Dadas dos arrays a[] y b[] , la tarea es encontrar el recuento de todos los pares (a[i], b[j]) de modo que a[i] + b[j] sea único entre todos los pares, es decir si dos pares tienen la misma suma, solo se contará uno en el resultado.
Ejemplos:
Entrada: a[] = {3, 3}, b[] = {3}
Salida: 1
Los dos pares posibles son (a[0], b[0]) y (a[1], b[0]) .
Par 1: 3 + 3 = 6
Par 2: 3 + 3 = 6
Entrada: a[] = {12, 2, 7}, b[] = {4, 3, 8}
Salida: 7
Enfoque: Inicialice count = 0 y ejecute dos bucles para considerar todos los pares posibles y almacene la suma de cada par en un conjunto_desordenado para comprobar si la suma se ha obtenido antes. Si es así, ignore el par actual; de lo contrario, incremente el conteo .
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 count
// of pairs with distinct sum
int countPairs(int a[], int b[], int n, int m)
{
// To store the required count
int cnt = 0;
// Set to store the sum
// obtained for each pair
unordered_set<int> s;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
// Sum of the current pair
int sum = a[i] + b[j];
// If the sum obtained is distinct
if (s.count(sum) == 0) {
// Increment the count
cnt++;
// Insert sum in the set
s.insert(sum);
}
}
}
return cnt;
}
// Driver code
int main()
{
int a[] = { 12, 2, 7 };
int n = sizeof(a) / sizeof(a[0]);
int b[] = { 4, 3, 8 };
int m = sizeof(b) / sizeof(b[0]);
cout << countPairs(a, b, n, m);
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the count
// of pairs with distinct sum
static int countPairs(int a[], int b[], int n, int m)
{
// To store the required count
int cnt = 0;
// Set to store the sum
// obtained for each pair
HashSet<Integer> s = new HashSet<Integer>();
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// Sum of the current pair
int sum = a[i] + b[j];
// If the sum obtained is distinct
if (s.contains(sum) == false)
{
// Increment the count
cnt++;
// Insert sum in the set
s.add(sum);
}
}
}
return cnt;
}
// Driver code
static public void main (String args[])
{
int a[] = { 12, 2, 7 };
int n = a.length;
int b[] = { 4, 3, 8 };
int m = b.length;
System.out.println(countPairs(a, b, n, m));
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 implementation of the approach # Function to return the count # of pairs with distinct sum def countPairs(a, b, n, m): # To store the required count cnt = 0 # Set to store the sum # obtained for each pair s=dict() for i in range(n): for j in range(m): # Sum of the current pair sum = a[i] + b[j] # If the sum obtained is distinct if (sum not in s.keys()): # Increment the count cnt+=1 # Insert sum in the set s[sum]=1 return cnt # Driver code a =[ 12, 2, 7] n = len(a) b =[ 4, 3, 8 ] m = len(b) print(countPairs(a, b, n, m)) # This code is contributed by mohit kumar 29
C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the count
// of pairs with distinct sum
static int countPairs(int []a, int []b,
int n, int m)
{
// To store the required count
int cnt = 0;
// Set to store the sum
// obtained for each pair
HashSet<int> s = new HashSet<int>();
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
// Sum of the current pair
int sum = a[i] + b[j];
// If the sum obtained is distinct
if (s.Contains(sum) == false)
{
// Increment the count
cnt++;
// Insert sum in the set
s.Add(sum);
}
}
}
return cnt;
}
// Driver code
static public void Main (String []args)
{
int []a = { 12, 2, 7 };
int n = a.Length;
int []b = { 4, 3, 8 };
int m = b.Length;
Console.WriteLine(countPairs(a, b, n, m));
}
}
// This code is contributed by PrinciRaj1992
Javascript
<script>
// Javascript implementation of the approach
// Function to return the count
// of pairs with distinct sum
function countPairs(a, b, n, m)
{
// To store the required count
let cnt = 0;
// Set to store the sum
// obtained for each pair
let s = new Set();
for (let i = 0; i < n; i++)
{
for (let j = 0; j < m; j++)
{
// Sum of the current pair
let sum = a[i] + b[j];
// If the sum obtained is distinct
if (s.has(sum) == false)
{
// Increment the count
cnt++;
// Insert sum in the set
s.add(sum);
}
}
}
return cnt;
}
// Driver code
let a = [ 12, 2, 7 ];
let n = a.length;
let b = [ 4, 3, 8 ];
let m = b.length;
document.write(countPairs(a, b, n, m));
// This code is contributed by susmitakundugoaldanga.
</script>
Producción:
7
Complejidad temporal: O(N * M).
Espacio Auxiliar : O(1).