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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 <bits/stdc++.h>
#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 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 steps
int countSteps(int curx, int cury, int n, int m,
               vector<pair<int, int> > 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<pair<int, int> > 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 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][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 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 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 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




<?php
// Php implementation of the above approach
 
// Function to return the count of
// possible steps in a single direction
function steps($cur, $x, $n)
{
 
    // It can cover infinite steps
    if ($x == 0)
        return PHP_INT_MAX;
 
    // We are approaching towards X = N
    if ($x > 0)
        return floor(abs(($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][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
    $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




<script>
 
// Javascript implementation of the above approach
  
    // Function to return the count of
    // possible steps in a single direction
    function steps(cur, x, n)
    {
 
        // It can cover infinite steps
        if (x == 0)
            return Number.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
    function countSteps(curx, cury, n, m, moves)
    {
        let count = 0;
        let k = moves.length;
        for (let i = 0; i < k; i++)
        {
            let x = moves[i][0];
            let y = moves[i][1];
 
            // Take the minimum of 
            // both moves independently
            let 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 Math.floor(count);
    }
 
// driver program
     
        let n = 4, m = 5, x = 1, y = 1;
        let moves = [[ 1, 1 ], [ 1, 1 ],
                          [ 0, -2 ]];
 
        document.write(countSteps(x, y, n, m, moves));
   
</script>
Output: 
4

 

Time Complexity: O(N)
 


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