计算给定总和的对

给定一个整数数组和一个数字“sum”，找出数组中总和等于“sum”的整数对的数量。

例子：

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

预期时间复杂度 O(n)

朴素的解决方案——一个简单的解决方案是遍历每个元素并检查数组中是否还有另一个数字可以添加到其中以得出总和。

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'
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;
}

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 program is contributed by Jyotsna

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

Javascript

 
// 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
 

输出

Count of pairs is 3

时间复杂度： O(n ² )
辅助空间： O(1)

高效的解决方案——
在 O(n) 时间内可能有更好的解决方案。下面是算法——

1. 创建一个映射来存储数组中每个数字的频率。 （需要单次遍历）
2. 在下一次遍历中，对于每个元素检查它是否可以与任何其他元素（除了它自己！）组合以给出所需的总和。相应地增加计数器。
3. 完成第二次遍历后，我们将在计数器中存储所需值的两倍，因为每一对都被计数两次。因此，将计数除以 2 并返回。

以下是上述想法的实现：

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'
int getPairsCount(int arr[], int n, int sum)
{
    unordered_map 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 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

蟒蛇3

# 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 hm
            = new Dictionary();
 
        // 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

输出

Count of pairs is 3

一个循环中更有效的解决方案：-

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'
int getPairsCount(int arr[], int n, int k)
{
    unordered_map 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;
}

输出

Count of pairs is 3

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。