Count possible moves in the given direction in a grid

• Last Updated : 11 Jun, 2021

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:
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

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 longusing namespace std; // Function to return the count of// possible steps in a single directionint steps(int cur, int x, int n){     // It can cover infinite 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 stepsint 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 codemain(){    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 approachclass 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 infinite 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];            int y = moves[i];             // 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 directiondef steps(cur, x, n):     # It can cover infinite 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 stepsdef countSteps(curx, cury, n, m, moves):     count = 0    k = len(moves)    for i in range(0, k):        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 codeif __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 approachusing 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 infinite 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 stepsfunction 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?>

Javascript


Output:
4

Time Complexity: O(N)

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