1155 Number of Dice Rolls With Target Sum - Medium

Problem:

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 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

Problem Analysis:

High-Level Strategy:

The solution uses a recursive approach with memoization (dynamic programming). The dfs function calculates the number of ways to obtain the target sum with a given number of dice and the remaining target value. The result is then memoized to avoid redundant calculations.

Complexity:

• Time Complexity: O(n * k * target), where n is the number of dice, k is the number of faces on each die, and target is the target sum.
• Space Complexity: O(n * target) for memoization.

Solutions:

class Solution:
    def numRollsToTarget(self, n: int, k: int, target: int) -> int:
        MOD = 10**9 + 7
        memo = {}

        def dfs(dice, remaining_target):
            if dice == 0:
                return 1 if remaining_target == 0 else 0

            if (dice, remaining_target) in memo:
                return memo[(dice, remaining_target)]

            ways = 0
            for face in range(1, k + 1):
                if remaining_target >= face:
                    ways = (ways + dfs(dice - 1, remaining_target - face)) % MOD

            memo[(dice, remaining_target)] = ways
            return ways

        return dfs(n, target) % MOD

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