具有奇数和的数组中的三元组数

给定一个包含N个整数的数组arr[] ，找出i 、 j和k三元组的数量，使得1<= i < j < k <= N并且arr[i] + arr[j] + arr[k]是奇怪的。

例子：

Input: arr[] = {1, 2, 3, 4, 5}
Output: 4
Explanation: The given array contains 4 triplets with an odd sum. They are {1, 2, 4}, {1, 3, 5}, {2, 3, 4} and {2, 4, 5}.

Input: arr[] ={4, 5, 6, 4, 5, 10, 1, 7}
Output: 28

朴素方法：给定的问题可以通过遍历数组中所有可能的无序三元组来解决，并跟踪三元组的数量，使它们的总和为奇数。
时间复杂度： O(N ³ )

有效的方法：上述方法可以使用整数的以下属性进行优化：

• 奇数 + 偶数 + 偶数 = 奇数
• 奇数 + 奇数 + 奇数 = 奇数

因此，要实现上述思想，我们可以统计数组中偶数和奇数的个数。假设奇数的个数是O ，偶数个数是E 。所以排列一个奇数和两个偶数的不同方法是^O C ₁ * ^E C ₂ -> (O * E * (E-1))/2排列三个奇数的不同方法是^O C ₃ - > (O * (O-1) * (O-2)) / 6 。最终答案将是上述两个值的总和。

下面是上述方法的实现：

C++

// C++ Program for the above approach
#include 
using namespace std;
 
// Function to count the number of
// unordered triplets such that their
// sum is an odd integer
int countTriplets(int arr[], int n)
{
    // Count the number of odd and
    // even integers in the array
    int odd = 0, even = 0;
    for (int i = 0; i < n; i++) {
        if (arr[i] & 1)
            odd++;
        else
            even++;
    }
 
    // Number of ways to create triplets
    // using one odd and two even integers
    int c1 = odd * (even * (even - 1)) / 2;
 
    // Number of ways to create triplets
    // using three odd integers
    int c2 = (odd * (odd - 1) * (odd - 2)) / 6;
 
    // Return answer
    return c1 + c2;
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 5, 6, 4, 5, 10, 1, 7 };
    int n = sizeof(arr) / sizeof(int);
 
    // Function Call
    int ans = countTriplets(arr, n);
 
    // Print Answer
    cout << ans;
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG {
 
// Function to count the number of
// unordered triplets such that their
// sum is an odd integer
static int countTriplets(int arr[], int n)
{
    // Count the number of odd and
    // even integers in the array
    int odd = 0, even = 0;
    for (int i = 0; i < n; i++) {
        if ((arr[i] & 1) != 0)
            odd++;
        else
            even++;
    }
 
    // Number of ways to create triplets
    // using one odd and two even integers
    int c1 = odd * (even * (even - 1)) / 2;
 
    // Number of ways to create triplets
    // using three odd integers
    int c2 = (odd * (odd - 1) * (odd - 2)) / 6;
 
    // Return answer
    return c1 + c2;
}
 
// Driver code
public static void main(String[] args)
{
     int arr[] = { 4, 5, 6, 4, 5, 10, 1, 7 };
    int n = arr.length;
 
    // Function Call
    int ans = countTriplets(arr, n);
 
    // Print Answer
    System.out.println(ans);
}
}
 
// This code is contributed by code_hunt.

Python3

# Python Program for the above approach
 
# Function to count the number of
# unordered triplets such that their
# sum is an odd integer
def countTriplets(arr, n):
     
    # Count the number of odd and
    # even integers in the array
    odd = 0
    even = 0
    for i in range(n):
        if (arr[i] & 1):
            odd+=1
        else:
            even+=1
     
    # Number of ways to create triplets
    # using one odd and two even integers
    c1 = odd * (even * (even - 1)) // 2
     
    # Number of ways to create triplets
    # using three odd integers
    c2 = (odd * (odd - 1) * (odd - 2)) // 6
     
    # Return answer
    return c1 + c2
     
     
# Driver Code
 
arr = [4, 5, 6, 4, 5, 10, 1, 7]
n = len(arr)
 
# Function Call
ans = countTriplets(arr, n)
 
# Print Answer
print(ans)
 
# This code is contributed by Shivani

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to count the number of unordered
// triplets such that their sum is an odd integer
static int countTriplets(int []arr, int n)
{
     
    // Count the number of odd and
    // even integers in the array
    int odd = 0, even = 0;
    for(int i = 0; i < n; i++)
    {
        if ((arr[i] & 1) != 0)
            odd++;
        else
            even++;
    }
 
    // Number of ways to create triplets
    // using one odd and two even integers
    int c1 = odd * (even * (even - 1)) / 2;
 
    // Number of ways to create triplets
    // using three odd integers
    int c2 = (odd * (odd - 1) * (odd - 2)) / 6;
 
    // Return answer
    return c1 + c2;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 4, 5, 6, 4, 5, 10, 1, 7 };
    int n = arr.Length;
 
    // Function Call
    int ans = countTriplets(arr, n);
 
    // Print Answer
    Console.Write(ans);
}
}
 
// This code is contributed by shivanisinghss2110.

Javascript

 
        // JavaScript program for the above approach;
 
        // Function to count the number of
        // unordered triplets such that their
        // sum is an odd integer
        function countTriplets(arr, n)
        {
         
            // Count the number of odd and
            // even integers in the array
            let odd = 0, even = 0;
            for (let i = 0; i < n; i++) {
                if (arr[i] & 1)
                    odd++;
                else
                    even++;
            }
 
            // Number of ways to create triplets
            // using one odd and two even integers
            let c1 = Math.floor(odd * (even * (even - 1)) / 2);
 
            // Number of ways to create triplets
            // using three odd integers
            let c2 = Math.floor((odd * (odd - 1) * (odd - 2)) / 6);
 
            // Return answer
 
            return c1 + c2;
        }
 
        // Driver Code
 
        let arr = [4, 5, 6, 4, 5, 10, 1, 7];
        let n = arr.length;
        // Function Call
        let ans = countTriplets(arr, n);
 
        // Print Answer
        document.write(ans);
 
   // This code is contributed by Potta Lokesh
    

输出

28

时间复杂度： O(N)
辅助空间： O(1)