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# Discover the variety of methods to fill the buckets with balls

Given N buckets and the infinite variety of balls. The utmost capability of every bucket is given in an array capability[], the duty is to seek out the quantity of the way to fill the buckets with balls such that every bucket has no less than 1 ball and all of the buckets have a distinct variety of balls in them. Return your reply to the modulo 10^9+7.

Examples:

Enter:Â  N = 1, capability[] = {6}
Output: 6
Rationalization: Since there is just oneÂ bucket. It might maintain any variety of ballsÂ starting from 1 to six.

Enter:Â N = 2, capability[] = {5, 8}
Output: 35
Rationalization: If the primary bucket has 1 ball in it then the second bucket cant have 1Â ball, so the second bucket has 8 â€“ 1 = 7 decisions. So, the primary bucket has 5 decisions, and for every alternative, the second bucket has 7 decisions. So whole there are 35 methods.

Strategy: This may be solved with the next thought:

Now we have to maintain every bucket with a distinct variety of balls. If 1st bucket comprises x variety of balls then buckets from 2nd to final canâ€™t have x variety of balls in them, which suggests they’ve their (capability â€“ 1) alternative. If observe rigorously, we will discover iâ€™th bucket has (capability of iâ€™th bucket â€“ i â€“ 1) decisions. Now we now have to rearrange the buckets in a manner such that the capability of the bucket is larger than (i â€“ 1).

Steps concerned within the implementation of code:

• Kind the capability vector.
• ans = (ans * (capability[i]-i)) % mod;
• return ans.

Beneath is the implementation of the above strategy:

## C++

 `#embrace ` `utilizing` `namespace` `std;` `Â `Â  `int` `totalWays(``int` `n, vector<``int``> capability)` `{` `Â Â Â Â ``lengthy` `lengthy` `int` `mod = 1000000007;` `Â `Â  `Â Â Â Â ` `Â Â Â Â ``kind(capability.start(), capability.finish());` `Â `Â  `Â Â Â Â ``bool` `flag = ``false``;` `Â Â Â Â ``lengthy` `lengthy` `int` `ans = 1;` `Â `Â  `Â Â Â Â ` `Â Â Â Â ``for` `(``int` `i = n - 1; i >= 0; i--) {` `Â Â Â Â Â Â Â Â ``if` `(capability[i] < i) {` `Â Â Â Â Â Â Â Â Â Â Â Â ``flag = ``true``;` `Â Â Â Â Â Â Â Â Â Â Â Â ``break``;` `Â Â Â Â Â Â Â Â ``}` `Â Â Â Â Â Â Â Â ``else` `Â Â Â Â Â Â Â Â Â Â Â Â ``ans = (ans % mod * (capability[i] - i) % mod)` `Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ``% mod;` `Â Â Â Â ``}` `Â `Â  `Â Â Â Â ``if` `(flag)` `Â Â Â Â Â Â Â Â ``return` `0;` `Â `Â  `Â Â Â Â ` `Â Â Â Â ``return` `int``(ans);` `}` `Â `Â  `int` `predominant()` `{` `Â `Â  `Â Â Â Â ``vector<``int``> capability = { 5, 8 };` `Â Â Â Â ``int` `n = 2;` `Â `Â  `Â Â Â Â ` `Â Â Â Â ``int` `ans = totalWays(n, capability);` `Â Â Â Â ``cout << ans;` `Â Â Â Â ``return` `0;` `}`

Time Complexity: O(n*log(n))
Auxilairy Area: O(1).

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