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Sliding Puzzle in C++
Suppose we have one 2x3 board, there are 5 tiles those are represented by the numbers 1 through 5, and one empty square is there, that is represented by 0.
Here a move means 0 and one adjacent number (top, bottom, left or right) and swapping it. This will be solved when the elements are arranged in this manner: [[1,2,3],[4,5,0]].
We have the puzzle board; we have to find the least number of moves required so that the state of the board is solved. If this is not possible to solve, then return -1.
So, if the input is like [[1,2,3],[0,4,5]], then the output will be 2, as we have to swap [0,4], then [0,5].
To solve this, we will follow these steps −
Define one function slidingPuzzle(), this will take board as input
if board is perfectly arranged then −
return 0
Define one queue q of 2d matrices
insert board into q
Define one set visited for 2d matrices
insert board into visited
for initialize lvl := 1, when not q is empty, update (increase lvl by 1), do −
sz := size of q
while sz is non-zero, decrease sz after each iteration, do −
Define one 2D array node = front element of q
delete element from q
dx := -1, y := -1
for initialize i := 0, when i < size of board, update (increase i by 1), do −
for initialize j := 0, when j < size of board[0], update (increase j by 1), do −
if node[i, j] is same as 0, then −
x := i
y := j
Come out from the loop
for initialize k := 0, when k < 4, update (increase k by 1), do −
if nx < 0 or ny < 0 or nx >= row count of board or ny >= column count of board, then −
Ignore following part, skip to the next iteration
exchange node[x, y] and node[nx, ny]
if node is in visited, then −
exchange node[x, y] and node[nx, ny]
Ignore following part, skip to the next iteration
insert node into visited
if node is perfect arrangemen of boards, then −
return lvl
insert node into q
exchange node[x, y] and node[nx, ny]
return -1
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
int dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
class Solution {
public:
bool ok(vector < vector <int> >& b){
return b[0][0] == 1 && b[0][1] == 2 && b[0][2] == 3 && b[1]
[0] == 4 && b[1][1] == 5;
}
int slidingPuzzle(vector<vector<int>>& board) {
if (ok(board))
return 0;
queue<vector<vector<int> > > q;
q.push(board);
set<vector<vector<int> > > visited;
visited.insert(board);
for (int lvl = 1; !q.empty(); lvl++) {
int sz = q.size();
while (sz--) {
vector<vector<int> > node = q.front();
q.pop();
int x = -1;
int y = -1;
for (int i = 0; i < board.size(); i++) {
for (int j = 0; j < board[0].size(); j++) {
if (node[i][j] == 0) {
x = i;
y = j;
break;
}
}
}
for (int k = 0; k < 4; k++) {
int nx = x + dir[k][0];
int ny = y + dir[k][1];
if (nx < 0 || ny < 0 || nx >= board.size() || ny
>= board[0].size())
continue;
swap(node[x][y], node[nx][ny]);
if (visited.count(node)) {
swap(node[x][y], node[nx][ny]);
continue;
}
visited.insert(node);
if (ok(node))
return lvl;
q.push(node);
swap(node[x][y], node[nx][ny]);
}
}
}
return -1;
}
};
main(){
Solution ob;
vector<vector<int>> v = {{1,2,3},{0,4,5}};
cout << (ob.slidingPuzzle(v));
}Input
{{1,2,3},{0,4,5}}Output
2
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