Identical grids

Question

You’re given two \$r×c\$ grids. Each cell contains either 0 or 1. What are the minimum number of swaps (between horizontally and vertically adjacent cell elements, no wrapping i.e no swapping between last and first element of a row) are required in the first grid for it to match the second. If the matched arrangement can never be achieved, output -1.

Constraints

\$1 \leq r \leq 100\$
\$1 \leq c \leq 100 \$

Examples

input:
00  
11  

01  
10  
output:
1  

input:
00  
11  

01  
00  
output:
-1  

input:
0011011  

0101101  
output:
2 

Answer 1

Python 3, 383 334 bytes

-6 bytes thanks to @Kevin Cruijssen

-43 bytes thanks to @Jitse

import numpy
z=len
d=lambda a,b,c,d:abs(a-c)+abs(b-d)
def g(l,m):
 if z(l)==1:yield d(*l[0],*m[0])
 for i in range(z(l)):
  for s in g(l[1:],m[:i]+m[i+1:]):yield d(*l[0],*m[i])+s
def c(a,b):
 e=a-b;k=z(a);f,*h=[],
 for x in range(k*z(a[0])):w=[(x%k,x//k)];v=e[x%k][x//k];f+=w*(v>0);h+=w*(v<0)
 return-1 if numpy.sum(e)else min(g(f,h))

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Takes input as two numpy arrays.

Explanation:

First, e=a-b determines which positions change from one grid to the other. The next line finds all of the differences and sorts them into the lists f and h. If the sum of differences is not zero, meaning there are more 1s in one grid than the other, this returns -1, or the minimum of all possible paths found by g.

This is the fun part. Essentially, in order to make this work, a 1 must move from one position to another by swapping. Therefore, the minimum number of swaps for each pair is the manhattan distance between them, found in d. g finds every possible pairing between start and end points and returns a list of the total distances between them. If there is only one start point, it returns the distance between that and the end point. Beyond that, it pairs the first start point with each end point iteratively and adds their distance to the total distance of the rest of the points, calculated recursively.

Answer 2

Based on @Hiatsu's answer.

Python 3, 262 bytes

e=enumerate
def f(r,s):
 if not r:yield 0
 for i,t in e(r):yield abs(t[0]-s[0][0])+abs(t[1]-s[0][1])+min(f(r[:i]+r[i+1:],s[1:]))
def g(x,y):
 z=x-y;n=len(x.T);v,*w=[],
 for i,p in e(z.flat):t=[(i%n,i//n)];v+=t*(p>0);w+=t*(p<0)
 return-1if z.sum()else min(f(v,w))

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