leetcode - 1155. Number of Dice Rolls With Target Sum

Description

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:

复制代码
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:

复制代码
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:

复制代码
Input: n = 30, k = 30, target = 500
Output: 222616187
Explanation: The answer must be returned modulo 109 + 7.

Constraints:

复制代码
1 <= n, k <= 30
1 <= target <= 1000

Solution

Recursive + memorization

The dp transformation equation is:
d p n t a r g e t = ∑ i = 1 k d p n − 1 t a r g e t − i dpntarget = \sum_{i=1}^{k}dpn-1target-i dpntarget=i=1∑kdpn−1target−i

So we could do recursive and memorization or dp.

Time complexity: o ( n ∗ t a r g e t ∗ k ) o(n*target*k) o(n∗target∗k)

Space complexity: o ( n ∗ t a r g e t ) o(n*target) o(n∗target)

DP

Feels top-bottom (recursive + memo) is easier to implement than bottom-top (dp here)

Code

Recursive + memorization

python3 复制代码
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)

DP

python3 复制代码
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]
相关推荐
其实防守也摸鱼12 小时前
个人学习 Pwn 思路:从基础到漏洞利用的进阶路径
大数据·前端·人工智能·chrome·算法·安全·网络安全
Nuanyt12 小时前
LeetCode Hot 100 链表专题(五)· 2 题(Java)
java·算法·leetcode·链表
w_zero_one15 小时前
递归算法(一)
java·开发语言·算法
圣保罗的大教堂16 小时前
leetcode 3736. 最小操作次数使数组元素相等 III 简单
leetcode
文丑颜不良啊16 小时前
LeetCode1881~1990题目解析
java·leetcode
风影同志7832417 小时前
20 个比特能装下多少行情?——实测拆解 Kronos 的 K 线分词器
算法
写的都是BUG17 小时前
APK 里的钥匙,一直就放在门口:聊聊加固真正该做什么
前端·算法
lsc_blog18 小时前
年底还在招的岗位,先确认缺口和到岗要求
git·缓存·ci/cd·职场和发展·github
地平线开发者18 小时前
CLIP 的零样本分类:一组文本如何变成图像分类器?
算法