算法 —— Count pairs with given sum ——dictionary的妙用，高，实在是高！！！

原文：https://www.geeksforgeeks.org/count-pairs-with-given-sum/

 

Given an array of integers, and a number ‘sum’, find the number of pairs of integers in the array whose sum is equal to ‘sum’.

Examples:  

Input  :  arr[] = {1, 5, 7, -1}, 
          sum = 6
Output :  2
Pairs with sum 6 are (1, 5) and (7, -1)

Input  :  arr[] = {1, 5, 7, -1, 5}, 
          sum = 6
Output :  3
Pairs with sum 6 are (1, 5), (7, -1) &
                     (1, 5)         

Input  :  arr[] = {1, 1, 1, 1}, 
          sum = 2
Output :  6
There are 3! pairs with sum 2.

Input  :  arr[] = {10, 12, 10, 15, -1, 7, 6, 
                   5, 4, 2, 1, 1, 1}, 
          sum = 11
Output :  9

Expected time complexity O(n)
 

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Naive Solution – A simple solution is to traverse each element and check if there’s another number in the array which can be added to it to give sum. 
 

 

 

// 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

Output

 

 

Count of pairs is 3

Time Complexity: O(n²) 
Auxiliary Space: O(1)
  

Efficient solution – 
A better solution is possible in O(n) time. Below is the Algorithm – 

1. Create a map to store frequency of each number in the array. (Single traversal is required)
2. In the next traversal, for every element check if it can be combined with any other element (other than itself!) to give the desired sum. Increment the counter accordingly.
3. After completion of second traversal, we’d have twice the required value stored in counter because every pair is counted two times. Hence divide count by 2 and return.

Below is the implementation of above idea : 
 

 

 

// 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

Output

Count of pairs is 3

This article is contributed by Ashutosh Kumar. 

 

修改错误code

        [Theory]
        [InlineData(new int[] { 6, 1, 3, 46 }, 47)]
        // Returns number of pairs in arr[0..n-1]
        // with sum equal to 'sum'
        public static int getPairsCount2(int[] arr, int sum)
        {
            int n = arr.Length;
            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.ContainsKey(sum - arr[i]) && 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;
        }

　　

 

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