Given two arrays A[] and B[] consisting of X and Y coordinates of N distinct points in a plane, and a positive integer K, the task is to check if there exists any point P in the plane such that the Manhattan distance between the point and all the given points is at most K. If there exists any such point P, then print “Yes”. Otherwise, print “No”.
Examples:
Input: A[] = {1, 0, 2, 1, 1}, B[] = {1, 1, 1, 0, 2}, K = 1
Output: Yes
Explanation:
Consider a point P(1, 1), then the Manhattan distance between P and all the given points are:
- Distance between P and (A[0], B[0]) is |1 – 1| + |1 – 1| = 0.
- Distance between P and (A[1], B[1]) is |1 – 0| + |1 – 1| = 1.
- Distance between P and (A[2], B[2]) is |1 – 2| + |1 – 1| = 1.
- Distance between P and (A[3], B[3]) is |1 – 1| + |1 – 0| = 1.
- Distance between P and (A[4], B[4]) is |1 – 1| + |1 – 2| = 1.
The distance between all the given points and P is at most K(= 1). Therefore, print “Yes”.
Input: A[] = {0, 3, 1}, B[] = {0, 3, 1}, K = 2
Output: No
Approach: The given problem can be solved by finding the Manhattan distance between every pair of N given points. After checking for all pairs of points, if the count of the distance between pairs of points is at most K, then print “Yes”. Otherwise, print “No”.
Below is the implementation of the above approach :
// C++ program for the above approach #include<bits/stdc++.h> using namespace std;
// Function to check if there // exists any point with at most // K distance from N given points string find( int a[], int b[], int N, int K)
{ // Traverse the given N points
for ( int i = 0; i < N; i++)
{
// Stores the count of pairs
// of coordinates having
// Manhattan distance <= K
int count = 0;
for ( int j = 0; j < N; j++)
{
// For the same coordinate
if (i == j)
{
continue ;
}
// Calculate Manhattan distance
long long int dis = abs (a[i] - a[j]) +
abs (b[i] - b[j]);
// If Manhattan distance <= K
if (dis <= K)
{
count++;
}
// If all coordinates
// can meet
if (count == N - 1)
{
return "Yes" ;
}
}
}
// If all coordinates can't meet
return "No" ;
} // Driver Code int main()
{ int N = 5;
int A[] = { 1, 0, 2, 1, 1 };
int B[] = { 1, 1, 1, 0, 2 };
int K = 1;
cout << find(A, B, N, K) << endl;
} // This code is contributed by bgangwar59 |
// Java program for the above approach import java.io.*;
class GFG {
// Function to check if there
// exists any point with at most
// K distance from N given points
public static String find(
int [] a, int [] b, int N, int K)
{
// Traverse the given N points
for ( int i = 0 ; i < N; i++) {
// Stores the count of pairs
// of coordinates having
// Manhattan distance <= K
int count = 0 ;
for ( int j = 0 ; j < N; j++) {
// For the same coordinate
if (i == j) {
continue ;
}
// Calculate Manhattan distance
long dis = Math.abs(a[i] - a[j])
+ Math.abs(b[i] - b[j]);
// If Manhattan distance <= K
if (dis <= K) {
count++;
}
// If all coordinates
// can meet
if (count == N - 1 ) {
return "Yes" ;
}
}
}
// If all coordinates can't meet
return "No" ;
}
// Driver Code
public static void main(String[] args)
{
int N = 5 ;
int [] A = { 1 , 0 , 2 , 1 , 1 };
int [] B = { 1 , 1 , 1 , 0 , 2 };
int K = 1 ;
System.out.println(
find(A, B, N, K));
}
} |
# Python3 program for the above approach # Function to check if there # exists any point with at most # K distance from N given points def find(a, b, N, K):
# Traverse the given n points
for i in range (N):
# Stores the count of pairs
# of coordinates having
# Manhattan distance <= K
count = 0
for j in range (N):
# For the same coordinate
if (i = = j):
continue
# Calculate Manhattan distance
dis = abs (a[i] - a[j]) + abs (b[i] - b[j])
# If Manhattan distance <= K
if (dis < = K):
count = count + 1
# If all coordinates
# can meet
if (count = = N - 1 ):
return "Yes"
# If all coordinates can't meet
return "No"
# Driver code N = 5
A = [ 1 , 0 , 2 , 1 , 1 ]
B = [ 1 , 1 , 1 , 0 , 2 ]
K = 1
print (find(A, B, N, K))
# This code is contributed by abhinavjain194 |
// C# program for the above approach using System;
class GFG{
// Function to check if there // exists any point with at most // K distance from N given points public static String find( int [] a, int [] b,
int N, int K)
{ // Traverse the given N points
for ( int i = 0; i < N; i++)
{
// Stores the count of pairs
// of coordinates having
// Manhattan distance <= K
int count = 0;
for ( int j = 0; j < N; j++)
{
// For the same coordinate
if (i == j)
{
continue ;
}
// Calculate Manhattan distance
long dis = Math.Abs(a[i] - a[j]) +
Math.Abs(b[i] - b[j]);
// If Manhattan distance <= K
if (dis <= K)
{
count++;
}
// If all coordinates
// can meet
if (count == N - 1)
{
return "Yes" ;
}
}
}
// If all coordinates can't meet
return "No" ;
} // Driver Code public static void Main( string [] args)
{ int N = 5;
int [] A = { 1, 0, 2, 1, 1 };
int [] B = { 1, 1, 1, 0, 2 };
int K = 1;
Console.WriteLine(find(A, B, N, K));
} } // This code is contributed by ukasp |
<script> // Javascript program for
// the above approach
// Function to check if there
// exists any point with at most
// K distance from N given points
function find(a, b, N, K) {
// Traverse the given N points
for (let i = 0; i < N; i++) {
// Stores the count of pairs
// of coordinates having
// Manhattan distance <= K
let count = 0;
for (let j = 0; j < N; j++) {
// For the same coordinate
if (i == j) {
continue ;
}
// Calculate Manhattan distance
let dis = Math.abs(a[i] - a[j]) +
Math.abs(b[i] - b[j]);
// If Manhattan distance <= K
if (dis <= K) {
count++;
}
// If all coordinates
// can meet
if (count == N - 1) {
return "Yes" ;
}
}
}
// If all coordinates can't meet
return "No" ;
}
// Driver Code
let N = 5;
let A = [ 1, 0, 2, 1, 1 ];
let B = [ 1, 1, 1, 0, 2 ];
let K = 1;
document.write(find(A, B, N, K));
// This code is contributed by Hritik
</script>
|
Yes
Time Complexity: O(N2)
Auxiliary Space: O(1)