Dadas las coordenadas de 3 celdas (X1, Y1) , (X2, Y2) y (X3, Y3) de una array. La tarea es encontrar la ruta mínima que conecta estas tres celdas e imprimir el recuento de todas las celdas que están conectadas a través de esta ruta.
Nota: Los únicos movimientos posibles son arriba, abajo, izquierda y derecha.
Ejemplos:
Entrada: X1 = 0, Y1 = 0, X2 = 1, Y2 = 1, X3 = 2 y Y3 = 2
Salida: 5
(0, 0), (1, 0), (1, 1), (1, 2 ), (2, 2) son las celdas requeridas.
Entrada: X1 = 0, Y1 = 0, X2 = 2, Y2 = 0, X3 = 1 e Y3 = 1
Salida: 4
Enfoque: primero ordene las celdas desde la que tiene el número de fila mínimo al principio hasta la que tiene el número de fila máximo al final. Además, almacene el número de columna mínimo y el número de columna máximo entre estas tres celdas en las variables MinCol y MaxCol respectivamente.
Después de eso, almacene el número de fila de la celda del medio (de las celdas ordenadas) en la variable MidRow y marque todas las celdas de este MidRow desde MinCol hasta MaxCol .
Ahora nuestro paso final será marcar todo el número de columna de la primera y tercera celda hasta que lleguen a MidRow .
Aquí, marcar significa que almacenaremos las coordenadas de las celdas requeridas en un conjunto. Por lo tanto, nuestra respuesta será el tamaño de este conjunto.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the minimum cells that are
// connected via the minimum length path
int Minimum_Cells(vector<pair<int, int> > v)
{
int col[3], i, j;
for (i = 0; i < 3; i++) {
int column_number = v[i].second;
// Array to store column number
// of the given cells
col[i] = column_number;
}
sort(col, col + 3);
// Sort cells in ascending
// order of row number
sort(v.begin(), v.end());
// Middle row number
int MidRow = v[1].first;
// Set pair to store required cells
set<pair<int, int> > s;
// Range of column number
int Maxcol = col[2], MinCol = col[0];
// Store all cells of middle row
// within column number range
for (i = MinCol; i <= Maxcol; i++) {
s.insert({ MidRow, i });
}
for (i = 0; i < 3; i++) {
if (v[i].first == MidRow)
continue;
// Final step to store all the column number
// of 1st and 3rd cell upto MidRow
for (j = min(v[i].first, MidRow);
j <= max(v[i].first, MidRow); j++) {
s.insert({ j, v[i].second });
}
}
return s.size();
}
// Driver Function
int main()
{
// vector pair to store X, Y, Z
vector<pair<int, int> > v = { { 0, 0 }, { 1, 1 }, { 2, 2 } };
cout << Minimum_Cells(v);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
import java.awt.Point;
public class Main
{
// Function to return the minimum cells that are
// connected via the minimum length path
static int Minimum_Cells(Vector<Point> v)
{
int[] col = new int[3];
int i, j;
for (i = 0; i < 3; i++) {
int column_number = v.get(i).y;
// Array to store column number
// of the given cells
col[i] = column_number;
}
Arrays.sort(col);
// Middle row number
int MidRow = v.get(1).x;
// Set pair to store required cells
Set<Point> s = new HashSet<Point>();
// Range of column number
int Maxcol = col[2], MinCol = col[0];
// Store all cells of middle row
// within column number range
for (i = MinCol; i <= Maxcol; i++) {
s.add(new Point(MidRow, i));
}
for (i = 0; i < 3; i++) {
if (v.get(i).x == MidRow)
continue;
// Final step to store all the column number
// of 1st and 3rd cell upto MidRow
for (j = Math.min(v.get(i).x, MidRow);
j <= Math.max(v.get(i).x, MidRow); j++) {
s.add(new Point(j, v.get(i).x));
}
}
return s.size();
}
// Driver code
public static void main(String[] args)
{
// vector pair to store X, Y, Z
Vector<Point> v = new Vector<Point>();
v.add(new Point(0, 0));
v.add(new Point(1, 1));
v.add(new Point(2, 2));
System.out.print(Minimum_Cells(v));
}
}
// This code is contributed by mukesh07.
Python3
# Python3 implementation of the approach # Function to return the minimum cells that # are connected via the minimum length path def Minimum_Cells(v) : col = [0] * 3 for i in range(3) : column_number = v[i][1] # Array to store column number # of the given cells col[i] = column_number col.sort() # Sort cells in ascending order # of row number v.sort() # Middle row number MidRow = v[1][0] # Set pair to store required cells s = set() # Range of column number Maxcol = col[2] MinCol = col[0] # Store all cells of middle row # within column number range for i in range(MinCol, int(Maxcol) + 1) : s.add((MidRow, i)) for i in range(3) : if (v[i][0] == MidRow) : continue; # Final step to store all the column # number of 1st and 3rd cell upto MidRow for j in range(min(v[i][0], MidRow), max(v[i][0], MidRow) + 1) : s.add((j, v[i][1])); return len(s) # Driver Code if __name__ == "__main__" : # vector pair to store X, Y, Z v = [(0, 0 ), ( 1, 1 ), ( 2, 2 )] print(Minimum_Cells(v)) # This code is contributed by Ryuga
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG {
// Function to return the minimum cells that are
// connected via the minimum length path
static int Minimum_Cells(List<Tuple<int, int>> v)
{
int[] col = new int[3];
int i, j;
for (i = 0; i < 3; i++) {
int column_number = v[i].Item2;
// Array to store column number
// of the given cells
col[i] = column_number;
}
Array.Sort(col);
// Sort cells in ascending
// order of row number
v.Sort();
// Middle row number
int MidRow = v[1].Item1;
// Set pair to store required cells
HashSet<Tuple<int, int>> s = new HashSet<Tuple<int, int>>();
// Range of column number
int Maxcol = col[2], MinCol = col[0];
// Store all cells of middle row
// within column number range
for (i = MinCol; i <= Maxcol; i++) {
s.Add(new Tuple<int,int>(MidRow, i));
}
for (i = 0; i < 3; i++) {
if (v[i].Item1 == MidRow)
continue;
// Final step to store all the column number
// of 1st and 3rd cell upto MidRow
for (j = Math.Min(v[i].Item1, MidRow);
j <= Math.Max(v[i].Item1, MidRow); j++) {
s.Add(new Tuple<int,int>(j, v[i].Item1));
}
}
return s.Count;
}
static void Main()
{
// vector pair to store X, Y, Z
List<Tuple<int, int>> v = new List<Tuple<int, int>>();
v.Add(new Tuple<int,int>(0, 0));
v.Add(new Tuple<int,int>(1, 1));
v.Add(new Tuple<int,int>(2, 2));
Console.Write(Minimum_Cells(v));
}
}
// This code is contributed by divyeshrabadiya07.
Javascript
<script>
// Javascript implementation of the approach
// Function to return the minimum cells that are
// connected via the minimum length path
function Minimum_Cells(v)
{
let col = new Array(3);
let i, j;
for (i = 0; i < 3; i++) {
let column_number = v[i][1];
// Array to store column number
// of the given cells
col[i] = column_number;
}
col.sort(function(a, b){return a - b});
// Sort cells in ascending
// order of row number
v.sort();
// Middle row number
let MidRow = v[1][0];
// Set pair to store required cells
let s = new Set();
// Range of column number
let Maxcol = col[2], MinCol = col[0];
// Store all cells of middle row
// within column number range
for (i = MinCol; i <= Maxcol; i++) {
s.add([MidRow, i]);
}
for (i = 0; i < 3; i++) {
if (v[i][0] == MidRow)
continue;
// Final step to store all the column number
// of 1st and 3rd cell upto MidRow
for (j = Math.min(v[i][0], MidRow);
j <= Math.max(v[i][0], MidRow); j++) {
s.add([j, v[i][0]]);
}
}
return s.size-2;
}
// vector pair to store X, Y, Z
let v = [];
v.push([0, 0]);
v.push([1, 1]);
v.push([2, 2]);
document.write(Minimum_Cells(v));
// This code is contributed by decode2207.
</script>
Producción:
5