在具有重复项的已排序数组中查找相等（或中间）点

给定一个 n 大小的排序数组，任务是查找数组中是否存在一个元素，其中较小元素的数量等于较大元素的数量。
如果 Equal Point 在输入数组中出现多次，则返回其第一次出现的索引。如果不存在，则返回-1。
例子 ：

Input  : arr[] = {1, 2, 3, 3, 3, 3}
Output : 1
Equal Point is arr[1] which is 2. Number of
elements smaller than 2 and greater than 2 
are same.

Input  : arr[] = {1, 2, 3, 3, 3, 3, 4, 4}
Output : Equal Point does not exist.

Input : arr[] = {1, 2, 3, 4, 4, 5, 6, 6, 6, 7}
Output : 3
First occurrence of equal point is arr[3]

一种朴素的方法是获取每个元素并计算有多少元素小于那个元素，然后计算更大的元素。然后比较两者是否相等。
一种有效的方法是创建一个辅助数组并将所有不同的元素存储在其中。如果不同元素的计数是偶数，则不存在 Equal Point。如果 count 是奇数，那么相等点就是辅助数组的中点。
下面是上述想法的实现。

C++

// C++ program to find Equal point in a sorted array
// which may have many duplicates.
#include 
using namespace std;
 
// Returns value of Equal point in a sorted array arr[]
// It returns -1 if there is no Equal Point.
int findEqualPoint(int arr[], int n)
{
     // To store first indexes of distinct elements of arr[]
     int distArr[n];
 
     // Traverse input array and store indexes of first
     // occurrences of distinct elements in distArr[]
     int i = 0, di = 0;
     while (i < n)
     {
        // This element must be first occurrence of a
        // number (this is made sure by below loop),
        // so add it to distinct array.
        distArr[di++] = i++;
 
        // Avoid all copies of arr[i] and move to next
        // distinct element.
        while (i>1] : -1;
}
 
// Driver code
int main()
{
    int arr[] = {1, 2, 3, 4, 4, 5, 6, 6, 6, 7};
    int n = sizeof(arr)/sizeof(arr[0]);
    int index = findEqualPoint(arr, n);
    if (index != -1)
        cout << "Equal Point = " << arr[index] ;
    else
        cout << "Equal Point does not exists";
 
    return 0;
}

Java

//Java program to find Equal point in a sorted array
// which may have many duplicates.
 
class Test
{
    // Returns value of Equal point in a sorted array arr[]
    // It returns -1 if there is no Equal Point.
    static int findEqualPoint(int arr[], int n)
    {
         // To store first indexes of distinct elements of arr[]
         int distArr[] = new int[n];
      
         // Traverse input array and store indexes of first
         // occurrences of distinct elements in distArr[]
         int i = 0, di = 0;
         while (i < n)
         {
            // This element must be first occurrence of a
            // number (this is made sure by below loop),
            // so add it to distinct array.
            distArr[di++] = i++;
      
            // Avoid all copies of arr[i] and move to next
            // distinct element.
            while (i>1] : -1;
    }
 
    // Driver method
    public static void main(String args[])
    {
        int arr[] = {1, 2, 3, 4, 4, 5, 6, 6, 6, 7};
         
        int index = findEqualPoint(arr, arr.length);
        System.out.println(index != -1 ? "Equal Point = " + arr[index]
                            : "Equal Point does not exists");
    }
}

Python 3

# Python 3 program to find
# Equal point in a sorted
# array which may have
# many duplicates.
 
# Returns value of Equal
# point in a sorted array
# arr[]. It returns -1 if
# there is no Equal Point.
def findEqualPoint(arr, n):
 
    # To store first indexes of
    # distinct elements of arr[]
    distArr = [0] * n
 
    # Traverse input array and
    # store indexes of first
    # occurrences of distinct
    # elements in distArr[]
    i = 0
    di = 0
    while (i < n):
     
        # This element must be
        # first occurrence of a
        # number (this is made
        # sure by below loop),
        # so add it to distinct array.
        distArr[di] = i
        di += 1
        i += 1
 
        # Avoid all copies of
        # arr[i] and move to
        # next distinct element.
        while (i < n and
               arr[i] == arr[i - 1]):
            i += 1
     
    # di now has total number
    # of distinct elements.
    # If di is odd, then equal
    # point exists and is at
    # di/2, otherwise return -1.
    return distArr[di >> 1] if (di & 1) else -1
 
# Driver code
arr = [1, 2, 3, 4, 4,
       5, 6, 6, 6, 7]
n = len(arr)
index = findEqualPoint(arr, n)
if (index != -1):
    print("Equal Point = " ,
                 arr[index])
else:
    print("Equal Point does " +
                  "not exists")
 
# This code is contributed
# by Smitha

C#

// C# program to find Equal
// point in a sorted array
// which may have many duplicates.
using System;
 
class GFG
{
    // Returns value of Equal point
    // in a sorted array arr[]
    // It returns -1 if there
    // is no Equal Point.
    static int findEqualPoint(int []arr,
                              int n)
    {
        // To store first indexes of
        // distinct elements of arr[]
        int []distArr = new int[n];
     
        // Traverse input array and
        // store indexes of first
        // occurrences of distinct
        // elements in distArr[]
        int i = 0, di = 0;
        while (i < n)
        {
            // This element must be
            // first occurrence of a
            // number (this is made
            // sure by below loop),
            // so add it to distinct array.
            distArr[di++] = i++;
     
            // Avoid all copies of
            // arr[i] and move to
            // next distinct element.
            while (i < n && arr[i] == arr[i - 1])
                i++;
        }
     
        // di now has total number
        // of distinct elements.
        // If di is odd, then equal
        // point exists and is at
        // di/2, otherwise return -1.
        return (di & 1) != 0 ?
            distArr[di >> 1] :
                           -1;
    }
 
    // Driver Code
    public static void Main()
    {
        int []arr = {1, 2, 3, 4, 4,
                     5, 6, 6, 6, 7};
         
        int index = findEqualPoint(arr, arr.Length);
        Console.Write(index != -1 ?
                      "Equal Point = " + arr[index] :
                      "Equal Point does not exists");
    }
}

PHP

>1] : -1;
}
 
// Driver code
$arr = array(1, 2, 3, 4, 4, 5, 6, 6, 6, 7);
$n = count($arr);
$index = findEqualPoint($arr, $n);
if ($index != -1)
    echo "Equal Point = " , $arr[$index] ;
else
    echo "Equal Point does not exists";
 
// This code is contributed by anuj_67.
?>

Javascript

 
    // JavaScript program to find Equal
    // point in a sorted array
    // which may have many duplicates.
     
    // Returns value of Equal point
    // in a sorted array arr[]
    // It returns -1 if there
    // is no Equal Point.
    function findEqualPoint(arr, n)
    {
        // To store first indexes of
        // distinct elements of arr[]
        let distArr = new Array(n);
        distArr.fill(0);
       
        // Traverse input array and
        // store indexes of first
        // occurrences of distinct
        // elements in distArr[]
        let i = 0, di = 0;
        while (i < n)
        {
            // This element must be
            // first occurrence of a
            // number (this is made
            // sure by below loop),
            // so add it to distinct array.
            distArr[di++] = i++;
       
            // Avoid all copies of
            // arr[i] and move to
            // next distinct element.
            while (i < n && arr[i] == arr[i - 1])
                i++;
        }
       
        // di now has total number
        // of distinct elements.
        // If di is odd, then equal
        // point exists and is at
        // di/2, otherwise return -1.
        return (di & 1) != 0 ?
            distArr[di >> 1] :
                           -1;
    }
     
    let arr = [1, 2, 3, 4, 4, 5, 6, 6, 6, 7];
           
    let index = findEqualPoint(arr, arr.length);
    document.write(index != -1 ?
                  "Equal Point = " + arr[index] :
                  "Equal Point does not exists");
     

输出：

Equal Point = 4

时间复杂度： O(n)
辅助空间： O(n)
空间优化：
我们可以通过遍历数组两次而不是一次来减少额外的空间。

1. 通过遍历输入数组来计算不同元素的总数。让这个计数为 distCount。
2. 如果 distCount 是偶数，则返回 -1。
3. 如果 distCount 为奇数，则再次遍历数组并在 distCount/2 处停止并返回此索引。