用和重复替换偶数对后的最小数组大小

给定一个大小为 N 的数组。从序列中选择一对随机元素，使它们的和为偶数，从序列中删除这两个元素并将它们的和插入序列中，以最小化数组的长度。最后，打印数组的最小可能大小。
例子：

Input : 88 98 1 7 3
Output : 2
By following above rules --[88, 98, 1, 10]
[98, 88, 1]---[186, 1]--we cannot move further since
186 + 1 = 187, which is not even. So, size = 2.

Input : 7 4 3 2 6
Output : 1
delete 7 and 3, insert 10---[10, 4, 2, 6]
repeating the process of deleting and inserting---[14, 8]--[22]
size of array becomes 1.

方法：
如果两个数字都是偶数或它们都是奇数，则一对数字可以加起来为偶数。因此，我们只需要计算给定数组中存在的奇数。答案可以是 2 或 1（仅此而已），具体取决于条件。如果数组中的总赔率是奇数，则打印 2，否则打印 1。

C++

// CPP Program to find minimum size of array
// after repeatedly replacing even sum pair
// with the sum
#include 
using namespace std;
 
// function to count and check
// total number of odds
bool check(int a[], int n)
{
    int i, c = 0;
 
    // loop to count total no. of
    // odd numbers in the array
    if (n == 1)
        return false;
 
    for (i = 0; i < n; i++)
        if (a[i] & 1)
            c++;   
 
    if (c & 1)
        return true;
 
    return false;
}
 
// Driver program to test
// above function
int main()
{
    int n, a[] = { 7, 4, 3, 2, 6 };
    n = sizeof(a) / sizeof(a[0]);
 
    // if in total there are
    // odd number of odds
    if (check(a, n))
        cout << 2;
    else
        cout << 1;
 
    return 0;
}

Java

// Java Program to find minimum size of array
// after repeatedly replacing even sum pair
// with the sum
import java.io.*;
import java.math.*;
 
 
class GFG {
     
     
    // function to count and check
    // total number of odds
    static boolean check(int a[], int n)
    {
        int i, c = 0;
     
        // loop to count total no. of
        // odd numbers in the array
        if (n == 1)
            return false;
     
        for (i = 0; i < n; i++)
            if ((a[i] & 1)==1)
                c++;
     
        if ((c & 1) == 1)
            return true;
     
        return false;
    }
 
    // Driver program to test
    // above function
    public static void main(String args[])
    {
        int n, a[] = { 7, 4, 3, 2, 6 };
        n = a.length;
 
        // if in total there are
        // odd number of odds
        if (check(a, n))
            System.out.println("2");
        else
            System.out.println("1");
    }
}
 
 
// This code is contributed by Nikita Tiwari.

Python 3

# Python 3 Program to illustrate
# the above approach
 
# function to count and check
# total number of odds
def check(a):
    c = 0
# if length is 1, then we return
# false in order to get output 1
    if len(a) == 1:
        return False
# loop to count number of odds
    for i in a:
        if i & 1:
            c += 1
# if c is odd
    if c & 1:
        return True
         
    return False
 
# Driver program to test above function
a = [7, 4, 3, 2, 6]
 
# if in total there are odd
# number of odd numbers
if(check(a)):
    print(2)
else:
    print(1)

C#

// C# Program to find minimum size of array
// after repeatedly replacing even sum pair
// with the sum
using System;
 
class GFG {
     
     
    // function to count and check
    // total number of odds
    static bool check(int []a, int n)
    {
        int i, c = 0;
     
        // loop to count total no. of
        // odd numbers in the array
        if (n == 1)
            return false;
     
        for (i = 0; i < n; i++)
            if ((a[i] & 1)==1)
                c++;
     
        if ((c & 1) == 1)
            return true;
     
        return false;
    }
 
    // Driver program
    public static void Main()
    {
        int n;
        int []a = { 7, 4, 3, 2, 6 };
        n = a.Length;
 
        // if in total there are
        // odd number of odds
        if (check(a, n))
            Console.WriteLine("2");
        else
            Console.WriteLine("1");
    }
}
 
 
// This code is contributed by vt_m.

PHP

Javascript

输出：

时间复杂度O(n)
空间复杂度O(1)。