Dada una array de enteros y un número ‘suma’, encuentra el número de pares de enteros en la array cuya suma es igual a ‘suma’.
Ejemplos:
C++
// C++ implementation of simple method to find count of
// pairs with given sum.
#include <bits/stdc++.h>
using namespace std;
// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
int getPairsCount(int arr[], int n, int sum)
{
int count = 0; // Initialize result
// Consider all possible pairs and check their sums
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (arr[i] + arr[j] == sum)
count++;
return count;
}
// Driver function to test the above function
int main()
{
int arr[] = { 1, 5, 7, -1, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
int sum = 6;
cout << "Count of pairs is "
<< getPairsCount(arr, n, sum);
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
C
// C implementation of simple method to find count of
// pairs with given sum.
#include <stdio.h>
// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
int getPairsCount(int arr[], int n, int sum)
{
int count = 0; // Initialize result
// Consider all possible pairs and check their sums
for (int i = 0; i < n; i++)
for (int j = i + 1; j < n; j++)
if (arr[i] + arr[j] == sum)
count++;
return count;
}
// Driver function to test the above function
int main()
{
int arr[] = { 1, 5, 7, -1, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
int sum = 6;
printf("Count of pairs is %d",getPairsCount(arr, n, sum));
return 0;
}
// This code is contributed by Aditya Kumar (adityakumar129)
Java
// Java implementation of simple method to find count of
// pairs with given sum.
public class find {
public static void main(String args[])
{
int[] arr = { 1, 5, 7, -1, 5 };
int sum = 6;
getPairsCount(arr, sum);
}
// Prints number of pairs in arr[0..n-1] with sum equal
// to 'sum'
public static void getPairsCount(int[] arr, int sum)
{
int count = 0; // Initialize result
// Consider all possible pairs and check their sums
for (int i = 0; i < arr.length; i++)
for (int j = i + 1; j < arr.length; j++)
if ((arr[i] + arr[j]) == sum)
count++;
System.out.printf("Count of pairs is %d", count);
}
}
// This code is contributed by Aditya Kumar (adityakumar129)
Python3
# Python3 implementation of simple method
# to find count of pairs with given sum.
# Returns number of pairs in arr[0..n-1]
# with sum equal to 'sum'
def getPairsCount(arr, n, sum):
count = 0 # Initialize result
# Consider all possible pairs
# and check their sums
for i in range(0, n):
for j in range(i + 1, n):
if arr[i] + arr[j] == sum:
count += 1
return count
# Driver function
arr = [1, 5, 7, -1, 5]
n = len(arr)
sum = 6
print("Count of pairs is",
getPairsCount(arr, n, sum))
# This code is contributed by Smitha Dinesh Semwal
C#
// C# implementation of simple
// method to find count of
// pairs with given sum.
using System;
class GFG {
public static void getPairsCount(int[] arr, int sum)
{
int count = 0; // Initialize result
// Consider all possible pairs
// and check their sums
for (int i = 0; i < arr.Length; i++)
for (int j = i + 1; j < arr.Length; j++)
if ((arr[i] + arr[j]) == sum)
count++;
Console.WriteLine("Count of pairs is " + count);
}
// Driver Code
static public void Main()
{
int[] arr = { 1, 5, 7, -1, 5 };
int sum = 6;
getPairsCount(arr, sum);
}
}
// This code is contributed
// by Sach_Code
PHP
<?php
// PHP implementation of simple
// method to find count of
// pairs with given sum.
// Returns number of pairs in
// arr[0..n-1] with sum equal
// to 'sum'
function getPairsCount($arr, $n, $sum)
{
// Initialize result
$count = 0;
// Consider all possible pairs
// and check their sums
for ($i = 0; $i < $n; $i++)
for ($j = $i + 1; $j < $n; $j++)
if ($arr[$i] + $arr[$j] == $sum)
$count++;
return $count;
}
// Driver Code
$arr = array(1, 5, 7, -1, 5) ;
$n = sizeof($arr);
$sum = 6;
echo "Count of pairs is "
, getPairsCount($arr, $n, $sum);
// This code is contributed by nitin mittal.
?>
Javascript
<script>
// Javascript implementation of simple method to find count of
// pairs with given sum.
// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
function getPairsCount(arr, n, sum)
{
let count = 0; // Initialize result
// Consider all possible pairs and check their sums
for (let i = 0; i < n; i++)
for (let j = i + 1; j < n; j++)
if (arr[i] + arr[j] == sum)
count++;
return count;
}
// Driver function to test the above function
let arr = [ 1, 5, 7, -1, 5 ];
let n = arr.length;
let sum = 6;
document.write("Count of pairs is "
+ getPairsCount(arr, n, sum));
// This code is contributed by Mayank Tyagi
</script>
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the count of pairs
int getPairsCount(int arr[], int n, int k)
{
sort(arr, arr + n);
int x = 0, c = 0, y, z;
for (int i = 0; i < n - 1; i++) {
x = k - arr[i];
// Lower bound from i+1
int y = lower_bound(arr + i + 1,
arr + n, x) - arr;
// Upper bound from i+1
int z = upper_bound(arr + i + 1,
arr + n, x) - arr;
c = c + z - y;
}
return c;
}
// Driver code
int main()
{
int arr[] = { 1, 5, 7, -1, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 6;
// Function call
cout << "Count of pairs is "
<< getPairsCount(arr, n, k);
return 0;
}
Python3
# Python code to implement the approach
import bisect
# Function to find the count of pairs
def getPairsCount(arr,n,k):
arr.sort()
x,c=0,0
for i in range(n-1):
x=k-arr[i]
# Lower bound from i+1
y = bisect.bisect_left(arr,x,i+1,n)
# Upper bound from i+1
z = bisect.bisect(arr,x,i+1,n)
c=c+z-y
return c
# Driver function
arr = [1, 5, 7, -1, 5]
n = len(arr)
k = 6
# Function call
print("Count of pairs is",getPairsCount(arr, n, k))
# This code is contributed by Pushpesh Raj
C++
// C++ implementation of simple method to find count of
// pairs with given sum.
#include <bits/stdc++.h>
using namespace std;
// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
int getPairsCount(int arr[], int n, int sum)
{
unordered_map<int, int> m;
// Store counts of all elements in map m
for (int i = 0; i < n; i++)
m[arr[i]]++;
int twice_count = 0;
// iterate through each element and increment the
// count (Notice that every pair is counted twice)
for (int i = 0; i < n; i++) {
twice_count += m[sum - arr[i]];
// if (arr[i], arr[i]) pair satisfies the condition,
// then we need to ensure that the count is
// decreased by one such that the (arr[i], arr[i])
// pair is not considered
if (sum - arr[i] == arr[i])
twice_count--;
}
// return the half of twice_count
return twice_count / 2;
}
// Driver function to test the above function
int main()
{
int arr[] = { 1, 5, 7, -1, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
int sum = 6;
cout << "Count of pairs is "
<< getPairsCount(arr, n, sum);
return 0;
}
Java
/* Java implementation of simple method to find count of
pairs with given sum*/
import java.util.HashMap;
class Test {
static int arr[] = new int[] { 1, 5, 7, -1, 5 };
// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
static int getPairsCount(int n, int sum)
{
HashMap<Integer, Integer> hm = new HashMap<>();
// Store counts of all elements in map hm
for (int i = 0; i < n; i++) {
// initializing value to 0, if key not found
if (!hm.containsKey(arr[i]))
hm.put(arr[i], 0);
hm.put(arr[i], hm.get(arr[i]) + 1);
}
int twice_count = 0;
// iterate through each element and increment the
// count (Notice that every pair is counted twice)
for (int i = 0; i < n; i++) {
if (hm.get(sum - arr[i]) != null)
twice_count += hm.get(sum - arr[i]);
// if (arr[i], arr[i]) pair satisfies the
// condition, then we need to ensure that the
// count is decreased by one such that the
// (arr[i], arr[i]) pair is not considered
if (sum - arr[i] == arr[i])
twice_count--;
}
// return the half of twice_count
return twice_count / 2;
}
// Driver method to test the above function
public static void main(String[] args)
{
int sum = 6;
System.out.println(
"Count of pairs is "
+ getPairsCount(arr.length, sum));
}
}
// This code is contributed by Gaurav Miglani
Python3
# Python 3 implementation of simple method
# to find count of pairs with given sum.
import sys
# Returns number of pairs in arr[0..n-1]
# with sum equal to 'sum'
def getPairsCount(arr, n, sum):
m = [0] * 1000
# Store counts of all elements in map m
for i in range(0, n):
m[arr[i]] += 1
twice_count = 0
# Iterate through each element and increment
# the count (Notice that every pair is
# counted twice)
for i in range(0, n):
twice_count += m[sum - arr[i]]
# if (arr[i], arr[i]) pair satisfies the
# condition, then we need to ensure that
# the count is decreased by one such
# that the (arr[i], arr[i]) pair is not
# considered
if (sum - arr[i] == arr[i]):
twice_count -= 1
# return the half of twice_count
return int(twice_count / 2)
# Driver function
arr = [1, 5, 7, -1, 5]
n = len(arr)
sum = 6
print("Count of pairs is", getPairsCount(arr,
n, sum))
# This code is contributed by
# Smitha Dinesh Semwal
C#
// C# implementation of simple method to
// find count of pairs with given sum
using System;
using System.Collections.Generic;
class GFG {
public static int[] arr = new int[] { 1, 5, 7, -1, 5 };
// Returns number of pairs in arr[0..n-1]
// with sum equal to 'sum'
public static int getPairsCount(int n, int sum)
{
Dictionary<int, int> hm
= new Dictionary<int, int>();
// Store counts of all elements
// in map hm
for (int i = 0; i < n; i++) {
// initializing value to 0,
// if key not found
if (!hm.ContainsKey(arr[i])) {
hm[arr[i]] = 0;
}
hm[arr[i]] = hm[arr[i]] + 1;
}
int twice_count = 0;
// iterate through each element and
// increment the count (Notice that
// every pair is counted twice)
for (int i = 0; i < n; i++) {
if (hm[sum - arr[i]] != 0) {
twice_count += hm[sum - arr[i]];
}
// if (arr[i], arr[i]) pair satisfies
// the condition, then we need to ensure
// that the count is decreased by one
// such that the (arr[i], arr[i])
// pair is not considered
if (sum - arr[i] == arr[i]) {
twice_count--;
}
}
// return the half of twice_count
return twice_count / 2;
}
// Driver Code
public static void Main(string[] args)
{
int sum = 6;
Console.WriteLine("Count of pairs is "
+ getPairsCount(arr.Length, sum));
}
}
// This code is contributed by Shrikant13
Javascript
<script>
/* javascript implementation of simple method to find count of
pairs with given sum*/
var arr = [ 1, 5, 7, -1, 5 ];
// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
function getPairsCount(n , sum) {
var hm = new Map();
// Store counts of all elements in map hm
for (var i = 0; i < n; i++) {
// initializing value to 0, if key not found
if (!hm.has(arr[i]))
hm.set(arr[i], 0);
hm.set(arr[i], hm.get(arr[i]) + 1);
}
var twice_count = 0;
// iterate through each element and increment the
// count (Notice that every pair is counted twice)
for (i = 0; i < n; i++) {
if (hm.get(sum - arr[i]) != null)
twice_count += hm.get(sum - arr[i]);
// if (arr[i], arr[i]) pair satisfies the
// condition, then we need to ensure that the
// count is decreased by one such that the
// (arr[i], arr[i]) pair is not considered
if (sum - arr[i] == arr[i])
twice_count--;
}
// return the half of twice_count
return twice_count / 2;
}
// Driver method to test the above function
var sum = 6;
document.write("Count of pairs is " + getPairsCount(arr.length, sum));
// This code is contributed by umadevi9616
</script>
C++
// C++ implementation of simple method to find count of
// pairs with given sum.
#include <bits/stdc++.h>
using namespace std;
// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
int getPairsCount(int arr[], int n, int k)
{
unordered_map<int, int> m;
int count = 0;
for (int i = 0; i < n; i++) {
if (m.find(k - arr[i]) != m.end()) {
count += m[k - arr[i]];
}
m[arr[i]]++;
}
return count;
}
// Driver function to test the above function
int main()
{
int arr[] = { 1, 5, 7, -1, 5};
int n = sizeof(arr) / sizeof(arr[0]);
int sum = 6;
cout << "Count of pairs is "
<< getPairsCount(arr, n, sum);
return 0;
}
Java
// Java implementation of simple method to find count of
// pairs with given sum.
import java.util.*;
class GFG{
// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
static int getPairsCount(int arr[], int n, int k)
{
HashMap<Integer,Integer> m = new HashMap<>();
int count = 0;
for (int i = 0; i < n; i++) {
if (m.containsKey(k - arr[i])) {
count += m.get(k - arr[i]);
}
if(m.containsKey(arr[i])){
m.put(arr[i], m.get(arr[i])+1);
}
else{
m.put(arr[i], 1);
}
}
return count;
}
// Driver function to test the above function
public static void main(String[] args)
{
int arr[] = { 1, 5, 7, -1, 5};
int n = arr.length;
int sum = 6;
System.out.print("Count of pairs is "
+ getPairsCount(arr, n, sum));
}
}
// This code is contributed by umadevi9616
Python3
# Python implementation of simple method to find count of
# pairs with given sum.
# Returns number of pairs in arr[0..n-1] with sum equal to 'sum'
def getPairsCount(arr, n, sum):
unordered_map = {}
count = 0
for i in range(n):
if sum - arr[i] in unordered_map:
count += unordered_map[sum - arr[i]]
if arr[i] in unordered_map:
unordered_map[arr[i]] += 1
else:
unordered_map[arr[i]] = 1
return count
# Driver code
arr = [1, 5, 7, -1, 5]
n = len(arr)
sum = 6
print('Count of pairs is', getPairsCount(arr, n, sum))
# This code is contributed by Manish Thapa
C#
// C# implementation of simple method to find count of
// pairs with given sum.
using System;
using System.Collections.Generic;
public class GFG {
// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
static int getPairsCount(int []arr, int n, int k) {
Dictionary<int, int> m = new Dictionary<int, int>();
int count = 0;
for (int i = 0; i < n; i++) {
if (m.ContainsKey(k - arr[i])) {
count += m[k - arr[i]];
}
if (m.ContainsKey(arr[i])) {
m[arr[i]] = m[arr[i]] + 1;
} else {
m.Add(arr[i], 1);
}
}
return count;
}
// Driver function to test the above function
public static void Main(String[] args) {
int []arr = { 1, 5, 7, -1, 5 };
int n = arr.Length;
int sum = 6;
Console.Write("Count of pairs is " + getPairsCount(arr, n, sum));
}
}
// This code is contributed by umadevi9616
Javascript
<script>
// javascript implementation of simple method to find count of
// pairs with given sum.
// Returns number of pairs in arr[0..n-1] with sum equal
// to 'sum'
function getPairsCount(arr , n , k) {
var m = new Map();
var count = 0;
for (var i = 0; i < n; i++) {
if (m.has(k - arr[i])) {
count += m.get(k - arr[i]);
}
if (m.has(arr[i])) {
m.set(arr[i], m.get(arr[i]) + 1);
} else {
m.set(arr[i], 1);
}
}
return count;
}
// Driver function to test the above function
var arr = [ 1, 5, 7, -1, 5 ];
var n = arr.length;
var sum = 6;
document.write("Count of pairs is " + getPairsCount(arr, n, sum));
// This code is contributed by umadevi9616
</script>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA