801. Minimum Swaps To Make Sequences Increasing

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].

 

Approach #1: Brute force. [C++][TEL]

class Solution {
public:
    int minSwap(vector<int>& A, vector<int>& B) {
        int ans = INT_MAX;
        dfs(A, B, 1, 0, ans);
        return ans;
    }
    
private:
    void dfs(vector<int>& A, vector<int>& B, int i, int c, int& ans) {
        if (c >= ans) return;
        if (i == A.size()) {
            ans = min(ans, c);
            return;
        }
        
        if (A[i] > A[i-1] && B[i] > B[i-1]) 
            dfs(A, B, i+1, c, ans);
        
        if (A[i] > B[i-1] && B[i] > A[i-1]) {
            swap(A[i], B[i]);
            dfs(A, B, i+1, c+1, ans);
            swap(A[i], B[i]);
        }
    }
};

　　

Approach #2: DP. [Java]

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]) {
                keep[i] = keep[i-1];
                swap[i] = swap[i-1] + 1;
            }
            
            if (A[i] > B[i-1] && B[i] > A[i-1]) {
                swap[i] = Math.min(swap[i], keep[i-1] + 1);
                keep[i] = Math.min(keep[i], swap[i-1]);
            }
        }
        
        return Math.min(keep[n-1], swap[n-1]);
    }
}

　　

Analysis:

This problem can be solved using dynamic programming, at each position, we can choose to swap or not. Since we want two sorted arrays, at each position, whether to swap or not depends on the choice at previous position, so we can form a recursive formula.

 

When A[0, i-1] and B[0, i-1] are sorted, since "It is guaranted that the given input always makes it possible.". there are two cases on index i:

 

They are both still sorted when add number at index i, A[i] > A[i-1] && B[i] > B[i-1] 

They are not sorted when add number at index i, in this case, only A[i] > B[i-1] && B[i] > A[i-1] can guarantee that "the given input always makes it possible".

 

swap[i] to represent the minimum swaps to make the A[0, i] and B[0, i] equences increasing for 0 <= i <= n in condition that we swap A[i] and B[i].

keep[i] torepresent the minimum swaps to make the A[0, i] and B[i] sequences increasing for 0 <= i <= n in condition that we don't swap A[i] and B[i].

 

Reference:

http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-801-minimum-swaps-to-make-sequences-increasing/

https://leetcode.com/problems/minimum-swaps-to-make-sequences-increasing/discuss/120516/C%2B%2B-solution-with-explanation