# Sliding Puzzle in C++

C++Server Side ProgrammingProgramming

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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 −

• exchange node[x, y] and node[nx, ny]

• if node is in visited, then −

• exchange node[x, y] and node[nx, ny]

• 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

Live Demo

#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
Updated on 08-Jun-2020 10:23:31