Given a point (X, Y) in a 2-D plane and an integer K, the task is to check whether it is possible to move from (0, 0) to the given point (X, Y) in exactly K moves. In a single move, the positions that are reachable from (X, Y) are (X, Y + 1), (X, Y – 1), (X + 1, Y) and (X – 1, Y).
Examples:
Input: X = 0, Y = 0, K = 2
Output: Yes
Move 1: (0, 0) -> (0, 1)
Move 2: (0, 1) -> (0, 0)
Input: X = 5, Y = 8, K = 20
Output: No
Approach: It is clear that the shortest path to reach (X, Y) from (0, 0) will be minMoves = (|X| + |Y|). So, if K < minMoves then it is impossible to reach (X, Y) but if K ? minMoves then after reaching (X, Y) in minMoves number of moves the remaining (K – minMoves) number of moves have to be even in order to remain at that point for the rest of the moves.
So it is possible to reach (X, Y) from (0, 0) only if K ? (|X| + |Y|) and (K – (|X| + |Y|)) % 2 = 0.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true if it is// possible to move from (0, 0) to// (x, y) in exactly k movesbool isPossible(int x, int y, int k){ // Minimum moves required int minMoves = abs(x) + abs(y); // If possible if (k >= minMoves && (k - minMoves) % 2 == 0) return true; return false;}// Driver codeint main(){ int x = 5, y = 8, k = 20; if (isPossible(x, y, k)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approach class GFG{ // Function that returns true if it is // possible to move from (0, 0) to // (x, y) in exactly k moves static boolean isPossible(int x, int y, int k) { // Minimum moves required int minMoves = Math.abs(x) + Math.abs(y); // If possible if (k >= minMoves && (k - minMoves) % 2 == 0) return true; return false; } // Driver code public static void main (String[] args) { int x = 5, y = 8, k = 20; if (isPossible(x, y, k)) System.out.println("Yes"); else System.out.println("No"); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach# Function that returns true if it is# possible to move from (0, 0) to# (x, y) in exactly k movesdef isPossible(x, y, k): # Minimum moves required minMoves = abs(x) + abs(y) # If possible if (k >= minMoves and (k - minMoves) % 2 == 0): return True return False# Driver codex = 5y = 8k = 20if (isPossible(x, y, k)): print("Yes")else: print("No")# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach using System;class GFG{ // Function that returns true if it is // possible to move from (0, 0) to // (x, y) in exactly k moves static bool isPossible(int x, int y, int k) { // Minimum moves required int minMoves = Math.Abs(x) + Math.Abs(y); // If possible if (k >= minMoves && (k - minMoves) % 2 == 0) return true; return false; } // Driver code public static void Main () { int x = 5, y = 8, k = 20; if (isPossible(x, y, k)) Console.Write("Yes"); else Console.Write("No"); } }// This code is contributed by Nidhi |
Javascript
<script>// javascript implementation of the approach // Function that returns true if it is // possible to move from (0, 0) to // (x, y) in exactly k moves function isPossible(x , y , k) { // Minimum moves required var minMoves = Math.abs(x) + Math.abs(y); // If possible if (k >= minMoves && (k - minMoves) % 2 == 0) return true; return false; } // Driver code var x = 5, y = 8, k = 20; if (isPossible(x, y, k)) document.write("Yes"); else document.write("No"); // This code contributed by shikhasingrajput </script> |
No
Time Complexity: O(1)
Auxiliary Space: O(1)
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