# 10.09.2023 [1359. Count All Valid Pickup and Delivery Options]

Count permutations of the `n` `pickup -> delivery` orders

10.09.2023

1359. Count All Valid Pickup and Delivery Options hard
blog post

Join me on Telegram

https://t.me/leetcode_daily_unstoppable/335

Problem TLDR

Count permutations of the n pickup -> delivery orders

Intuition

Let’s look at how orders can be placed and draw the picture:

      // 1: p1 d1            variantsCount = 1
      // 2:                  length = 2
      // "___p1____d1_____": vacantPlaces = 3
      //              p2 d2
      //        p2       d2
      // p2              d2
      //        p2 d2
      // p2        d2
      // p2 d2
      //                                variantsCount = 6
      // 3:                             length = 4
      // "___p1____d1____p2____d2____": vacantPlaces = 5
      //                         p3 d3 
      //                    p3      d3
      //              p3            d3
      //        p3                  d3
      // p3                         d3
      //                    p3 d3
      //              p3       d3             x6
      //        p3             d3
      // p3                    d3
      //              p3 d3
      //        p3       d3
      // p3              d3
      //        p3 d3
      // p3        d3
      // p3 d3 

In this example, we can see the pattern:

• the number of vacant places grows by 2 each round

• inside each round there are repeating parts of arithmetic sum, that can be reused

Approach

• use Long to avoid overflow

Complexity

• Time complexity:
  O(n)

• Space complexity:
  O(1)

Code

    fun countOrders(n: Int): Int {
      var variantsCount = 1L
      var currSum = 1L
      var item = 1L
      val m = 1_000_000_007L
      repeat(n - 1) {
        item = (item + 1L) % m
        currSum = (currSum + item) % m
        item = (item + 1L) % m
        currSum = (currSum + item) % m
        variantsCount = (variantsCount * currSum) % m
      }
      return variantsCount.toInt()
    }