# Check if any point exists in a plane whose Manhattan distance is at most K from N given points

• Last Updated : 22 Apr, 2021

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++

 `// C++ program for the above approach``#include``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

 `// 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

 `# 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#

 `// 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`

## Javascript

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Output:

`Yes`

Time Complexity: O(N2)
Auxiliary Space: O(1)

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