给定N个正整数的数组arr [] ，任务是找到奇数和的对数和偶数的对数。
例子：
Input: arr[] = {1, 2, 3, 4, 5}
Output:
Odd pairs = 6
Even pairs = 4
Input: arr[] = {7, 4, 3, 2}
Output:
Odd pairs = 4
Even pairs = 2
天真的方法：
解决此问题的幼稚方法是遍历数组中的每对元素，检查它们的和，然后计算具有奇数和偶数和的对的数量。此方法将花费O(N * N)时间。
高效方法：
- 计算数组中奇数和偶数元素的数量，并将它们存储在变量cntEven和cntOdd中。
- 为了使偶数和为偶数，所有偶数元素仅与偶数元素配对，所有奇数元素仅与奇数元素配对，计数为((cntEven *(cntEven – 1))/ 2 )+((cntOdd *(cntOdd – 1))/ 2)
- 现在，要使总和为奇数，该对中的一个元素必须为偶数，另一个元素必须为奇数，所需的计数将为cntEven * cntOdd 。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to find the count of pairs
// with odd sum and the count
// of pairs with even sum
void findPairs(int arr[], int n)
{
// To store the count of even and
// odd number from the array
int cntEven = 0, cntOdd = 0;
for (int i = 0; i < n; i++) {
// If the current element is even
if (arr[i] % 2 == 0)
cntEven++;
// If it is odd
else
cntOdd++;
}
// To store the count of
// pairs with even sum
int evenPairs = 0;
// All the even elements will make
// pairs with each other and the
// sum of the pair will be even
evenPairs += ((cntEven * (cntEven - 1)) / 2);
// All the odd elements will make
// pairs with each other and the
// sum of the pair will be even
evenPairs += ((cntOdd * (cntOdd - 1)) / 2);
// To store the count of
// pairs with odd sum
int oddPairs = 0;
// All the even elements will make pairs
// with all the odd element and the
// sum of the pair will be odd
oddPairs += (cntEven * cntOdd);
cout << "Odd pairs = " << oddPairs << endl;
cout << "Even pairs = " << evenPairs;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = sizeof(arr) / sizeof(int);
findPairs(arr, n);
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Function to find the count of pairs
// with odd sum and the count
// of pairs with even sum
static void findPairs(int arr[], int n)
{
// To store the count of even and
// odd number from the array
int cntEven = 0, cntOdd = 0;
for (int i = 0; i < n; i++)
{
// If the current element is even
if (arr[i] % 2 == 0)
cntEven++;
// If it is odd
else
cntOdd++;
}
// To store the count of
// pairs with even sum
int evenPairs = 0;
// All the even elements will make
// pairs with each other and the
// sum of the pair will be even
evenPairs += ((cntEven * (cntEven - 1)) / 2);
// All the odd elements will make
// pairs with each other and the
// sum of the pair will be even
evenPairs += ((cntOdd * (cntOdd - 1)) / 2);
// To store the count of
// pairs with odd sum
int oddPairs = 0;
// All the even elements will make pairs
// with all the odd element and the
// sum of the pair will be odd
oddPairs += (cntEven * cntOdd);
System.out.println("Odd pairs = " + oddPairs);
System.out.println("Even pairs = " + evenPairs);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 2, 3, 4, 5 };
int n = arr.length;
findPairs(arr, n);
}
}
// This code is contributed by Rajput-JiPython3
# Python3 implementation of the approach
# Function to find the count of pairs
# with odd sum and the count
# of pairs with even sum
def findPairs(arr, n) :
# To store the count of even and
# odd number from the array
cntEven = 0; cntOdd = 0;
for i in range(n) :
# If the current element is even
if (arr[i] % 2 == 0) :
cntEven += 1;
# If it is odd
else :
cntOdd += 1;
# To store the count of
# pairs with even sum
evenPairs = 0;
# All the even elements will make
# pairs with each other and the
# sum of the pair will be even
evenPairs += ((cntEven * (cntEven - 1)) // 2);
# All the odd elements will make
# pairs with each other and the
# sum of the pair will be even
evenPairs += ((cntOdd * (cntOdd - 1)) // 2);
# To store the count of
# pairs with odd sum
oddPairs = 0;
# All the even elements will make pairs
# with all the odd element and the
# sum of the pair will be odd
oddPairs += (cntEven * cntOdd);
print("Odd pairs = ", oddPairs);
print("Even pairs = ", evenPairs);
# Driver code
if __name__ == "__main__" :
arr = [ 1, 2, 3, 4, 5 ];
n = len(arr);
findPairs(arr, n);
# This code is contributed by kanugargngC#
// C# implementation of the approach
using System;
class GFG
{
// Function to find the count of pairs
// with odd sum and the count
// of pairs with even sum
static void findPairs(int []arr, int n)
{
// To store the count of even and
// odd number from the array
int cntEven = 0, cntOdd = 0;
for (int i = 0; i < n; i++)
{
// If the current element is even
if (arr[i] % 2 == 0)
cntEven++;
// If it is odd
else
cntOdd++;
}
// To store the count of
// pairs with even sum
int evenPairs = 0;
// All the even elements will make
// pairs with each other and the
// sum of the pair will be even
evenPairs += ((cntEven * (cntEven - 1)) / 2);
// All the odd elements will make
// pairs with each other and the
// sum of the pair will be even
evenPairs += ((cntOdd * (cntOdd - 1)) / 2);
// To store the count of
// pairs with odd sum
int oddPairs = 0;
// All the even elements will make pairs
// with all the odd element and the
// sum of the pair will be odd
oddPairs += (cntEven * cntOdd);
Console.WriteLine("Odd pairs = " + oddPairs);
Console.WriteLine("Even pairs = " + evenPairs);
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 2, 3, 4, 5 };
int n = arr.Length;
findPairs(arr, n);
}
}
// This code is contributed by Rajput-Ji输出：
Odd pairs = 6
Even pairs = 4
复杂度分析：
- 时间复杂度： O(N)。
在for循环的每次迭代中，将处理来自任一数组的元素。因此，时间复杂度为O(N)。 - 辅助空间： O(1)。
不需要任何额外的空间，因此空间复杂度是恒定的。
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