Count All Valid Pickup and Delivery Options - Math - Permutation [JS]

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Description

Solution: Math - Permutations

pickups: n!
deliveries: (1 * 3 * 5 * 7 * ...) n times

e.g: P1 P2 D1 D2

deliveries:
1 choice:
P1 P2 -> put D2: only 1 choice -> P1 P2 D2
P2 P1 -> put D1: only 1 choice -> P2 P1 D1

3 choices:
P1 P2 D2 -> put D1: 3 choices -> (P1 D1 P2 D2), (P1 P2 D1 D2), (P1 P2 D2 D1)
P2 P1 D1 -> put D2: 3 choices -> (P2 D2 P1 D1), (P2 P1 D2 D1), (P2 P1 D1 D2)

It can be proven that the total number of valid pickups/deliveries is n! * (1 * 3 * 5 * 7 * ...) n times

deliveries:
0 -> 1
1 -> 3
2 -> 5
3 -> 7
The formula is (i * 2 + 1) for the deliveries

Time Complexity: O(n) 101ms
Space Complexity: O(1) 42.6MB

var countOrders = function(n) {
  let ans = 1, mod = 10 ** 9 + 7;
  for (let i = 0; i < n; i++) {
    ans = (ans * (i + 1)) % mod;
    ans = (ans * (i * 2 + 1)) % mod;
  }
  return ans;
};

javascript
Location: Auckland, New Zealand

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