Given a source and destination in a matrix[][] of infinite rows and columns, the task is to find whether it is possible to reach the destination from the source in an even number of steps. Also, you can only move up, down, left, and right.
Examples:
Input: Source = {2, 1}, Destination = {1, 4}
Output: YesInput: Source = {2, 2}, Destination = {1, 4}
Output: No
Observation:
The observation is that if the steps required to reach the destination in the shortest path is even, then steps required in every other route to reach it will always be even. Also, there can be an infinite number of ways to reach the target point. Some paths to reach (4, 1) from (1, 2) in a 4 x 5 matrix are given below :
Minimum number of steps required =4
So our problem is reduced in finding the minimum number of steps required to reach the destination from source in a matrix, which can be calculated easily by simply taking the sum of the absolute values of the difference between the X coordinates and Y coordinates.
Below is the implementation of the above approach:
C++
// C++ Program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check destination can be// reached from source in even number of// stepsvoid IsEvenPath(int Source[], int Destination[]){ // Coordinates differences int x_dif = abs(Source[0] - Destination[0]); int y_dif = abs(Source[1] - Destination[1]); // minimum number of steps required int minsteps = x_dif + y_dif; // Minsteps is even if (minsteps % 2 == 0) cout << "Yes"; // Minsteps is odd else cout << "No";}// Driver Codeint main(){ // Given Input int Source[] = { 2, 1 }; int Destination[] = { 1, 4 }; // Function Call IsEvenPath(Source, Destination); return 0;} |
Java
// Java program for the above approachimport java.lang.*;import java.util.*;class GFG{// Function to check destination can be// reached from source in even number of// stepsstatic void IsEvenPath(int Source[], int Destination[]){ // Coordinates differences int x_dif = Math.abs(Source[0] - Destination[0]); int y_dif = Math.abs(Source[1] - Destination[1]); // Minimum number of steps required int minsteps = x_dif + y_dif; // Minsteps is even if (minsteps % 2 == 0) System.out.println("Yes"); // Minsteps is odd else System.out.println("No");}// Driver codepublic static void main(String[] args){ // Given Input int Source[] = { 2, 1 }; int Destination[] = { 1, 4 }; // Function Call IsEvenPath(Source, Destination);}}// This code is contributed by sanjoy_62 |
Python3
# Python3 program for the above approach# Function to check destination can be# reached from source in even number of# stepsdef IsEvenPath(Source, Destination): # Coordinates differences x_dif = abs(Source[0] - Destination[0]) y_dif = abs(Source[1] - Destination[1]) # Minimum number of steps required minsteps = x_dif + y_dif # Minsteps is even if (minsteps % 2 == 0): print("Yes") # Minsteps is odd else: print("No")# Driver Codeif __name__ == '__main__': # Given Input Source = [ 2, 1 ] Destination = [ 1, 4 ] # Function Call IsEvenPath(Source, Destination)# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{// Function to check destination can be// reached from source in even number of// stepsstatic void IsEvenPath(int[] Source, int[] Destination){ // Coordinates differences int x_dif = Math.Abs(Source[0] - Destination[0]); int y_dif = Math.Abs(Source[1] - Destination[1]); // Minimum number of steps required int minsteps = x_dif + y_dif; // Minsteps is even if (minsteps % 2 == 0) Console.WriteLine("Yes"); // Minsteps is odd else Console.WriteLine("No");}// Driver codepublic static void Main(string[] args){ // Given Input int[] Source = { 2, 1 }; int[] Destination = { 1, 4 }; // Function Call IsEvenPath(Source, Destination);}}// This code is contributed by code_hunt. |
Javascript
<script> // JavaScript Program for the above approach // Function to check destination can be // reached from source in even number of // steps function IsEvenPath(Source, Destination) { // Coordinates differences let x_dif = Math.abs(Source[0] - Destination[0]); let y_dif = Math.abs(Source[1] - Destination[1]); // minimum number of steps required let minsteps = x_dif + y_dif; // Minsteps is even if (minsteps % 2 == 0) document.write("Yes"); // Minsteps is odd else document.write("No"); } // Driver Code // Given Input let Source = [2, 1]; let Destination = [1, 4]; // Function Call IsEvenPath(Source, Destination); // This code is contributed by Potta Lokesh </script> |
Yes
Time Complexity: O(1) // since no loop is used the algorithm takes up constant time to perform the operations
Auxiliary Space: O(1) // since no extra array is used so the space taken by the algorithm is constant
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