Cuente pares en una array cuya diferencia absoluta sea divisible por K – Part 1

Posted on by Rudeus Greyrat

Dada una array arr[] y un entero positivo K . La tarea es contar el número total de pares en el arreglo cuya diferencia absoluta es divisible por K. 

Ejemplos: 

Entrada: arr[] = {1, 2, 3, 4}, K = 2 
Salida: 2 
Explicación: Hay 
un total de 2 pares en la array con una diferencia absoluta divisible por 2. 
Los pares son: (1, 3), (2 , 4).

Entrada: arr[] = {3, 3, 3}, K = 3 
Salida: 3 
Explicación: Hay 
un total de 3 pares en esta array con una diferencia absoluta divisible por 3. 
Los pares son: (3, 3), (3, 3 ), (3, 3). 
 

Enfoque ingenuo: la idea es verificar cada par de la array uno por uno y contar el número total de pares cuya diferencia absoluta es divisible por K.  

C++

#include <bits/stdc++.h>
using namespace std;
 
// function to count pairs in an array
// whose absolute difference is
// divisible by k
void countPairs(int arr[], int n, int k)
{
 
    // initialize count as zero.
    int i, j, cnt = 0;
 
    // loop to count the valid pair
    for (i = 0; i < n - 1; i++) {
        for (j = i + 1; j < n; j++) {
            if ((arr[i] - arr[j] + k) % k == 0)
                cnt += 1;
        }
    }
    cout << cnt << endl;
}
 
// Driver code
int main()
{
    // input array
    int arr[] = {3, 3, 3};
    int k = 3;
 
    // calculate the size of array
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // function to count the valid pair
    countPairs(arr, n, k);
    return 0;
}

Java

import java.util.*;
 
class GFG
{
 
// function to count pairs in an array
// whose absolute difference is
// divisible by k
static void countPairs(int arr[], int n, int k)
{
 
    // initialize count as zero.
    int i, j, cnt = 0;
 
    // loop to count the valid pair
    for (i = 0; i < n - 1; i++)
    {
        for (j = i + 1; j < n; j++)
        {
            if ((arr[i] - arr[j] + k) % k == 0)
                cnt += 1;
        }
    }
    System.out.print(cnt +"\n");
}
 
// Driver code
public static void main(String[] args)
{
    // input array
    int arr[] = {3, 3, 3};
    int k = 3;
 
    // calculate the size of array
    int n = arr.length;
 
    // function to count the valid pair
    countPairs(arr, n, k);
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 Code implementation of the above approach
 
# function to count pairs in an array
# whose absolute difference is
# divisible by k
def countPairs(arr, n, k) :
 
    # initialize count as zero.
    cnt = 0;
 
    # loop to count the valid pair
    for i in range(n - 1) :
        for j in range(i + 1, n) :
            if ((arr[i] - arr[j] + k) % k == 0) :
                cnt += 1;
     
    print(cnt) ;
 
# Driver code
if __name__ == "__main__" :
 
    # input array
    arr = [3, 3, 3];
    k = 3;
 
    # calculate the size of array
    n = len(arr);
 
    # function to count the valid pair
    countPairs(arr, n, k);
     
# This code is contributed by AnkitRai01

C#

using System;
 
class GFG
{
 
// function to count pairs in an array
// whose absolute difference is
// divisible by k
static void countPairs(int []arr, int n, int k)
{
 
    // initialize count as zero.
    int i, j, cnt = 0;
 
    // loop to count the valid pair
    for (i = 0; i < n - 1; i++)
    {
        for (j = i + 1; j < n; j++)
        {
            if ((arr[i] - arr[j] + k) % k == 0)
                cnt += 1;
        }
    }
    Console.Write(cnt +"\n");
}
 
// Driver code
public static void Main(String[] args)
{
    // input array
    int []arr = {3, 3, 3};
    int k = 3;
 
    // calculate the size of array
    int n = arr.Length;
 
    // function to count the valid pair
    countPairs(arr, n, k);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// Function to count pairs in an array
// whose absolute difference is
// divisible by k
function countPairs(arr, n, k)
{
     
    // Initialize count as zero.
    var i, j, cnt = 0;
 
    // Loop to count the valid pair
    for(i = 0; i < n - 1; i++)
    {
        for(j = i + 1; j < n; j++)
        {
            if ((arr[i] - arr[j] + k) % k == 0)
                cnt += 1;
        }
    }
    document.write(cnt + "\n");
}
 
// Driver code
 
// Input array
var arr = [ 3, 3, 3 ];
var k = 3;
 
// Calculate the size of array
var n = arr.length;
 
// Function to count the valid pair
countPairs(arr, n, k);
 
// This code is contributed by umadevi9616
 
</script>
Producción: 

3

 

Complejidad de tiempo: O( N 2 ) 
Complejidad de espacio: O(1)

Enfoque eficiente:  

Algoritmo: 
 

  1. Convierta cada elemento (A[i]) de la array en ((A[i]+K)%K)
  2. Use la técnica hash para almacenar la cantidad de veces que (A[i]%K) ocurre en la array
  3. Ahora, si un elemento A[i] aparece x veces en la array, agregue x*(x-1)/2 (elegir 2 elementos cualquiera de x elementos) en el par de conteo donde 1<=i<=n .Esto se debe a que el valor de cada elemento de la array se encuentra entre 0 y K-1, por lo que la diferencia absoluta es divisible solo si el valor de ambos elementos de un par es igual

C++

// Write CPP code here
#include <bits/stdc++.h>
using namespace std;
 
// function to Count pairs in an array whose
// absolute difference is divisible by k
void countPair(int arr[], int n, int k)
{
 
    // initialize the count
    int cnt = 0;
 
    // making every element of arr in
    // range 0 to k - 1
    for (int i = 0; i < n; i++) {
        arr[i] = (arr[i] + k) % k;
    }
 
    // create an array hash[]
    int hash[k] = { 0 };
 
    // store to count of element of arr
    // in hash[]
    for (int i = 0; i < n; i++) {
        hash[arr[i]]++;
    }
 
    // count the pair whose absolute
    // difference is divisible by k
    for (int i = 0; i < k; i++) {
        cnt += (hash[i] * (hash[i] - 1)) / 2;
    }
 
    // print the value of count
    cout << cnt << endl;
}
 
// Driver Code
int main()
{
    // input array
    int arr[] = {1, 2, 3, 4};
    int k = 2;
 
    // calculate the size of array
    int n = sizeof(arr) / sizeof(arr[0]);
    countPair(arr, n, k);
    return 0;
}

Java

// JAVA Implementation of above approach
import java.util.*;
 
class GFG
{
 
// function to Count pairs in an array whose
// absolute difference is divisible by k
static void countPair(int arr[], int n, int k)
{
 
    // initialize the count
    int cnt = 0;
 
    // making every element of arr in
    // range 0 to k - 1
    for (int i = 0; i < n; i++)
    {
        arr[i] = (arr[i] + k) % k;
    }
 
    // create an array hash[]
    int hash[] = new int[k];
 
    // store to count of element of arr
    // in hash[]
    for (int i = 0; i < n; i++)
    {
        hash[arr[i]]++;
    }
 
    // count the pair whose absolute
    // difference is divisible by k
    for (int i = 0; i < k; i++)
    {
        cnt += (hash[i] * (hash[i] - 1)) / 2;
    }
 
    // print the value of count
    System.out.print(cnt +"\n");
}
 
// Driver Code
public static void main(String[] args)
{
    // input array
    int arr[] = {1, 2, 3, 4};
    int k = 2;
 
    // calculate the size of array
    int n = arr.length;
    countPair(arr, n, k);
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 Implementation of above approach
 
# function to Count pairs in an array whose
# absolute difference is divisible by k
def countPair(arr, n, k):
 
    # initialize the count
    cnt = 0;
 
    # making every element of arr in
    # range 0 to k - 1
    for i in range(n):
        arr[i] = (arr[i] + k) % k;
 
    # create an array hash
    hash = [0]*k;
 
    # store to count of element of arr
    # in hash
    for i in range(n):
        hash[arr[i]] += 1;
 
    # count the pair whose absolute
    # difference is divisible by k
    for i in range(k):
        cnt += (hash[i] * (hash[i] - 1)) / 2;
 
    # print value of count
    print(int(cnt));
 
# Driver Code
if __name__ == '__main__':
 
    # input array
    arr = [1, 2, 3, 4];
    k = 2;
 
    # calculate the size of array
    n = len(arr);
    countPair(arr, n, k);
 
# This code is contributed by 29AjayKumar

C#

// C# Implementation of above approach
using System;
 
class GFG
{
 
// function to Count pairs in an array whose
// absolute difference is divisible by k
static void countPair(int []arr, int n, int k)
{
 
    // initialize the count
    int cnt = 0;
 
    // making every element of arr in
    // range 0 to k - 1
    for (int i = 0; i < n; i++)
    {
        arr[i] = (arr[i] + k) % k;
    }
 
    // create an array hash[]
    int []hash = new int[k];
 
    // store to count of element of arr
    // in hash[]
    for (int i = 0; i < n; i++)
    {
        hash[arr[i]]++;
    }
 
    // count the pair whose absolute
    // difference is divisible by k
    for (int i = 0; i < k; i++)
    {
        cnt += (hash[i] * (hash[i] - 1)) / 2;
    }
 
    // print the value of count
    Console.Write(cnt +"\n");
}
 
// Driver Code
public static void Main(String[] args)
{
    // input array
    int []arr = {1, 2, 3, 4};
    int k = 2;
 
    // calculate the size of array
    int n = arr.Length;
    countPair(arr, n, k);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

<script>
 
// Javascript Implementation of above approach
 
// function to Count pairs in an array whose
// absolute difference is divisible by k
function countPair(arr, n, k)
{
    // let initialize the count
    let cnt = 0;
  
    // making every element of arr in
    // range 0 to k - 1
    for (let i = 0; i < n; i++)
    {
        arr[i] = (arr[i] + k) % k;
    }
  
    // create an array hash[]
    let hash =  Array.from({length: k}, (_, i) => 0);
  
    // store to count of element of arr
    // in hash[]
    for (let i = 0; i < n; i++)
    {
        hash[arr[i]]++;
    }
  
    // count the pair whose absolute
    // difference is divisible by k
    for (let i = 0; i < k; i++)
    {
        cnt += (hash[i] * (hash[i] - 1)) / 2;
    }
  
    // print the value of count
    document.write(cnt +"<br/>");
}
       
// Driver code
     
      // input array
    let arr = [1, 2, 3, 4];
    let k = 2;
  
    // calculate the size of array
    let n = arr.length;
    countPair(arr, n, k);
                                                                                      
</script>
Producción: 

2

 

Complejidad Temporal: O( n+k ) 
Espacio Auxiliar: O( k )
 

Publicación traducida automáticamente

Artículo escrito por rrlinus y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

Deja una respuesta

Tu dirección de correo electrónico no será publicada. Los campos obligatorios están marcados con *