给定整数数组arr [] ，任务是检查是否有可能从数组中获得K个元素的子序列，以使它们的和为奇数。如果可能，请打印“是” 。否则，打印No。
例子：
Input: arr[] = { 2, 5, 6, 7, 4 }, K = 3
Output: Yes
Explanation:
Subsequences {2, 5, 6}, {2, 6, 7} and {2, 7, 4} have odd sum
Input: arr[] = { 1, 5, 7, 11 }, K = 4
Output: No
Explanation:
Only subsequence of length 4 is {1, 5, 7, 11} which has even sum (24). Hence no such subsequence exists.
天真的方法：
解决该问题的最简单方法是生成所有长度为K的子序列，并检查这些子序列中是否有一个奇数和。这种方法的时间复杂度将是指数级的，因此效率很低。
高效方法：
解决上述问题的有效方法是对数组中奇数元素的数量进行计数，然后在不可能找到具有奇数和的子序列时，仅检查所有边缘情况。
无法生成这种子序列时要考虑的边缘情况如下：
- 如果数组中没有奇数元素，则任何子序列将仅包含偶数元素，并且将获得偶数和。因此，不可能生成具有奇数和的子序列。
- 如果K为偶数，并且数组中不存在偶数元素，则奇数和的子序列是不可能的。
对于所有其他情况，将可能生成具有奇数和的子序列。
下面是上述方法的实现：
C++
// C++ program to check if a
// subsequence of length K
// with odd sum exists in the
// given array
#include
using namespace std;
// Function to check if any required
// subsequence exists or not
bool isSubseqPossible(int arr[], int N, int K)
{
int i;
// Store count of odd and
// even elements in the array
int odd = 0, even = 0;
// Calculate the count of
// odd and even elements
for (i = 0; i < N; i++) {
if (arr[i] % 2 == 1)
odd++;
else
even++;
}
// If no odd elements exists
// or no even elements exists
// when K is even
if (odd == 0
|| (even == 0 && K % 2 == 0))
// Subsequence is not possible
return false;
// Possible otherwise
return true;
}
// Driver Code
int main()
{
int arr[] = { 2, 3, 5, 7, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 3;
cout << (isSubseqPossible(arr, N, K)
? "Yes"
: "No");
return 0;
} Java
// Java program to check if a
// subsequence of length K
// with odd sum exists in the
// given array
class GFG{
// Function to check if any required
// subsequence exists or not
static boolean isSubseqPossible(int []arr,
int N, int K)
{
int i;
// Store count of odd and
// even elements in the array
int odd = 0, even = 0;
// Calculate the count of
// odd and even elements
for (i = 0; i < N; i++)
{
if (arr[i] % 2 == 1)
odd++;
else
even++;
}
// If no odd elements exists
// or no even elements exists
// when K is even
if (odd == 0 || (even == 0 && K % 2 == 0))
// Subsequence is not possible
return false;
// Possible otherwise
return true;
}
// Driver Code
public static void main(String args[])
{
int []arr = { 2, 3, 5, 7, 4 };
int N = arr.length;
int K = 3;
System.out.print(isSubseqPossible(arr, N, K) ?
"Yes" : "No");
}
}
// This code is contributed by Code_MechPython3
# Python3 program to check if a subsequence
# of length K with odd sum exists in the
# given array
# Function to check if any required
# subsequence exists or not
def isSubseqPossible(arr, N, K):
i = 0
# Store count of odd and
# even elements in the array
odd = 0
even = 0
# Calculate the count of
# odd and even elements
for i in range(N):
if (arr[i] % 2 == 1):
odd += 1
else:
even += 1
# If no odd element exists or no
# even element exists when K even
if (odd == 0 or (even == 0 and K % 2 == 0)):
# Subsequence is not possible
return False
# Otherwise possible
return True
# Driver code
if __name__ == '__main__':
arr = [ 2, 3, 5, 7, 4 ]
N = len(arr)
K = 3
print("Yes" if isSubseqPossible(arr, N, K) else "No")
# This code is contributed by himanshu77C#
// C# program to check if a
// subsequence of length K
// with odd sum exists in the
// given array
using System;
class GFG{
// Function to check if any required
// subsequence exists or not
static bool isSubseqPossible(int []arr,
int N, int K)
{
int i;
// Store count of odd and
// even elements in the array
int odd = 0, even = 0;
// Calculate the count of
// odd and even elements
for (i = 0; i < N; i++)
{
if (arr[i] % 2 == 1)
odd++;
else
even++;
}
// If no odd elements exists
// or no even elements exists
// when K is even
if (odd == 0 || (even == 0 && K % 2 == 0))
// Subsequence is not possible
return false;
// Possible otherwise
return true;
}
// Driver Code
public static void Main()
{
int []arr = { 2, 3, 5, 7, 4 };
int N = arr.Length;
int K = 3;
Console.Write(isSubseqPossible(arr, N, K) ?
"Yes" : "No");
}
}
// This code is contributed by Code_Mech输出：
Yes
时间复杂度： O(N)
辅助空间： O(1)