動機

這個真的想不到!!

  • 用dp紀錄狀態
  • 從dp(swap)找出另一個dp(fix)

Problem

You are given two integer arrays of the same length nums1 and nums2. In one operation, you are allowed to swap nums1[i] with nums2[i].

  • For example, if nums1 = [1,2,3,8], and nums2 = [5,6,7,4], you can swap the element at i = 3 to obtain nums1 = [1,2,3,4] and nums2 = [5,6,7,8].

Return the minimum number of needed operations to make nums1 and nums2 strictly increasing. The test cases are generated so that the given input always makes it possible.

An array arr is strictly increasing if and only if arr[0] < arr[1] < arr[2] < ... < arr[arr.length - 1].

 

Example 1:

Input: nums1 = [1,3,5,4], nums2 = [1,2,3,7]Output: 1Explanation: Swap nums1[3] and nums2[3]. Then the sequences are:nums1 = [1, 3, 5, 7] and nums2 = [1, 2, 3, 4]which are both strictly increasing.

Example 2:

Input: nums1 = [0,3,5,8,9], nums2 = [2,1,4,6,9]Output: 1

 

Constraints:

  • 2 <= nums1.length <= 105
  • nums2.length == nums1.length
  • 0 <= nums1[i], nums2[i] <= 2 * 105

Sol

fix是這一點沒有被swap的情況下成為increasing需要幾次swap swap是這一點被swap後的情況下成為increasing需要幾次swap

class Solution:
    def minSwap(self, nums1: List[int], nums2: List[int]) -> int:
        fix = 0
        swap = 1
        for i in range(1,len(nums1)):
            fixa = fixb = swapa = swapb = len(nums1)
            if nums1[i-1] < nums1[i] and nums2[i-1] < nums2[i]:
                fixa = fix
                swapa = swap+1
            if nums1[i-1] < nums2[i] and nums2[i-1] < nums1[i]:
                fixb = swap
                swapb = fix+1
            fix, swap = min(fixa,fixb), min(swapa,swapb)
        return min(fix,swap)