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📜  在数组中查找总和在数组中不存在的对

📅  最后修改于: 2021-10-27 06:57:48             🧑  作者: Mango

给定一个由N 个正整数组成的数组arr[] ,任务是打印给定数组中不存在总和的所有数组元素对。如果不存在这样的对,则打印“-1”
例子:

天真的方法:
解决这个问题的最简单的方法是一个一个地生成所有可能的对,并通过遍历数组检查它的和是否存在于数组中。如果发现任何对的总和存在于数组中,则打印该对。否则,移动到下一对。
时间复杂度: O(N 3 )
辅助空间: O(N)
有效的方法:
这个问题可以使用 HashSet 来解决。请按照以下步骤解决问题:

  • 将数组中的元素存储在HashSet 中
  • 现在,遍历所有数组元素并生成所有可能的对。
  • 对于每一对,检查该对的总和是否存在于 HashSet 中。如果是,请打印该对。否则,移动到下一对。

下面是上述方法的实现:

C++
// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to print all pairs
// with sum not present in the array
void findPair(int arr[], int n)
{
    int i, j;
 
    // Corner Case
    if (n < 2)
    {
        cout << "-1" << endl;
    }
 
    // Stores the distinct array
    // elements
    set  hashMap;
 
    for(int k = 0; k < n; k++)
    {
        hashMap.insert(arr[k]);
    }
 
    // Generate all possible pairs
    for(i = 0; i < n - 1; i++)
    {
        for(j = i + 1; j < n; j++)
        {
             
            // Calculate sum of current pair
            int sum = arr[i] + arr[j];
 
            // Check if the sum exists in
            // the HashSet or not
            if (hashMap.find(sum) == hashMap.end())
            {
     
                cout << "(" << arr[i] << ", "
                    << arr[j] << ")" << endl;
            }
        }
    }
}
 
// Driver code
int main()
{
    int arr[] = { 2, 4, 2, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    findPair(arr, n);
    return 0;
}
 
// This code is contributed by divyeshrabadiya07


Java
// Java program to implement
// the above approach
import java.util.*;
 
class GFG {
 
    // Function to print all pairs
    // with sum not present in the array
    public static void findPair(
        int[] arr, int n)
    {
        int i, j;
 
        // Corner Case
        if (n < 2) {
            System.out.println("-1");
        }
 
        // Stores the distinct array
        // elements
        HashSet hashMap
            = new HashSet();
 
        for (Integer k : arr) {
            hashMap.add(k);
        }
 
        // Generate all possible pairs
        for (i = 0; i < n - 1; i++) {
 
            for (j = i + 1; j < n; j++) {
 
                // Calculate sum of current pair
                int sum = arr[i] + arr[j];
 
                // Check if the sum exists in
                // the HashSet or not
                if (!hashMap.contains(sum)) {
 
                    System.out.println(
                        "(" + arr[i] + ", "
                        + arr[j] + ")");
                }
            }
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 2, 4, 2, 6 };
        int n = arr.length;
 
        findPair(arr, n);
    }
}


Python3
# Python3 program to implement
# the above approach
 
# Function to print all pairs
# with sum not present in the array
def findPair(arr, n):
 
    # Corner Case
    if (n < 2):
        print("-1")
 
    # Stores the distinct array
    # elements
    hashMap = []
 
    for k in arr:
        hashMap.append(k)
 
    # Generate all possible pairs
    for i in range (n - 1):
        for j in range (i + 1, n):
            # Calculate sum of current pair
            sum = arr[i] + arr[j]
 
            # Check if the sum exists in
            # the HashSet or not
            if sum not in hashMap:
                print("(", arr[i] , ", ",
                        arr[j] , ")")
 
# Driver Code
if __name__ == "__main__":
     
    arr = [ 2, 4, 2, 6 ]
    n = len(arr)
         
    findPair(arr, n)
 
# This code is contributed by ChitraNayal


C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to print all pairs
// with sum not present in the array
public static void findPair(int[] arr, int n)
{
    int i, j;
 
    // Corner Case
    if (n < 2)
    {
        Console.Write("-1");
    }
 
    // Stores the distinct array
    // elements
    HashSet hashMap = new HashSet();
 
    foreach(int k in arr)
    {
        hashMap.Add(k);
    }
 
    // Generate all possible pairs
    for(i = 0; i < n - 1; i++)
    {
        for(j = i + 1; j < n; j++)
        {
             
            // Calculate sum of current pair
            int sum = arr[i] + arr[j];
 
            // Check if the sum exists in
            // the HashSet or not
            if (!hashMap.Contains(sum))
            {
                Console.Write("(" + arr[i] +
                             ", " + arr[j] +
                             ")\n");
            }
        }
    }
}
 
// Driver Code
public static void Main(string[] args)
{
    int[] arr = { 2, 4, 2, 6 };
    int n = arr.Length;
 
    findPair(arr, n);
}
}
 
// This code is contributed by rutvik_56


Javascript


输出:
(2, 6)
(4, 6)
(2, 6)

时间复杂度: O(N 2 )
辅助空间: O(N)

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