Cuente pares de rectángulos similares posibles de una array dada

Dada una array 2D A[][2] de tamaño N (1 ≤ N ≤ 10 ³ ) , donde A[i][0] y A[i][1] denotan el largo y el ancho del rectángulo i respectivamente.

Dos rectángulos i y j donde (i < j) son similares si la razón de su largo y ancho es igual

A[i][0] / A[i][1] = A[j][0] / A[j][1]

La tarea es contar el par de rectángulos que son casi similares. 

Ejemplos: 

Entrada : A[][2] = {{4, 8}, {15, 30}, {3, 6}, {10, 20}}
Salida: 6
Explicación: Los pares de rectángulos similares son (0, 1), {0, 2), (0, 3), (1, 2), (1, 3), (2, 3). Para cada rectángulo, la relación de largo: ancho es 1: 2

Entrada : A[][2] = {{2, 3}, {4, 5}, {7, 8}}
Salida: 0
Explicación: No existe ningún par de rectángulos similares.

Método 1 (método de comparación simple)

Enfoque: Siga los pasos para resolver el problema 

• Atraviesa la array .
• Para cada par tal que ( i < j ), verifique si los rectángulos son similares o no verificando si la condición A[i][0] / A[i][1] = A[j][0] / A[ j][1] se cumple o no.
• Si se encuentra que es cierto, incremente el conteo.
• Finalmente, imprima el conteo obtenido.

A continuación se muestra la implementación del enfoque anterior: 

// C++ Program for the above approach
 
#include <iostream>
using namespace std;
 
// Function to calculate the count
// of similar rectangles
int getCount(int rows,
             int columns, int A[][2])
{
    int res = 0;
 
    for (int i = 0; i < rows; i++) {
        for (int j = i + 1; j < rows; j++) {
 
            if (A[i][0] * 1LL * A[j][1]
                == A[i][1] * 1LL * A[j][0]) {
 
                res++;
            }
        }
    }
    return res;
}
 
// Driver Code
int main()
{
    // Input
    int A[][2]
        = { { 4, 8 }, { 10, 20 }, { 15, 30 }, { 3, 6 } };
    int columns = 2;
    int rows = sizeof(A) / sizeof(A[0]);
 
    cout << getCount(rows, columns, A);
 
    return 0;
}

// Java program for the above approach
import java.util.*;
class GFG{
  
// Function to calculate the count
// of similar rectangles
static int getCount(int rows, int columns,
                    int[][] A)
{
    int res = 0;  
    for(int i = 0; i < rows; i++)
    {
        for(int j = i + 1; j < rows; j++)
        {
            if (A[i][0] * A[j][1] ==
                A[i][1] * A[j][0])
            {
                res++;
            }
        }
    }
    return res;
}
  
// Driver Code
public static void main(String[] args)
{
   
    // Input
    int[][] A = { { 4, 8 }, { 10, 20 },
                 { 15, 30 }, { 3, 6 } };
    int columns = 2;
    int rows = 4;
   
    System.out.print(getCount(rows, columns, A));
}
}
 
// This code is contributed by code_hunt.

# Python3 program for the above approach
 
# Function to calculate the count
# of similar rectangles
def getCount(rows, columns, A):
     
    res = 0
 
    for i in range(rows):
        for j in range(i + 1, rows, 1):
            if (A[i][0] * A[j][1] ==
                A[i][1] * A[j][0]):
                res += 1
                 
    return res
 
# Driver Code
if __name__ == '__main__':
     
    # Input
    A =  [ [ 4, 8 ], [ 10, 20 ],
           [ 15, 30 ], [ 3, 6 ] ]
    columns = 2
    rows =  len(A)
 
    print(getCount(rows, columns, A))
 
# This code is contributed by SURENDRA_GANGWAR

// C# program for the above approach
using System;
 
class GFG{
 
// Function to calculate the count
// of similar rectangles
static int getCount(int rows, int columns,
                    int[,] A)
{
    int res = 0;
   
    for(int i = 0; i < rows; i++)
    {
        for(int j = i + 1; j < rows; j++)
        {
            if (A[i, 0] * A[j, 1] ==
                A[i, 1] * A[j, 0])
            {
                res++;
            }
        }
    }
    return res;
}
 
// Driver code
static void Main()
{
     
    // Input
    int[,] A = { { 4, 8 }, { 10, 20 },
                 { 15, 30 }, { 3, 6 } };
    int columns = 2;
    int rows = 4;
   
    Console.Write(getCount(rows, columns, A));
}
}
 
// This code is contributed by divyesh072019

<script>
 
// Javascript program for the above approach
 
// Function to calculate the count
// of similar rectangles
function getCount(rows, columns, A)
{
    var res = 0;  
    for(var i = 0; i < rows; i++)
    {
        for(var j = i + 1; j < rows; j++)
        {
            if (A[i][0] * A[j][1] ==
                A[i][1] * A[j][0])
            {
                res++;
            }
        }
    }
    return res;
}
 
// Driver Code
 
// Input
var A = [ [ 4, 8 ], [ 10, 20 ],
          [ 15, 30 ], [ 3, 6 ] ];
var columns = 2;
var rows = 4;
 
document.write(getCount(rows, columns, A));
 
// This code is contributed by kirti
 
</script>

6

Complejidad de Tiempo : O(N ² )
Espacio Auxiliar : O(1)

Método 2 (usando HashMap)

En lugar de comparar directamente la proporción de un rectángulo con la de otro rectángulo, podemos almacenar todas las proporciones calculadas en el HashMap. Dado que HashMap tiene el costo de acceso de O (1), eso permitiría una búsqueda rápida de otros rectángulos con la misma proporción y almacenarlos juntos. El número de pares de rectángulos similares se puede derivar del número de rectángulos con la misma razón usando \frac{N \times (N – 1)}{2}. La razón por la que uno puede usar la ecuación mencionada para encontrar el número de pares de rectángulos similares es que el número de pares aumenta   cada vez que se agrega un rectángulo con la misma proporción.

// C++ Program for the hashmap Approach
#include <iostream>
#include <unordered_map>
using namespace std;
 
// Get the count of all pairs of similar rectangles
 
int getCount(int rows, int columns, int sides[][2])
{
    // Initialize the result value and
    // map to store the ratio to the rectangles
    int ans = 0;
    unordered_map<double, int> umap;
 
    // Calculate the rectangular ratio and save them
    for (int i = 0; i < rows; i++)
    {
        double ratio = (double)sides[i][0] / sides[i][1];
        umap[ratio]++;
    }
 
    // Calculate pairs of similar rectangles from its common ratio
    for (auto x : umap)
    {
        int value = x.second;
        if (value > 1)
        {
            ans += (value * (value - 1)) / 2;
        }
    }
    return ans;
}
 
int main()
{
    int sides[4][2] = {{4, 8}, {10, 20}, {15, 30}, {3, 6}};
    int rows = 4;
    int columns = 2;
    cout << getCount(rows, columns, sides);
    return 0;
 
    //this code is contributed by krishna_6431
}

import java.util.*;
 
class GFG {
 
      // Get the count of all pairs of similar rectangles
    public static int getCount(int rows, int columns,
                                int[][] sides)
    {
          // Initialize the result value and
          // map to store the ratio to the rectangles
        int res = 0;
        Map<Double, Integer> ratio
            = new HashMap<Double, Integer>();
 
          // Calculate the rectangular ratio and save them
          for (int i = 0; i < rows; i++) {
            double rectRatio = (double) sides[i][0] /
                  sides[i][1];
              if (ratio.get(rectRatio) == null) {
                  ratio.put(rectRatio, 0);
            }
              ratio.put(rectRatio, ratio.get(rectRatio) + 1);
        }
 
          // Calculate pairs of similar rectangles from
          // its common ratio
        for (double key : ratio.keySet()) {
            int val = ratio.get(key);
            if (val > 1) {
                res += (val * (val - 1)) / 2;
            }
        }
 
        return res;
    }
  
      public static void main(String[] args) {
          int[][] A = {{4, 8}, {10, 20}, {15, 30}, {3, 6}};
          int columns = 2;
          int rows = 4;
           
          System.out.println(getCount(rows, columns, A));
    }
}
 
// This code is contributed by nathnet

# Python 3 Program for the hashmap Approach
from collections import defaultdict
 
# Get the count of all pairs of similar rectangles
def getCount(rows,  columns, sides):
 
    # Initialize the result value and
    # map to store the ratio to the rectangles
    ans = 0
    umap = defaultdict(int)
 
    # Calculate the rectangular ratio and save them
    for i in range(rows):
 
        ratio = sides[i][0] / sides[i][1]
        umap[ratio] += 1
 
    # Calculate pairs of similar rectangles from its common ratio
    for x in umap:
 
        value = umap[x]
        if (value > 1):
            ans += (value * (value - 1)) / 2
 
    return ans
 
# Driver code
if __name__ == "__main__":
 
    sides = [[4, 8], [10, 20], [15, 30], [3, 6]]
    rows = 4
    columns = 2
    print(int(getCount(rows, columns, sides)))
 
    # This code is contributed by ukasp.

using System;
using System.Collections.Generic;
 
class GFG {
 
    // Get the count of all pairs of similar rectangles
    public static int getCount(int rows, int columns,
                               int[, ] sides)
    {
        // Initialize the result value and
        // map to store the ratio to the rectangles
        int res = 0;
        Dictionary<double, int> ratio
            = new Dictionary<double, int>();
 
        // Calculate the rectangular ratio and save them
        for (int i = 0; i < rows; i++) {
            double rectRatio
                = (double)sides[i, 0] / sides[i, 1];
            if (!ratio.ContainsKey(rectRatio)) {
                ratio[rectRatio] = 0;
            }
            ratio[rectRatio] = ratio[rectRatio] + 1;
        }
 
        // Calculate pairs of similar rectangles from
        // its common ratio
        foreach(KeyValuePair<double, int> p in ratio)
        {
            int val = p.Value;
            if (val > 1) {
                res += (val * (val - 1)) / 2;
            }
        }
 
        return res;
    }
 
  // Driver code
    public static void Main(string[] args)
    {
        int[, ] A = {
            { 4, 8 }, { 10, 20 }, { 15, 30 }, { 3, 6 }
        };
        int columns = 2;
        int rows = 4;
 
        Console.WriteLine(getCount(rows, columns, A));
    }
}
 
// This code is contributed by ukasp.

<script>
 
// JavaScript Program for the hashmap Approach
 
// Get the count of all pairs of similar rectangles
function getCount(rows, columns, sides)
{
    // Initialize the result value and
    // map to store the ratio to the rectangles
    let ans = 0;
    let umap = new Map();
 
    // Calculate the rectangular ratio and save them
    for (let i = 0; i < rows; i++)
    {
        let ratio = sides[i][0] / sides[i][1];
        if(umap.has(ratio)){
             umap.set(ratio,umap.get(ratio)+1);
        }
        else umap.set(ratio,1);
    }
 
    // Calculate pairs of similar rectangles from its common ratio
    for (let [x,y] of umap)
    {
        let value = y;
        if (value > 1)
        {
            ans += (value * (value - 1)) / 2;
        }
    }
    return ans;
}
 
// driver code
let sides = [[4, 8], [10, 20], [15, 30], [3, 6]];
let rows = 4;
let columns = 2;
document.write(getCount(rows, columns, sides));
 
// This code is contributed by shinjanpatra
 
</script>

6

Complejidad de tiempo : O(N)

Espacio Auxiliar : O(N)