# Count possible moves in the given direction in a grid

Given a grid N x M size and initial position (X, Y) and a series of K moves.
Each move contains 2 integers Dx and Dy.
A single move consists of the following operation that can also be performed as many times as possible:

X = X + Dx and Y = Y + Dy.

The task is to count the number of moves performed.
Note that at every point of time, the current position must be inside the grid.

Examples:

Input: N = 4, M = 5, X = 1, Y = 1, k[] = {{1, 1}, {1, 1}, {0, -2}}
Output: 4
The values of (X, Y) are as follows:
Steps in Move 1: (1, 1) -> (2, 2) -> (3, 3) -> (4, 4) i.e. 3 moves
Move 2 is not possible as we cannot move to (5, 5) which is out of the bounds of the grid
Steps in Move 3: (4, 4) -> (4, 2) i.e. a single move and (4, 0) will be out of the bound so no further moves possible.
Total moves = 3 + 1 = 4

Input: N = 10, M = 10, X = 3, Y = 3, k[] = {{1, 1}, {-1, -1}}
Output: 16

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: It can be observed that we can handle X and Y independently, and merge them as number of steps in current move= minimum of (number of steps with X, number of steps with Y). To find number of moves with X, we can use the sign of Dx to know which end we are approaching and find the distance to that end, with speed being Dx, we can find out the number of steps it would take.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `#define int long long ` `using` `namespace` `std; ` ` `  `// Function to return the count of ` `// possible steps in a single direction ` `int` `steps(``int` `cur, ``int` `x, ``int` `n) ` `{ ` ` `  `    ``// It can cover infinte steps ` `    ``if` `(x == 0) ` `        ``return` `INT_MAX; ` ` `  `    ``// We are approaching towards X = N ` `    ``if` `(x > 0) ` `        ``return` `abs``((n - cur) / x); ` ` `  `    ``// We are approaching towards X = 1 ` `    ``else` `        ``return` `abs``((cur - 1) / x); ` `} ` ` `  `// Function to return the count of steps ` `int` `countSteps(``int` `curx, ``int` `cury, ``int` `n, ``int` `m, ` `               ``vector > moves) ` `{ ` `    ``int` `count = 0; ` `    ``int` `k = moves.size(); ` `    ``for` `(``int` `i = 0; i < k; i++) { ` `        ``int` `x = moves[i].first; ` `        ``int` `y = moves[i].second; ` ` `  `        ``// Take the minimum of both moves independently ` `        ``int` `stepct ` `            ``= min(steps(curx, x, n), steps(cury, y, m)); ` ` `  `        ``// Update count and current positions ` `        ``count += stepct; ` `        ``curx += stepct * x; ` `        ``cury += stepct * y; ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `main() ` `{ ` `    ``int` `n = 4, m = 5, x = 1, y = 1; ` `    ``vector > moves = { { 1, 1 }, ` `                                      ``{ 1, 1 }, ` `                                      ``{ 0, -2 } }; ` `    ``int` `k = moves.size(); ` `    ``cout << countSteps(x, y, n, m, moves); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach ` `class` `GFG ` `{ ` ` `  `    ``// Function to return the count of ` `    ``// possible steps in a single direction ` `    ``static` `int` `steps(``int` `cur, ``int` `x, ``int` `n)  ` `    ``{ ` ` `  `        ``// It can cover infinte steps ` `        ``if` `(x == ``0``) ` `            ``return` `Integer.MAX_VALUE; ` ` `  `        ``// We are approaching towards X = N ` `        ``if` `(x > ``0``) ` `            ``return` `Math.abs((n - cur) / x); ` ` `  `        ``// We are approaching towards X = 1 ` `        ``else` `            ``return` `Math.abs((cur - ``1``) / x); ` `    ``} ` ` `  `    ``// Function to return the count of steps ` `    ``static` `int` `countSteps(``int` `curx, ``int` `cury, ` `                             ``int` `n, ``int` `m,  ` `                             ``int``[][] moves)  ` `    ``{ ` `        ``int` `count = ``0``; ` `        ``int` `k = moves.length; ` `        ``for` `(``int` `i = ``0``; i < k; i++) ` `        ``{ ` `            ``int` `x = moves[i][``0``]; ` `            ``int` `y = moves[i][``1``]; ` ` `  `            ``// Take the minimum of   ` `            ``// both moves independently ` `            ``int` `stepct = Math.min(steps(curx, x, n),  ` `                                  ``steps(cury, y, m)); ` ` `  `            ``// Update count and current positions ` `            ``count += stepct; ` `            ``curx += stepct * x; ` `            ``cury += stepct * y; ` `        ``} ` `        ``return` `count; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``4``, m = ``5``, x = ``1``, y = ``1``; ` `        ``int``[][] moves = { { ``1``, ``1` `}, { ``1``, ``1` `}, ` `                          ``{ ``0``, -``2` `} }; ` ` `  `        ``System.out.print(countSteps(x, y, n, m, moves)); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

## Python3

 `# Python3 implementation of the approach  ` ` `  `# Function to return the count of  ` `# possible steps in a single direction  ` `def` `steps(cur, x, n):  ` ` `  `    ``# It can cover infinte steps  ` `    ``if` `x ``=``=` `0``: ` `        ``return` `float``(``'inf'``)  ` ` `  `    ``# We are approaching towards X = N  ` `    ``elif` `x > ``0``: ` `        ``return` `abs``((n ``-` `cur) ``/``/` `x)  ` ` `  `    ``# We are approaching towards X = 1  ` `    ``else``: ` `        ``return` `abs``(``int``((cur ``-` `1``) ``/` `x))  ` ` `  `# Function to return the count of steps  ` `def` `countSteps(curx, cury, n, m, moves):  ` ` `  `    ``count ``=` `0` `    ``k ``=` `len``(moves)  ` `    ``for` `i ``in` `range``(``0``, k):  ` `        ``x ``=` `moves[i][``0``]  ` `        ``y ``=` `moves[i][``1``] ` ` `  `        ``# Take the minimum of both moves  ` `        ``# independently  ` `        ``stepct ``=` `min``(steps(curx, x, n),  ` `                     ``steps(cury, y, m))  ` ` `  `        ``# Update count and current positions  ` `        ``count ``+``=` `stepct  ` `        ``curx ``+``=` `stepct ``*` `x  ` `        ``cury ``+``=` `stepct ``*` `y ` `     `  `    ``return` `count  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``n, m, x, y ``=` `4``, ``5``, ``1``, ``1` `    ``moves ``=` `[[``1``, ``1``], [``1``, ``1``], [``0``, ``-``2``]] ` `     `  `    ``print``(countSteps(x, y, n, m, moves)) ` ` `  `# This code is contributed  ` `# by Rituraj Jain `

## C#

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to return the count of ` `    ``// possible steps in a single direction ` `    ``static` `int` `steps(``int` `cur, ``int` `x, ``int` `n)  ` `    ``{ ` ` `  `        ``// It can cover infinte steps ` `        ``if` `(x == 0) ` `            ``return` `int``.MaxValue; ` ` `  `        ``// We are approaching towards X = N ` `        ``if` `(x > 0) ` `            ``return` `Math.Abs((n - cur) / x); ` ` `  `        ``// We are approaching towards X = 1 ` `        ``else` `            ``return` `Math.Abs((cur - 1) / x); ` `    ``} ` ` `  `    ``// Function to return the count of steps ` `    ``static` `int` `countSteps(``int` `curx, ``int` `cury, ` `                          ``int` `n, ``int` `m,  ` `                          ``int``[,] moves)  ` `    ``{ ` `        ``int` `count = 0; ` `        ``int` `k = moves.GetLength(0); ` `        ``for` `(``int` `i = 0; i < k; i++) ` `        ``{ ` `            ``int` `x = moves[i, 0]; ` `            ``int` `y = moves[i, 1]; ` ` `  `            ``// Take the minimum of  ` `            ``// both moves independently ` `            ``int` `stepct = Math.Min(steps(curx, x, n),  ` `                                  ``steps(cury, y, m)); ` ` `  `            ``// Update count and current positions ` `            ``count += stepct; ` `            ``curx += stepct * x; ` `            ``cury += stepct * y; ` `        ``} ` `        ``return` `count; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int` `n = 4, m = 5, x = 1, y = 1; ` `        ``int``[,] moves = { { 1, 1 }, { 1, 1 }, ` `                         ``{ 0, -2 } }; ` ` `  `        ``Console.Write(countSteps(x, y, n, m, moves)); ` `    ``} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## PHP

 ` 0)  ` `        ``return` `floor``(``abs``((``\$n` `- ``\$cur``) / ``\$x``));  ` ` `  `    ``// We are approaching towards X = 1  ` `    ``else` `        ``return` `floor``(``abs``((``\$cur` `- 1) / ``\$x``));  ` `}  ` ` `  `// Function to return the count of steps  ` `function` `countSteps(``\$curx``, ``\$cury``, ``\$n``, ``\$m``, ``\$moves``)  ` `{  ` `    ``\$count` `= 0;  ` `    ``\$k` `= sizeof(``\$moves``);  ` `    ``for` `(``\$i` `= 0; ``\$i` `< ``\$k``; ``\$i``++)  ` `    ``{  ` `        ``\$x` `= ``\$moves``[``\$i``];  ` `        ``\$y` `= ``\$moves``[``\$i``];  ` ` `  `        ``// Take the minimum of both moves independently  ` `        ``\$stepct` `= min(steps(``\$curx``, ``\$x``, ``\$n``), steps(``\$cury``, ``\$y``, ``\$m``));  ` ` `  `        ``// Update count and current positions  ` `        ``\$count` `+= ``\$stepct``;  ` `        ``\$curx` `+= ``\$stepct` `* ``\$x``;  ` `        ``\$cury` `+= ``\$stepct` `* ``\$y``;  ` `    ``}  ` `    ``return` `\$count``;  ` `}  ` ` `  `    ``// Driver code  ` `    ``\$n` `= 4 ; ` `    ``\$m` `= 5 ; ` `    ``\$x` `= 1 ; ` `    ``\$y` `= 1 ;  ` `    ``\$moves` `= ``array``( ``array``( 1, 1 ),  ` `                        ``array``( 1, 1 ),  ` `                        ``array``( 0, -2 ) );  ` `    ``\$k` `= sizeof(``\$moves``); ` `     `  `    ``echo` `countSteps(``\$x``, ``\$y``, ``\$n``, ``\$m``, ``\$moves``);  ` `     `  `// This code is contributed by Ryuga ` `?> `

Output:

```4
```

Time Complexity: O(N) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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