1155 - Number of Dice Rolls With Target Sum (Medium)
Problem Link
https://leetcode.com/problems/number-of-dice-rolls-with-target-sum/
Problem Statement
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
Approach 1: Dynamic Programming
class Solution {
public:
int numRollsToTarget(int n, int k, int target) {
int M = 1e9 + 7;
// dp[i][j]: number of ways to reach target j using i dice
vector<vector<int>> dp(n + 1, vector<int>(target + 1));
// there is 1 way to reach 0 using 0 dice
dp[0][0] = 1;
// iterate each dice
for (int i = 1; i <= n; i++) {
// iterate each target
for (int j = 1; j <= target; j++) {
// iterate each face
for (int f = 1; f <= k; f++) {
// we can use this f only if we add it
// the sum won't exceed target
if (j - f >= 0) {
(dp[i][j] += dp[i - 1][j - f]) %= M;
}
}
}
}
return dp[n][target];
}
};