# 1155. Number of Dice Rolls With Target Sum

## 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 <= 30
• 1 <= 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:

$f[i][j] = \sum_{h=1}^{\min(j, k)} f[i-1][j-h]$

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 $O(n \times k \times target)$, and the space complexity is $O(n \times target)$.

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 $O(target)$.

• 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];
}
}

• 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];
}
};

• 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]


• 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]
}

• 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];
}


• 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]
}
}