**Dice Throw - Amazon Top Interview Questions**

### Problem Statement :

Given integers n, faces, and total, return the number of ways it is possible to throw n dice with faces faces each to get total. Mod the result by 10 ** 9 + 7. Constraints 1 ≤ n, faces, total ≤ 100 Example 1 Input n = 2 faces = 6 total = 7 Output 6 Explanation There are 6 ways to make 7 with 2 6-sided dice: 1 and 6 6 and 1 2 and 5 5 and 2 3 and 4 4 and 3

### Solution :

` ````
Solution in C++ :
const int MOD = 1e9 + 7;
int solve(int n, int faces, int total) {
vector<int> dp(total + 1);
for (int i = 1; i <= min(faces, total); ++i) dp[i] = 1;
vector<int> pre(total + 1);
for (int turn = 2; turn <= n; ++turn) {
for (int i = 1; i <= total; ++i) pre[i] = (pre[i - 1] + dp[i]) % MOD;
for (int i = 1; i <= total; ++i) {
int l = max(0, i - faces - 1);
dp[i] = (pre[i - 1] - pre[l] + MOD) % MOD;
}
}
return dp[total];
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
int[][] dp;
int mod = (int) 1e9 + 7;
public int solve(int n, int faces, int total) {
dp = new int[n + 1][total + 1];
for (int i = 0; i <= n; i++) {
Arrays.fill(dp[i], -1);
}
return solve_(n, faces, total, 0);
}
public int solve_(int n, int faces, int total, int sum) {
if (n == 0 && total == sum) {
return 1;
}
if (n < 0 || sum > total) {
return 0;
}
if (dp[n][sum] != -1) {
return dp[n][sum];
}
int count = 0;
for (int i = 1; i <= faces; i++) {
count += solve_(n - 1, faces, total, sum + i) % mod;
count %= mod;
}
return dp[n][sum] = count;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, n, faces, total):
if total < n:
return 0
if total > n * faces:
return 0
previous = [1]
for i in range(n):
max = min(len(previous) + faces, total + 1)
current = []
running = 0
for j in range(max):
current.append(running)
if j < len(previous):
running += previous[j]
if j >= faces:
running -= previous[j - faces]
previous = current
return current[total] % (10 ** 9 + 7)
```

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