You have n dice, and each die 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 10^9 + 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
The dp transformation equation is:
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dp[n][target] = \sum_{i=1}^{k}dp[n-1][target-i]
dp[n][target]=i=1∑k?dp[n?1][target?i]
So we could do recursive and memorization or dp.
Time complexity:
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o(n*target*k)
o(n?target?k)
Space complexity:
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o(n*target)
o(n?target)
Feels top-bottom (recursive + memo) is easier to implement than bottom-top (dp here)
class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
def helper(n: int, target: int) -> int:
if (n, target) in memo:
return memo[(n, target)]
if n == 1:
if 0 < target <= k:
memo[(n, target)] = 1
else:
memo[(n, target)] = 0
return memo[(n, target)]
res = 0
for i in range(1, k + 1):
res += helper(n - 1, target - i)
res %= mod_val
memo[(n, target)] = res
return memo[(n, target)]
mod_val = 1000000007
memo = {}
return helper(n, target)
class Solution:
def numRollsToTarget(self, n: int, k: int, target: int) -> int:
mod_val = 1000000007
dp = [[0] * (target + 1) for _ in range(n + 1)]
# init
for pseudo_target in range(1, min(target + 1, k + 1)):
dp[1][pseudo_target] = 1
for pseudo_n in range(2, n + 1):
for pseudo_t in range(1, target + 1):
for i in range(1, k + 1):
if pseudo_t - i >= 0:
dp[pseudo_n][pseudo_t] += dp[pseudo_n - 1][pseudo_t - i]
dp[pseudo_n][pseudo_t] %= mod_val
return dp[-1][-1]