801. Minimum Swaps To Make Sequences Increasing

[LLancelot]

https://leetcode.com/problems/minimum-swaps-to-make-sequences-increasing/

题目

We have two integer sequences A and B of the same non-zero length.

We are allowed to swap elements A[i] and B[i]. Note that both elements are in the same index position in their respective sequences.

At the end of some number of swaps, A and B are both strictly increasing. (A sequence is strictly increasing if and only if A[0] < A[1] < A[2] < ... < A[A.length - 1].)

Given A and B, return the minimum number of swaps to make both sequences strictly increasing. It is guaranteed that the given input always makes it possible.

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

A, B are arrays with the same length, and that length will be in the range [1, 1000].
A[i], B[i] are integer values in the range [0, 2000].

代码

class Solution {
    public int minSwap(int[] A, int[] B) {
        int n = A.length;
        int[] keep = new int[n];
        int[] swap = new int[n];
        Arrays.fill(keep, Integer.MAX_VALUE);
        Arrays.fill(swap, Integer.MAX_VALUE);
        keep[0] = 0;
        swap[0] = 1;
        for (int i = 1; i < n; i++) {
            if (A[i] > A[i-1] && B[i] > B[i-1]) {
                // Good case, no swapping needed.
                keep[i] = keep[i-1];

                // Swapped A[i-1] / B[i-1], swap A[i], B[i] as well
                swap[i] = swap[i-1] + 1;
            }
            
            if (A[i] > B[i-1] && B[i] > A[i-1]) {
                // A[i-1] / B[i-1] weren't swapped.
                keep[i] = Math.min(keep[i], swap[i-1]);
                
                // Swapped A[i-1] / B[i-1], no swap needed for A[i] / B[i]
                swap[i] = Math.min(swap[i], keep[i-1] + 1);
            }
        }
        
        return Math.min(keep[n-1], swap[n-1]);
    }
}