You have d dice, and each die has f faces numbered 1, 2, ..., f.
Return the number of possible ways (out of fd total ways) modulo 10^9 + 7 to roll the dice so the sum of the face up numbers equals target.
Example 1:
Input: d = 1, f = 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: d = 2, f = 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: d = 2, f = 5, target = 10
Output: 1
Explanation:
You throw two dice, each with 5 faces. There is only one way to get a sum of 10: 5+5.
Example 4:
Input: d = 1, f = 2, target = 3
Output: 0
Explanation:
You throw one die with 2 faces. There is no way to get a sum of 3.
Example 5:
Input: d = 30, f = 30, target = 500
Output: 222616187
Explanation:
The answer must be returned modulo 10^9 + 7.
Constraints:
1 <= d, f <= 30
1 <= target <= 1000
// O(d*target) Space & O(d*f*target) Time
class Solution {
public int numRollsToTarget(int d, int f, int target) {
int MOD = 1000000007;
int[][] dp = new int[d + 1][target + 1];
dp[0][0] = 1;
// dp[i][j] -> how many ways can i dices sum up to j;
for (int i = 1; i <= d; i++) {
for (int j = 1; j <= target; j++) {
if (j > i * f)
continue; // If j is larger than largest possible sum of i dices, there is no possible
// ways.
else
for (int currentFace = 1; currentFace <= f && currentFace <= j; currentFace++)
dp[i][j] = (dp[i][j] + dp[i - 1][j - currentFace]) % MOD;
}
}
return dp[d][target];
}
}
// O(target) Space Solution
class Solution {
public int numRollsToTarget(int d, int f, int target) {
int MOD = 1000000007;
int[] dp = new int[target + 1];
dp[0] = 1;
// dp[j] -> how many ways to get sum up to j;
for (int i = 1; i <= d; i++) {
int[] dp2 = new int[target + 1];
for (int j = 1; j <= target; j++) {
if (j > i * f)
continue; // If j is larger than largest possible sum of i dices, there is no possible
// ways.
else
for (int currentFace = 1; currentFace <= f && currentFace <= j; currentFace++)
dp2[j] = (dp2[j] + dp[j - currentFace]) % MOD;
}
dp = dp2;
}
return dp[target];
}
}