🔴 Hot News: dsa/check if it is possible to move from x to z using point y

Given three points X, Y, and Z of a 2D plane, the task is to check if it is possible to move from X to Z using point Y with at most one turn of 90 degrees. If possible return "YES" else return "NO".

Note: Diagonal moves are not allowed, only vertical or horizontal moves are allowed.

Examples:

Input: X = {1, 3}, Y{4, 3}, Z = {4, 5}
Output: YES
Explanation:

Explanation of first test case

Input: X = {1, 1}, Y = {2, 2}, Z = {3, 3}
Output: NO
Explanation: It can be verified that we can't reach point Z from X using point Y.

Brute Force Approach :

We can check all possible paths from X to Z that involve at most one turn of 90 degrees at Y. We can start by moving from X to Y in a straight line, and then from Y to Z in a straight line, and check if this path is valid. If not, we can check all paths that involve one turn of 90 degrees at Y by considering all possible directions of the turn (left, right, up, or down), and checking if each path is valid. If we find a valid path, we can return "YES", otherwise we return "NO".

Below is the Implementation of the above approach :

C++

#include <iostream>
using namespace std;

struct Point {
    int x, y;
};

// function to check if two points are on the same horizontal or vertical line
bool isOnLine(Point p1, Point p2) {
    return (p1.x == p2.x || p1.y == p2.y);
}

// function to check if it is possible to move from point X to point Z
// using point Y with at most one turn of 90 degrees
string isPossible(Point X, Point Y, Point Z) {
    // check if there is a straight line path from X to Z
    if (isOnLine(X, Z)) {
        return "YES";
    }
    // check all possible paths that involve one turn of 90 degrees at Y
    // by considering all possible directions of the turn
    for (int dx = -1; dx <= 1; dx++) {
        for (int dy = -1; dy <= 1; dy++) {
            // check if the turn direction is valid
            if (dx == 0 || dy == 0 || dx == dy || dx == -dy) {
                // calculate the intermediate point after the turn
                Point p = {Y.x + dx, Y.y + dy};
                // check if the path from X to p and from p to Z is valid
                if (isOnLine(X, p) && isOnLine(p, Z)) {
                    return "YES";
                }
            }
        }
    }
    // if no valid path is found, return "NO"
    return "NO";
}

int main() {
    Point X = {1, 3};  // example points
    Point Y = {4, 3};
    Point Z = {4, 5};
    cout << isPossible(X, Y, Z) << endl;  // output: YES
    return 0;
}

Java

public class Main {

    static class Point {
        int x, y;
        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }

    // function to check if two points are on the same horizontal or vertical line
    static boolean isOnLine(Point p1, Point p2) {
        return (p1.x == p2.x || p1.y == p2.y);
    }

    // function to check if it is possible to move from point X to point Z
    // using point Y with at most one turn of 90 degrees
    static String isPossible(Point X, Point Y, Point Z) {
        // check if there is a straight line path from X to Z
        if (isOnLine(X, Z)) {
            return "YES";
        }
        // check all possible paths that involve one turn of 90 degrees at Y
        // by considering all possible directions of the turn
        for (int dx = -1; dx <= 1; dx++) {
            for (int dy = -1; dy <= 1; dy++) {
                // check if the turn direction is valid
                if (dx == 0 || dy == 0 || dx == dy || dx == -dy) {
                    // calculate the intermediate point after the turn
                    Point p = new Point(Y.x + dx, Y.y + dy);
                    // check if the path from X to p and from p to Z is valid
                    if (isOnLine(X, p) && isOnLine(p, Z)) {
                        return "YES";
                    }
                }
            }
        }
        // if no valid path is found, return "NO"
        return "NO";
    }

    public static void main(String[] args) {
        Point X = new Point(1, 3); // example points
        Point Y = new Point(4, 3);
        Point Z = new Point(4, 5);
        System.out.println(isPossible(X, Y, Z)); // output: YES
    }
}
// This code is contributed by Prajwal Kandekar

Python3

class Point:
    def __init__(self, x, y):
        self.x = x
        self.y = y
  
# function to check if two points are on the same horizontal or vertical line     
def isOnLine(p1, p2):
    return (p1.x == p2.x or p1.y == p2.y)

# function to check if it is possible to move from point X to point Z
# using point Y with at most one turn of 90 degrees
def isPossible(X, Y, Z):
    
    # check if there is a straight line path from X to Z
    if isOnLine(X, Z):
        return "YES"
    
    # check all possible paths that involve one turn of 90 degrees at Y
    # by considering all possible directions of the turn
    for dx in range(-1, 2):
        for dy in range(-1, 2):
            
            # check if the turn direction is valid
            if dx == 0 or dy == 0 or dx == dy or dx == -dy:
                
                # calculate the intermediate point after the turn
                p = Point(Y.x + dx, Y.y + dy)
                
                # check if the path from X to p and from p to Z is valid
                if isOnLine(X, p) and isOnLine(p, Z):
                    return "YES"
                    
    # if no valid path is found, return "NO"            
    return "NO"

X = Point(1, 3) # example points
Y = Point(4, 3)
Z = Point(4, 5)
print(isPossible(X, Y, Z)) # output: YES

using System;

public class Program {
    static void Main(string[] args)
    {
        Point X = new Point(1, 3); // example points
        Point Y = new Point(4, 3);
        Point Z = new Point(4, 5);
        Console.WriteLine(
            isPossible(X, Y, Z)); // output: YES
    }

    class Point {
        public int x, y;
        public Point(int x, int y)
        {
            this.x = x;
            this.y = y;
        }
    }

    // function to check if two points are on the same
    // horizontal or vertical line
    static bool isOnLine(Point p1, Point p2)
    {
        return (p1.x == p2.x || p1.y == p2.y);
    }

    // function to check if it is possible to move from
    // point X to point Z using point Y with at most one
    // turn of 90 degrees
    static string isPossible(Point X, Point Y, Point Z)
    {
        // check if there is a straight line path from X to
        // Z
        if (isOnLine(X, Z)) {
            return "YES";
        }
        // check all possible paths that involve one turn of
        // 90 degrees at Y by considering all possible
        // directions of the turn
        for (int dx = -1; dx <= 1; dx++) {
            for (int dy = -1; dy <= 1; dy++) {
                // check if the turn direction is valid
                if (dx == 0 || dy == 0 || dx == dy
                    || dx == -dy) {
                    // calculate the intermediate point
                    // after the turn
                    Point p = new Point(Y.x + dx, Y.y + dy);
                    // check if the path from X to p and
                    // from p to Z is valid
                    if (isOnLine(X, p) && isOnLine(p, Z)) {
                        return "YES";
                    }
                }
            }
        }
        // if no valid path is found, return "NO"
        return "NO";
    }
}

// This code is contributed by Tapesh(tapeshdua420)

JavaScript

function Point(x, y) {
  this.x = x;
  this.y = y;
}

// function to check if two points are 
// on the same horizontal or vertical line
function isOnLine(p1, p2) {
  return (p1.x == p2.x || p1.y == p2.y);
}

// function to check if it is possible to move from point X to point Z
// using point Y with at most one turn of 90 degrees
function isPossible(X, Y, Z) {
  // check if there is a straight line path from X to Z
  if (isOnLine(X, Z)) {
    return "YES";
  }
  // check all possible paths that involve one turn of 90 degrees at Y
  // by considering all possible directions of the turn
  for (let dx = -1; dx <= 1; dx++) {
    for (let dy = -1; dy <= 1; dy++) {
      // check if the turn direction is valid
      if (dx == 0 || dy == 0 || dx == dy || dx == -dy) {
        // calculate the intermediate point after the turn
        let p = new Point(Y.x + dx, Y.y + dy);
        // check if the path from X to p and from p to Z is valid
        if (isOnLine(X, p) && isOnLine(p, Z)) {
          return "YES";
        }
      }
    }
  }
  
  // if no valid path is found, return "NO"
  return "NO";
}

let X = new Point(1, 3);  // example points
let Y = new Point(4, 3);
let Z = new Point(4, 5);
console.log(isPossible(X, Y, Z));  // output: YES

Output :

YES

Complexity Analysis :

The time complexity of this solution is O(1) for the initial check if X and Z are on the same line, and O(9) for the nested loops that consider all possible paths with one turn at Y. Therefore, the overall time complexity is O(1) + O(9) = O(1).

The auxiliary space of this solution is O(1) as it only uses a constant amount of additional space to store the three points and the intermediate point after the turn.

Approach: Implement the idea below to solve the problem:

The problem is observation based and can be solved via implementing those observations. For more clarification see the Concept of approach section below.

Concept of approach:

It should be noted that Reaching from X to Z is only possible when Y is an intermediate point between X and Z. We can move in the horizontal or vertical direction, So that either the Y should be in between x coordinate of X and Z or y coordinate of X and Z also. So the conditions at which reaching is possible are:

x2 ? min(x1, x3) && x2 ? max(x1, x3) && (y2 == y1 || y2 == y3)
y2 ? min(y1, y3) && y2 ? max(y1, y3) && (x2 == x1 || x2 == x3)

Where X = {x1, y1}, Y = {x2, y2}, Z = {x3, y3}. All other cases will have no path with at most one turn.

Follow the below steps to solve the problem:

Check for the conditions:
- if (x2 ? min(x1, x3) && x2 ? max(x1, x3) && (y2 == y1 ||y2 == y3))
- else if (y2 ? min(y1, y3) && y2 ? max(y1, y3) && (x2 == x1 || x2 == x3)), Only at these conditions a path will exist.
Otherwise, no path is possible.

Below is the Implementation of the above approach:

C++

// C++ code to implement the approach
#include <algorithm>
#include <iostream>

using namespace std;

int main()
{
  // Inputs of X, Y and Z
  long x1 = 1, x2 = 2, x3 = 2, y1 = 4, y2 = 4, y3 = 5;

  // Minimum X coordinate from x1 and x3
  long smallx = min(x1, x3);

  // Maximum X coordinate from x1 and x3
  long bigx = max(x1, x3);

  // Minimum Y coordinate from y1 and y3
  long bigy = max(y1, y3);

  // Maximum Y coordinate from y1 and y3
  long smally = min(y1, y3);

  // Checking for conditions
  if (x2 >= smallx && x2 <= bigx
      && (y2 == y1 || y2 == y3)) {
    cout << "YES" << endl;
  }
  else if (y2 >= smally && y2 <= bigy
           && (x2 == x1 || x2 == x3)) {
    cout << "YES" << endl;
  }
  else {
    cout << "NO" << endl;
  }
  return 0;
}

Java

// Java code to implement the approach

public class GFG {

    // Driver Function
    public static void main(String[] args)
    {

        // Inputs of X, Y and Z
        long x1 = 1, x2 = 2, x3 = 2, y1 = 4, y2 = 4, y3 = 5;

        // Minimum X coordinate from x1 and x3
        long smallx = Math.min(x1, x3);

        // Maximum X coordinate from x1 and x3
        long bigx = Math.max(x1, x3);

        // Minimum Y coordinate from y1 and y3
        long bigy = Math.max(y1, y3);

        // Maximum Y coordinate from y1 and y3
        long smally = Math.min(y1, y3);

        // Checking for conditions
        if (x2 >= smallx && x2 <= bigx
            && (y2 == y1 || y2 == y3)) {
            System.out.println("YES");
        }
        else if (y2 >= smally && y2 <= bigy
                 && (x2 == x1 || x2 == x3)) {
            System.out.println("YES");
        }
        else
            System.out.println("NO");
    }
}

Python3

# Python code to implement the approach
import sys

# Inputs of X, Y and Z
x1, x2, x3, y1, y2, y3 = 1, 2, 2, 4, 4, 5

# Minimum X coordinate from x1 and x3
smallx = min(x1, x3)

# Maximum X coordinate from x1 and x3
bigx = max(x1, x3)

# Minimum Y coordinate from y1 and y3
bigy = max(y1, y3)

# Maximum Y coordinate from y1 and y3
smally = min(y1, y3)

# Checking for conditions
if x2 >= smallx and x2 <= bigx and (y2 == y1 or y2 == y3):
    print("YES")
elif y2 >= smally and y2 <= bigy and (x2 == x1 or x2 == x3):
    print("YES")
else:
    print("NO")
# This code is contributed by prasad264

// C# code to implement the approach
using System;

public class GFG {
  public static void Main(string[] args)
  {
    
    // Inputs of X, Y and Z
    long x1 = 1, x2 = 2, x3 = 2, y1 = 4, y2 = 4, y3 = 5;

    // Minimum X coordinate from x1 and x3
    long smallx = Math.Min(x1, x3);

    // Maximum X coordinate from x1 and x3
    long bigx = Math.Max(x1, x3);

    // Minimum Y coordinate from y1 and y3
    long bigy = Math.Max(y1, y3);

    // Maximum Y coordinate from y1 and y3
    long smally = Math.Min(y1, y3);

    // Checking for conditions
    if (x2 >= smallx && x2 <= bigx
        && (y2 == y1 || y2 == y3)) {
      Console.WriteLine("YES");
    }
    else if (y2 >= smally && y2 <= bigy
             && (x2 == x1 || x2 == x3)) {
      Console.WriteLine("YES");
    }
    else {
      Console.WriteLine("NO");
    }
  }
}

// This code is contributed by prasad264

JavaScript

// Inputs of X, Y and Z
let x1 = 1, x2 = 2, x3 = 2, y1 = 4, y2 = 4, y3 = 5;

// Minimum X coordinate from x1 and x3
let smallx = Math.min(x1, x3);

// Maximum X coordinate from x1 and x3
let bigx = Math.max(x1, x3);

// Minimum Y coordinate from y1 and y3
let bigy = Math.max(y1, y3);

// Maximum Y coordinate from y1 and y3
let smally = Math.min(y1, y3);

// Checking for conditions
if (x2 >= smallx && x2 <= bigx
    && (y2 == y1 || y2 == y3)) {
  console.log("YES");
}
else if (y2 >= smally && y2 <= bigy
         && (x2 == x1 || x2 == x3)) {
  console.log("YES");
}
else {
  console.log("NO");
}

Output

YES

Time Complexity: O(1)
Auxiliary Space: O(1)

Check if it is possible to move from X to Z using point Y

Explore