CareerCup

It could be similar to 8 queens problem.

HashMap map = new Map(); // Map to keep track of all numbers already used.
void placeNumber(int pos) {
      if (pos = 8)
               return;

      for (int i = 1; i < 9; i++) {
            int existingPos = map.get(i);
            if (existingPos == pos) {
                      // You need to remove old entry from map, if you reach back there again by backtracking
                       map.remove(i);
             }
            if (!map.exists(i) && isPossible(pos, i) {
                      map.put(i, pos);
                      placeNumber(pos+1)
            }
      }

Sorry formatting is not correct figure was like
YXY
XXX
XXX
YXY
assume figure like above pattern where X are square blocks and Y are empty spaces where figure does not exist... so you have fill 1 -8 in all X places according to above conditions... hope m clear now

Y7Y 
314
586
Y2Y

More outputs and many more
Y7Y Y2Y Y2Y Y3Y
413 685 314 547
685 413 586 261
Y2Y Y7Y Y7Y Y8Y

The logic behind this is,
Start filling any number in first row. say Y5Y. and proceed further with the given rule

b[k] ={1,2,3,4,5,6,7,8}
m = 3
n = 3

V()
	for each i from 0 to n-1
		for each j from 0 to m-1
			if((a[i][j] != a[i+1][j] +/- 1) && (a[i][j] != a[i][j+1] +/- 1) && (a[i][j] != a[i-1][j] +/- 1) && (a[i][j] != a[i][j-1] +/- 1))
				return true
			else 
				return false
				
SS(k)
	if(k == 7)
		if(V())
			for each i from 0 to n-1
				for each j from o to m-1
					print a[i][j]
	for each i from 0 to n-1
		for each i from 0 to m-1
			if(!((i == 0 && j ==0 ) || (i == 0 && j ==2) || (i == 2 && j ==0) || (i == 2 && j ==2)))
				a[i][j] = b[k]
				SS(k+1)

b[k] ={1,2,3,4,5,6,7,8}
m = 3
n = 3

V()
for each i from 0 to n-1
for each j from 0 to m-1
if((a[i][j] != a[i+1][j] +/- 1) && (a[i][j] != a[i][j+1] +/- 1) && (a[i][j] != a[i-1][j] +/- 1) && (a[i][j] != a[i][j-1] +/- 1))
return true
else
return false

SS(k)
if(k == 7)
if(V())
for each i from 0 to n-1
for each j from o to m-1
print a[i][j]
return
for each i from 0 to n-1
for each i from 0 to m-1
if(!((i == 0 && j ==0 ) || (i == 0 && j ==2) || (i == 2 && j ==0) || (i == 2 && j ==2)))
a[i][j] = b[k]
SS(k+1)