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
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 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<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 codemain(){ 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 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][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 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
<?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)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
