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# 2850. Minimum Moves to Spread Stones Over Grid
## Problem Statement
You are given a **0-indexed** 2D integer matrix `grid` of size `3 * 3`, representing the number of stones in each cell. The grid contains exactly `9` stones, and there can be **multiple** stones in a single cell.
In one move, you can move a single stone from its current cell to any other cell if the two cells share a side.
Return *the **minimum number of moves** required to place one stone in each cell*.
**Example 1:**
Input: grid = [[1,1,0],[1,1,1],[1,2,1]]
Output: 3
Explanation: One possible sequence of moves to place one stone in each cell is:
1- Move one stone from cell (2,1) to cell (2,2).
2- Move one stone from cell (2,2) to cell (1,2).
3- Move one stone from cell (1,2) to cell (0,2).
In total, it takes 3 moves to place one stone in each cell of the grid.
It can be shown that 3 is the minimum number of moves required to place one stone in each cell.
**Example 2:**
Input: grid = [[1,3,0],[1,0,0],[1,0,3]]
Output: 4
Explanation: One possible sequence of moves to place one stone in each cell is:
1- Move one stone from cell (0,1) to cell (0,2).
2- Move one stone from cell (0,1) to cell (1,1).
3- Move one stone from cell (2,2) to cell (1,2).
4- Move one stone from cell (2,2) to cell (2,1).
In total, it takes 4 moves to place one stone in each cell of the grid.
It can be shown that 4 is the minimum number of moves required to place one stone in each cell.
**Constraints:**
* `grid.length == grid[i].length == 3`
* `0 <= grid[i][j] <= 9`
* Sum of `grid` is equal to `9`.
## Intuition
```
Here BFS wont work, as stone collection is dependent on other
So we use Recursion and try all ways
```
### Links
### Video Links
### Approach 1:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
class Solution {
public:
vector> xtra,zero;
int ret;
void solve(int i, int count){
if(i>=zero.size()){
ret = min(ret, count);
return;
}
for(int k=0; k>& grid) {
for(int i=0; i<3; i++){
for(int j=0; j<3; j++){
if(grid[i][j] == 0)
zero.push_back({i,j});
else if(grid[i][j] > 1)
xtra.push_back({i,j,grid[i][j]-1});
}
}
if(zero.size() == 0) return 0;
ret = INT_MAX;
solve(0,0);
return ret;
}
};
```
{% endcode %}
### Approach 2:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Approach 3:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Approach 4:
```
```
{% code title="C++" lineNumbers="true" %}
```cpp
```
{% endcode %}
### Similar Problems
###
---
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