int = 1e9 + 7
f := make([][]int, n+1)
for i := range f {
f[i] = make([]int, target+1)
}
f[0][0] = 1
for i := 1; i <= n; i++ {
for j := 1; j <= min(target, i*k); j++ {
for h := 1; h <= min(j, k); h++ {
f[i][j] = (f[i][j] + f[i-1][j-h]) % mod
}
}
}
return f[n][target]
}(code-box)
function numRollsToTarget(n: number, k: number, target: number): number {
const f = Array.from({ length: n + 1 }, () => Array(target + 1).fill(0));
f[0][0] = 1;
const mod = 1e9 + 7;
for (let i = 1; i <= n; ++i) {
for (let j = 1; j <= Math.min(i * k, target); ++j) {
for (let h = 1; h <= Math.min(j, k); ++h) {
f[i][j] = (f[i][j] + f[i - 1][j - h]) % mod;
}
}
}
return f[n][target];
}(code-box)
impl Solution {
pub fn num_rolls_to_target(n: i32, k: i32, target: i32) -> i32 {
let _mod = 1_000_000_007;
let n = n as usize;
let k = k as usize;
let target = target as usize;
let mut f = vec![vec![0; target + 1]; n + 1];
f[0][0] = 1;
for i in 1..=n {
for j in 1..=target.min(i * k) {
for h in 1..=j.min(k) {
f[i][j] = (f[i][j] + f[i - 1][j - h]) % _mod;
}
}
}
f[n][target]
}
}(code-box)
Solution 2
PythonJavaC++GoTypeScriptRust
class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
f = [1] + [0] * target
mod = 10**9 + 7
for i in range(1, n + 1):
g = [0] * (target + 1)
for j in range(1, min(i * k, target) + 1):
for h in range(1, min(j, k) + 1):
g[j] = (g[j] + f[j - h]) % mod
f = g
return f[target](code-box)
class Solution {
public int numRollsToTarget(int n, int k, int target) {
final int mod = (int) 1e9 + 7;
int[] f = new int[target + 1];
f[0] = 1;
for (int i = 1; i <= n; ++i) {
int[] g = new int[target + 1];
for (int j = 1; j <= Math.min(target, i * k); ++j) {
for (int h = 1; h <= Math.min(j, k); ++h) {
g[j] = (g[j] + f[j - h]) % mod;
}
}
f = g;
}
return f[target];
}
}(code-box)
class Solution {
public:
int numRollsToTarget(int n, int k, int target) {
const int mod = 1e9 + 7;
vector<int> f(target + 1);
f[0] = 1;
for (int i = 1; i <= n; ++i) {
vector<int> g(target + 1);
for (int j = 1; j <= min(target, i * k); ++j) {
for (int h = 1; h <= min(j, k); ++h) {
g[j] = (g[j] + f[j - h]) % mod;
}
}
f = move(g);
}
return f[target];
}
};(code-box)
func numRollsToTarget(n int, k int, target int) int {
const mod int = 1e9 + 7
f := make([]int, target+1)
f[0] = 1
for i := 1; i <= n; i++ {
g := make([]int, target+1)
for j := 1; j <= min(target, i*k); j++ {
for h := 1; h <= min(j, k); h++ {
g[j] = (g[j] + f[j-h]) % mod
}
}
f = g
}
return f[target]
}(code-box)
function numRollsToTarget(n: number, k: number, target: number): number {
const f = Array(target + 1).fill(0);
f[0] = 1;
const mod = 1e9 + 7;
for (let i = 1; i <= n; ++i) {
const g = Array(target + 1).fill(0);
for (let j = 1; j <= Math.min(i * k, target); ++j) {
for (let h = 1; h <= Math.min(j, k); ++h) {
g[j] = (g[j] + f[j - h]) % mod;
}
}
f.splice(0, target + 1, ...g);
}
return f[target];
}(code-box)
impl Solution {
pub fn num_rolls_to_target(n: i32, k: i32, target: i32) -> i32 {
let _mod = 1_000_000_007;
let n = n as usize;
let k = k as usize;
let target = target as usize;
let mut f = vec![0; target + 1];
f[0] = 1;
for i in 1..=n {
let mut g = vec![0; target + 1];
for j in 1..=target {
for h in 1..=j.min(k) {
g[j] = (g[j] + f[j - h]) % _mod;
}
}
f = g;
}
f[target]
}
}(code-box)
