# Minimum cells traversed to reach corner where every cell represents jumps

Suppose A is at position (0, 0) of a 2-D grid containing ‘m’ rows and ‘n’ columns. His aim is to reach the bottom right point of this grid traveling through as minimum number of cells as possible.

Each cell of the grid contains a positive integer that defines the number of cells A can jump either in the right or the downward direction when he reaches that cell.

Find the minimum no of cells that need to be touched in order to reach bottom right corner.

Examples:

```Input : 2 4 2
5 3 8
1 1 1
Output :
So following two paths exist to reach (2, 2) from (0, 0)
(0, 0) => (0, 2) => (2, 2)
(0, 0) => (2, 0) => (2, 1) => (2, 2)

Hence the output for this test case should be 3
```

Following is a Breadth First Search(BFS) solution of the problem:

1. Think of this matrix as tree and (0, 0) as root and apply BFS using level order traversal.
2. Push the coordinates and no of jumps in a queue.
3. Pop the queue after every level of tree.
4. Add the value at cell to the coordinates while traversing right and downward direction.
5. Return no of cells touched while jumping when it reaches bottom right corner.
`// C++ program to reach bottom right corner using`
`// minimum jumps.`
`#include <bits/stdc++.h>`
`using` `namespace` `std;`
`#define R 3`
`#define C 3`
`// function to check coordinates are in valid range.`
`bool` `safe(``int` `x, ``int` `y)`
`{`
`    ``if` `(x < R && y < C && x >= 0 && y >= 0)`
`        ``return` `true``;`
`    ``return` `false``;`
`}`
`// function to return minimum no of cells to reach`
`// bottom right cell.`
`int` `matrixJump(``int` `M[R][C], ``int` `R1, ``int` `C1)`
`{`
`    ``queue<pair<``int``, pair<``int``, ``int``> > > q;`
`    ``// push the no of cells and coordinates in a queue.`
`    ``q.push(make_pair(1, make_pair(R1, C1)));`
`    ``while` `(!q.empty()) {`
`        ``int` `x = q.front().second.first; ``// x coordinate`
`        ``int` `y = q.front().second.second; ``// y coordinate`
`        ``int` `no_of_cells = q.front().first; ``// no of cells`
`        ``q.pop();`
`        ``// when it reaches bottom right return no of cells`
`        ``if` `(x == R - 1 && y == C - 1)            `
`            ``return` `no_of_cells;`
`        ``int` `v = M[x][y];`
`        ``if` `(safe(x + v, y))`
`            ``q.push(make_pair(no_of_cells + 1, make_pair(x + v, y)));`
`        ``if` `(safe(x, y + v))`
`            ``q.push(make_pair(no_of_cells + 1, make_pair(x, y + v)));`
`    ``}`
`    ``// when destination cannot be reached`
`    ``return` `-1;`
`}`
`// driver function`
`int` `main()`
`{`
`    ``int` `M[R][C] = { { 2, 4, 2 },`
`                    ``{ 5, 3, 8 },`
`                    ``{ 1, 1, 1 } };`
`    ``cout << matrixJump(M, 0, 0);`
`    ``return` `0;`
`}`

Output:

``` 3

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rakesh