Number of cells in matrix which are equidistant from given two points
Last Updated :
24 Feb, 2022
Given a matrix of N rows and M columns, given two points on the matrix; the task is to count the number of cells that are equidistant from given two points. Any traversal either in the horizontal direction or vertical direction or both ways is considered valid but the diagonal path is not valid.
Examples:
Input: 5 5
2 4
5 3
Output: 5
Explanation:
Out of all cells, these are the points (3, 1);(3, 2);(3, 3);(4, 4);(4, 5)
which satisfy given condition.
Input: 4 3
2 3
4 1
Output: 4
Explanation:
Out of all cells, these are the points (1, 1);(2, 1);(3, 2);(4, 3)
which satisfy given condition.
Approach:
- Every cell of the matrix is traversed.
- Let ‘A’ be the distance between current cell and first point and similarly ‘B’ be the distance between current cell and second point.
- Distance between two points is calculated using Manhattan distance. If A & B are equal, the count is incremented.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int numberOfPoints( int N, int M, int x1,
int y1, int x2, int y2)
{
int count = 0;
for ( int i = 1; i <= N; i++) {
for ( int j = 1; j <= M; j++) {
if ( abs (i - x1) + abs (j - y1)
== abs (i - x2) + abs (j - y2))
count++;
}
}
return count;
}
int main()
{
int n = 5;
int m = 5;
int x1 = 2;
int y1 = 4;
int x2 = 5;
int y2 = 3;
cout << numberOfPoints(n, m, x1, y1, x2, y2);
}
|
Java
import java.util.*;
import java.lang.Math;
public class GFG {
public static int numberOfPoints( int N, int M, int x1,
int y1, int x2, int y2)
{
int count = 0 , i, j;
for (i = 1 ; i <= N; i++) {
for (j = 1 ; j <= M; j++) {
if (Math.abs(i - x1) + Math.abs(j - y1)
== Math.abs(i - x2) + Math.abs(j - y2))
count += 1 ;
}
}
return count;
}
public static void main(String[] args)
{
int n = 5 ;
int m = 5 ;
int x1 = 2 ;
int y1 = 4 ;
int x2 = 5 ;
int y2 = 3 ;
System.out.println(numberOfPoints(n, m, x1, y1, x2, y2));
}
}
|
Python
def numberPoints(N, M, x1, y1, x2, y2):
count = 0
for i in range ( 1 , N + 1 ):
for j in range ( 1 , M + 1 ):
if ( abs (i - x1) + abs (j - y1)) = = ( abs (i - x2) + abs (j - y2)):
count + = 1
return count
N = 5
M = 5
x1 = 2
y1 = 4
x2 = 5
y2 = 3
print (numberPoints(N, M, x1, y1, x2, y2))
|
C#
using System;
class GFG
{
static int numberOfPoints( int N, int M, int x1,
int y1, int x2, int y2)
{
int count = 0, i, j;
for (i = 1; i <= N; i++)
{
for (j = 1; j <= M; j++)
{
if (Math.Abs(i - x1) + Math.Abs(j - y1)
== Math.Abs(i - x2) + Math.Abs(j - y2))
count += 1;
}
}
return count;
}
public static void Main()
{
int n = 5;
int m = 5;
int x1 = 2;
int y1 = 4;
int x2 = 5;
int y2 = 3;
Console.WriteLine(numberOfPoints(n, m, x1, y1, x2, y2));
}
}
|
Javascript
<script>
function numberOfPoints(N , M , x1 , y1 , x2 , y2)
{
var count = 0, i, j;
for (i = 1; i <= N; i++) {
for (j = 1; j <= M; j++) {
if (Math.abs(i - x1) + Math.abs(j - y1) == Math.abs(i - x2) + Math.abs(j - y2))
count += 1;
}
}
return count;
}
var n = 5;
var m = 5;
var x1 = 2;
var y1 = 4;
var x2 = 5;
var y2 = 3;
document.write(numberOfPoints(n, m, x1, y1, x2, y2));
</script>
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Time Complexity: O(N * M)
Auxiliary Space: O(1)
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