查找与所有其他数组元素的奇偶性不同的元素的索引

给定大小为N (N> 3)的数组arr [] ，任务是查找相对于所有其他数组元素而言奇偶性(奇/偶)不同的元素的位置。
注意：保证始终有一个与所有其他元素的奇偶校验数字不同的数字。

例子：

Input: arr[] = {2, 4, 7, 8, 10}
Output: 2
Explanation: The only odd element in the array is 7 (= arr[2]). Therefore, required output is 2.

Input: arr[] = {2, 1, 1}
Output: 0

天真的方法：解决此问题的最简单方法是将所有偶数和奇数及其索引存储在Multimap中，并打印存在于Map中大小为1的元素的索引。请按照以下步骤解决问题：

1. 初始化两个Multimap ，以获取偶数和奇数。
2. 遍历数组并将数组元素及其索引存储在其各自的Multimap中。
3. 现在找到大小等于1的Multimap 。打印存储在该Multimap中的数组元素的索引。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to print the array
// element which differs in parity
// with the remaining array elements
int OddOneOut(int arr[], int N)
{
 
    // Multimaps to store even
    // and odd numbers along
    // with their indices
    multimap e, o;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // If array element is even
        if (arr[i] % 2 == 0) {
            e.insert({ arr[i], i });
        }
 
        // Otherwise
        else {
            o.insert({ arr[i], i });
        }
    }
 
    // If only one even element
    // is present in the array
    if (e.size() == 1) {
        cout << e.begin()->second;
    }
 
    // If only one odd element
    // is present in the array
    else {
        cout << o.begin()->second;
    }
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 4, 7, 8, 10 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    OddOneOut(arr, N);
 
    return 0;
}

Python3

# Python3 program for the above approach
 
# Function to print the array
# element which differs in parity
# with the remaining array elements
def OddOneOut(arr, N) :
 
    # Multimaps to store even
    # and odd numbers along
    # with their indices
    e, o = {}, {}
 
    # Traverse the array
    for i in range(N) :
 
        # If array element is even
        if (arr[i] % 2 == 0) :
            e[arr[i]] = i
 
        # Otherwise
        else :
            o[arr[i]] = i
 
    # If only one even element
    # is present in the array
    if (len(e) == 1) :
        print(list(e.values())[0] )
 
    # If only one odd element
    # is present in the array
    else :
        print(list(o.values())[0] )
 
# Given array
arr = [ 2, 4, 7, 8, 10 ]
 
# Size of the array
N = len(arr)
 
OddOneOut(arr, N)
 
# This code is contributed by divyesh072019.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Linq;
class GFG {
 
  // Function to print the array
  // element which differs in parity
  // with the remaining array elements
  static void OddOneOut(int[] arr, int N)
  {
 
    // Multimaps to store even
    // and odd numbers along
    // with their indices
    Dictionary e = new Dictionary();
    Dictionary o = new Dictionary();
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
      // If array element is even
      if (arr[i] % 2 == 0) {
        e[arr[i]] = i;
      }
 
      // Otherwise
      else {
        o[arr[i]] = i;
      }
    }
 
    // If only one even element
    // is present in the array
    if (e.Count == 1) {
      Console.Write(e.First().Value);
    }
 
    // If only one odd element
    // is present in the array
    else {
      Console.Write(o.First().Value);
    }
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
 
    // Given array
    int[] arr = { 2, 4, 7, 8, 10 };
 
    // Size of the array
    int N = arr.Length;
 
    OddOneOut(arr, N);
  }
}
 
// This code is contributed by chitranayal.

输出：

2

时间复杂度： O(N * logN)
辅助空间： O(N)

高效的方法：为了优化上述方法，我们的想法是在两个变量中保留一个偶数和奇数数组元素的计数，并检查哪个计数等于1 。请按照以下步骤解决问题：

1. 维护四个变量even，odd以保持数组中偶数和奇数数组元素的数量，并保留lastOdd，lastEven来存储遇到的最后一个奇数和偶数数组元素的索引。
2. 遍历数组后，如果发现奇数为1 ，则打印lastOdd 。
3. 否则，打印lastEven 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to print the element
// which differs in parity
int oddOneOut(int arr[], int N)
{
    // Stores the count of odd and
    // even array elements encountered
    int odd = 0, even = 0;
 
    // Stores the indices of the last
    // odd and even array elements encountered
    int lastOdd = 0, lastEven = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // If array element is even
        if (arr[i] % 2 == 0) {
            even++;
            lastEven = i;
        }
 
        // Otherwise
        else {
            odd++;
            lastOdd = i;
        }
    }
 
    // If only one odd element
    // is present in the array
    if (odd == 1) {
        cout << lastOdd << endl;
    }
 
    // If only one even element
    // is present in the array
    else {
        cout << lastEven << endl;
    }
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 4, 7, 8, 10 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    oddOneOut(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
  
// Function to print the element
// which differs in parity
static void oddOneOut(int arr[], int N)
{
   
    // Stores the count of odd and
    // even array elements encountered
    int odd = 0, even = 0;
 
    // Stores the indices of the last
    // odd and even array elements encountered
    int lastOdd = 0, lastEven = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // If array element is even
        if (arr[i] % 2 == 0) {
            even++;
            lastEven = i;
        }
 
        // Otherwise
        else {
            odd++;
            lastOdd = i;
        }
    }
 
    // If only one odd element
    // is present in the array
    if (odd == 1) {
        System.out.println(lastOdd);
    }
 
    // If only one even element
    // is present in the array
    else {
       System.out.println(lastEven);
    }
}
   
// Driver Code
public static void main(String args[])
{
   
    // Given array
    int arr[] = { 2, 4, 7, 8, 10 };
 
    // Size of the array
    int N = arr.length;
    oddOneOut(arr, N);
}
}
 
// This code is contributed by jana_sayantan.

Python3

# Python program for the above approach
 
# Function to prthe element
# which differs in parity
def oddOneOut(arr, N) :
     
    # Stores the count of odd and
    # even array elements encountered
    odd = 0
    even = 0
 
    # Stores the indices of the last
    # odd and even array elements encountered
    lastOdd = 0
    lastEven = 0
 
    # Traverse the array
    for i in range(N):
 
        # If array element is even
        if (arr[i] % 2 == 0) :
            even += 1
            lastEven = i
         
        # Otherwise
        else :
            odd += 1
            lastOdd = i
         
    # If only one odd element
    # is present in the array
    if (odd == 1) :
        print(lastOdd)
     
    # If only one even element
    # is present in the array
    else :
        print(lastEven)
     
# Driver Code
 
# Given array
arr = [ 2, 4, 7, 8, 10 ]
 
# Size of the array
N = len(arr)
 
oddOneOut(arr, N)
 
# This code is contributed by susmitakundugoaldanga.

C＃

// C# program for the above approach
using System;
class GFG
{
     
    // Function to print the element
    // which differs in parity
    static void oddOneOut(int[] arr, int N)
    {
       
        // Stores the count of odd and
        // even array elements encountered
        int odd = 0, even = 0;
      
        // Stores the indices of the last
        // odd and even array elements encountered
        int lastOdd = 0, lastEven = 0;
      
        // Traverse the array
        for (int i = 0; i < N; i++)
        {
      
            // If array element is even
            if (arr[i] % 2 == 0)
            {
                even++;
                lastEven = i;
            }
      
            // Otherwise
            else
            {
                odd++;
                lastOdd = i;
            }
        }
      
        // If only one odd element
        // is present in the array
        if (odd == 1)
        {
            Console.WriteLine(lastOdd);
        }
      
        // If only one even element
        // is present in the array
        else
        {
            Console.WriteLine(lastEven);
        }
    }
 
  // Driver code
  static void Main()
  {
     
    // Given array
    int[] arr = { 2, 4, 7, 8, 10 };
  
    // Size of the array
    int N = arr.Length;
    oddOneOut(arr, N);
  }
}
 
// This code is contributed by divyeshrabadiya07.

输出：

2

时间复杂度： O(N)
辅助空间： O(1)