打印给定总和的所有对

给定一个整数数组和一个数字“ sum”，则打印该数组中所有总和等于“ sum”的对。

Examples :
Input  :  arr[] = {1, 5, 7, -1, 5}, 
          sum = 6
Output : (1, 5) (7, -1) (1, 5)

Input  :  arr[] = {2, 5, 17, -1}, 
          sum = 7
Output :  (2, 5)

一个简单的解决方案是遍历每个元素，并检查数组中是否可以添加另一个数字以求和。

C++

// C++ implementation of simple method to
// find print pairs with given sum.
#include 
using namespace std;
 
// Returns number of pairs in arr[0..n-1]
// with sum equal to 'sum'
int printPairs(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)
                cout << "(" << arr[i] << ", "
                     << arr[j] << ")" << endl;
}
 
// 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;
    printPairs(arr, n, sum);
    return 0;
}

Java

// Java implementation of
// simple method to find
// print pairs with given sum.
 
class GFG {
 
    // Returns number of pairs
    // in arr[0..n-1] with sum
    // equal to 'sum'
    static void printPairs(int arr[],
                           int n, int sum)
    {
        // int count = 0;
 
        // 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)
                    System.out.println("(" + arr[i] + ", " + arr[j] + ")");
    }
 
    // Driver Code
    public static void main(String[] arg)
    {
        int arr[] = { 1, 5, 7, -1, 5 };
        int n = arr.length;
        int sum = 6;
        printPairs(arr, n, sum);
    }
}
 
// This code is contributed
// by Smitha

Python3

# Python 3 implementation
# of simple method to find
# print pairs with given sum.
 
# Returns number of pairs
# in arr[0..n-1] with sum
# equal to 'sum'
def printPairs(arr, n, sum):
 
    # count = 0
 
    # 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):
                print("(", arr[i],
                      ", ", arr[j],
                      ")", sep = "")
 
 
# Driver Code
arr = [1, 5, 7, -1, 5]
n = len(arr)
sum = 6
printPairs(arr, n, sum)
 
# This code is contributed
# by Smitha

C#

// C# implementation of simple
// method to find print pairs
// with given sum.
using System;
 
class GFG {
    // Returns number of pairs
    // in arr[0..n-1] with sum
    // equal to 'sum'
    static void printPairs(int[] arr,
                           int n, int sum)
    {
        // int count = 0;
 
        // 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)
                    Console.Write("(" + arr[i] + ", " + arr[j] + ")"
                                  + "\n");
    }
 
    // Driver Code
    public static void Main()
    {
        int[] arr = { 1, 5, 7, -1, 5 };
        int n = arr.Length;
        int sum = 6;
        printPairs(arr, n, sum);
    }
}
 
// This code is contributed
// by Smitha

Javascript

 
// JavaScript implementation of simple method to
// find print pairs with given sum.
 
// Returns number of pairs in arr[0..n-1]
// with sum equal to 'sum'
function printPairs(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)
                 document.write("(" + arr[i] + ", "
                    + arr[j] + ")" + "<br>");
}
 
// Driver function to test the above function
 
    let arr = [ 1, 5, 7, -1, 5 ];
    let n = arr.length;
    let sum = 6;
    printPairs(arr, n, sum);
 
 
// This code is contributed by Surbhi Tyagi
 

输出 ：

(1, 5)
(1, 5)
(7, -1)

方法2(使用哈希) 。
我们创建一个空的哈希表。现在，我们遍历数组并检查哈希表中的对。如果找到匹配的元素，我们将打印出等于该匹配元素出现次数的对数。
请注意，此解决方案的时间复杂度最坏的情况是O(c + n) ，其中c是具有给定总和的对的计数。

C++

// C++ implementation of simple method to
// find count of pairs with given sum.
#include 
using namespace std;
 
// Returns number of pairs in arr[0..n-1]
// with sum equal to 'sum'
void printPairs(int arr[], int n, int sum)
{
    // Store counts of all elements in map m
    unordered_map m;
 
    // Traverse through all elements
    for (int i = 0; i < n; i++) {
 
        // Search if a pair can be formed with
        // arr[i].
        int rem = sum - arr[i];
        if (m.find(rem) != m.end()) {
            int count = m[rem];
            for (int j = 0; j < count; j++)
                cout << "(" << rem << ", "
                     << arr[i] << ")" << endl;
        }
        m[arr[i]]++;
    }
}
 
// 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;
    printPairs(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 void printPairs(int arr[], int n,
                       int sum)
{
     
    // Store counts of all elements in map m
    HashMap mp = new HashMap();
 
    // Traverse through all elements
    for(int i = 0; i < n; i++)
    {
         
        // Search if a pair can be formed with
        // arr[i].
        int rem = sum - arr[i];
        if (mp.containsKey(rem))
        {
            int count = mp.get(rem);
             
            for(int j = 0; j < count; j++)
                System.out.print("(" + rem +
                                ", " + arr[i] +
                                 ")" + "\n");
        }
        if (mp.containsKey(arr[i]))
        {
            mp.put(arr[i], mp.get(arr[i]) + 1);
        }
        else
        {
            mp.put(arr[i], 1);
        }
    }
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 5, 7, -1, 5 };
    int n = arr.length;
    int sum = 6;
     
    printPairs(arr, n, sum);
}
}
 
// This code is contributed by Princi Singh

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 printPairs(arr, n, sum):
     
    # Store counts of all elements
    # in a dictionary
    mydict = dict()
 
    # Traverse through all the elements
    for i in range(n):
         
        # Search if a pair can be
        # formed with arr[i]
        temp = sum - arr[i]
         
        if temp in mydict:
            count = mydict[temp]
            for j in range(count):
                print("(", temp, ", ", arr[i],
                      ")", sep = "", end = '\n')
                       
        if arr[i] in mydict:
            mydict[arr[i]] += 1
        else:
            mydict[arr[i]] = 1
 
# Driver code
if __name__ == '__main__':
     
    arr = [ 1, 5, 7, -1, 5 ]
    n = len(arr)
    sum = 6
 
    printPairs(arr, n, sum)
 
# This code is contributed by MuskanKalra1

C＃

// C# implementation of simple method to
// find count of pairs with given sum.
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
  
// Returns number of pairs in arr[0..n-1]
// with sum equal to 'sum'
static void printPairs(int []arr, int n, int sum)
{
     
    // Store counts of all elements in map m
    Dictionary m = new Dictionary();
                                        
    // Traverse through all elements
    for(int i = 0; i < n; i++)
    {
         
        // Search if a pair can be formed with
        // arr[i].
        int rem = sum - arr[i];
         
        if (m.ContainsKey(rem))
        {
            int count = m[rem];
             
            for(int j = 0; j < count; j++)
            {
                Console.Write("(" + rem + ", " +
                           arr[i] + ")" + "\n");
            }
        }
         
        if (m.ContainsKey(arr[i]))
        {
            m[arr[i]]++;
        }
        else
        {
            m[arr[i]] = 1;
        }
    }
}
 
// Driver code
public static void Main(string[] args)
{
    int []arr = { 1, 5, 7, -1, 5 };
    int n = arr.Length;
    int sum = 6;
     
    printPairs(arr, n, sum);
}
}
 
// This code is contributed by rutvik_56

输出 ：

(1, 5)
(7, -1)
(1, 5)

方法3 。
给出另一种打印具有给定总和的所有对的方法，如下所示：

C++

// C++ code to implement
// the above approach
#include
using namespace std;
 
void pairedElements(int arr[],
                    int sum, int n)
{
  int low = 0;
  int high = n - 1;
 
  while (low < high)
  {
    if (arr[low] + arr[high] == sum)
    {
      cout << "The pair is : (" <<
               arr[low] << ", " <<
               arr[high] << ")" << endl;
    }
    if (arr[low] + arr[high] > sum)
    {
      high--;
    }
    else
    {
      low++;
    }
  }
}
 
// Driver code
int  main()
{
  int arr[] = {2, 3, 4, -2,
               6, 8, 9, 11};
  int n = sizeof(arr) / sizeof(arr[0]);
  sort(arr, arr + n);
  pairedElements(arr, 6, n);
}
 
// This code is contributed by Rajput-Ji

Java

import java.util.Arrays;
 
/**
 * Created by sampat.
 */
public class SumOfPairs {
 
    public void pairedElements(int arr[], int sum)
    {
        int low = 0;
        int high = arr.length - 1;
 
        while (low < high) {
            if (arr[low] + arr[high] == sum) {
                System.out.println("The pair is : ("
                                   + arr[low] + ", " + arr[high] + ")");
            }
            if (arr[low] + arr[high] > sum) {
                high--;
            }
            else {
                low++;
            }
        }
    }
 
    public static void main(String[] args)
    {
        int arr[] = { 2, 3, 4, -2, 6, 8, 9, 11 };
        Arrays.sort(arr);
 
        SumOfPairs sp = new SumOfPairs();
        sp.pairedElements(arr, 6);
    }
}

Python3

# Python3 program for the
# above approach
def pairedElements(arr, sum):
   
    low = 0;
    high = len(arr) - 1;
 
    while (low < high):
        if (arr[low] +
            arr[high] == sum):
            print("The pair is : (", arr[low],
                  ", ", arr[high], ")");
        if (arr[low] + arr[high] > sum):
            high -= 1;
        else:
            low += 1;
 
# Driver code
if __name__ == '__main__':
   
    arr = [2, 3, 4, -2,
           6, 8, 9, 11];
    arr.sort();
    pairedElements(arr, 6);
 
# This code contributed by shikhasingrajput

C＃

// C# program to find triplets in a given
// array whose sum is equal to given sum.
using System;
 
public class SumOfPairs
{
 
    public void pairedElements(int []arr, int sum)
    {
        int low = 0;
        int high = arr.Length - 1;
 
        while (low < high)
        {
            if (arr[low] + arr[high] == sum)
            {
                Console.WriteLine("The pair is : ("
                                + arr[low] + ", " + arr[high] + ")");
            }
            if (arr[low] + arr[high] > sum)
            {
                high--;
            }
            else
            {
                low++;
            }
        }
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = { 2, 3, 4, -2, 6, 8, 9, 11 };
        Array.Sort(arr);
 
        SumOfPairs sp = new SumOfPairs();
        sp.pairedElements(arr, 6);
    }
}
 
// This code is contributed by Princi Singh

Java脚本

 
        // Javascript code to implement
        // the above approach
        function pairedElements(arr, sum, n) {
            var low = 0;
            var high = n - 1;
 
            while (low < high) {
                if (arr[low] + arr[high] == sum) {
                    document.write("The pair is : (" +
                        arr[low] + ", " +
                        arr[high] + ")<br>");
                }
                if (arr[low] + arr[high] > sum) {
                    high--;
                }
                else {
                    low++;
                }
            }
        }
 
        // Driver code
        var arr = [ 2, 3, 4, -2, 6, 8, 9, 11]
        var n = arr.length;
        arr.sort(function(a,b){return a-b;});
        pairedElements(arr, 6, n);
 
 
    

输出 ：

The pair is : (-2, 8)
The pair is : (2, 4)