在 O(n) 时间和 O(1) 额外空间中找到最大重复数

给定一个大小为n的数组，该数组包含范围从 0 到k-1的数字，其中k是一个正整数并且k <= n。查找此数组中的最大重复数。例如，设k为 10，给定数组arr[] = {1, 2, 2, 2, 0, 2, 0, 2, 3, 8, 0, 9, 2, 3}，最大重复次数为为 2. 预期时间复杂度为O(n)并且允许的额外空间为O(1) 。允许修改数组。

天真的方法是运行两个循环，外循环一个接一个地选取一个元素，内循环计算所选取元素的出现次数。最后返回具有最大计数的元素。这种方法的时间复杂度是O(n^2) 。
更好的方法是创建一个大小为 k 的 count 数组并将count[]的所有元素初始化为 0。遍历输入数组的所有元素，并为每个元素arr[i]递增count[arr[i]] 。最后，遍历count[]并返回最大值的索引。这种方法需要 O(n) 时间，但需要 O(k) 空间。
以下是O(n)时间和O(1)额外空间的方法。让我们通过一个简单的例子来理解这种方法，其中arr[] = {2, 3, 3, 5, 3, 4, 1, 7}, k = 8, n = 8（arr[] 中的元素数）。
1)遍历输入数组arr[] ，对于每个元素arr[i] ，将arr[arr[i]%k]增加k （ arr[]变为 {2, 11, 11, 29, 11, 12, 1, 15 })
2）在修改后的数组中找到最大值（最大值为29）。最大值的索引是最大重复元素（索引 29 是 3）。
3）如果我们想取回原始数组，我们可以再遍历一次数组并执行arr[i] = arr[i] % k其中i从 0 到n-1变化。
上述算法是如何工作的？由于我们使用arr[i]%k作为索引并在索引arr[i]%k处添加值k ，因此等于最大重复元素的索引最终将具有最大值。请注意，在等于最大重复元素的索引处添加k的最大次数，并且所有数组元素都小于k。
以下是上述算法的 C++ 实现。

C++

// C++ program to find the maximum repeating number
 
#include
using namespace std;
 
// Returns maximum repeating element in arr[0..n-1].
// The array elements are in range from 0 to k-1
int maxRepeating(int* arr, int n, int k)
{
    // Iterate though input array, for every element
    // arr[i], increment arr[arr[i]%k] by k
    for (int i = 0; i< n; i++)
        arr[arr[i]%k] += k;
 
    // Find index of the maximum repeating element
    int max = arr[0], result = 0;
    for (int i = 1; i < n; i++)
    {
        if (arr[i] > max)
        {
            max = arr[i];
            result = i;
        }
    }
 
    /* Uncomment this code to get the original array back
       for (int i = 0; i< n; i++)
          arr[i] = arr[i]%k; */
 
    // Return index of the maximum element
    return result;
}
 
// Driver program to test above function
int main()
{
    int arr[] = {2, 3, 3, 5, 3, 4, 1, 7};
    int n = sizeof(arr)/sizeof(arr[0]);
    int k = 8;
 
    cout << "The maximum repeating number is " <<
         maxRepeating(arr, n, k) << endl;
 
    return 0;
}

Java

// Java program to find the maximum repeating number
import java.io.*;
 
class MaxRepeating {
 
    // Returns maximum repeating element in arr[0..n-1].
    // The array elements are in range from 0 to k-1
    static int maxRepeating(int arr[], int n, int k)
    {
        // Iterate though input array, for every element
        // arr[i], increment arr[arr[i]%k] by k
        for (int i = 0; i< n; i++)
            arr[(arr[i]%k)] += k;
 
        // Find index of the maximum repeating element
        int max = arr[0], result = 0;
        for (int i = 1; i < n; i++)
        {
            if (arr[i] > max)
            {
                max = arr[i];
                result = i;
            }
        }
 
        /* Uncomment this code to get the original array back
        for (int i = 0; i< n; i++)
          arr[i] = arr[i]%k; */
 
        // Return index of the maximum element
        return result;
    }
 
    /*Driver function to check for above function*/
    public static void main (String[] args)
    {
 
        int arr[] = {2, 3, 3, 5, 3, 4, 1, 7};
        int n = arr.length;
        int k=8;
        System.out.println("Maximum repeating element is: " +
                           maxRepeating(arr,n,k));
    }
}
/* This code is contributed by Devesh Agrawal */

Python3

# Python program to find the maximum repeating number
 
# Returns maximum repeating element in arr[0..n-1].
# The array elements are in range from 0 to k-1
def maxRepeating(arr, n,  k):
 
    # Iterate though input array, for every element
    # arr[i], increment arr[arr[i]%k] by k
    for i in range(0,  n):
        arr[arr[i]%k] += k
 
    # Find index of the maximum repeating element
    max = arr[0]
    result = 0
    for i in range(1, n):
     
        if arr[i] > max:
            max = arr[i]
            result = i
 
    # Uncomment this code to get the original array back
    #for i in range(0, n):
    #    arr[i] = arr[i]%k
 
    # Return index of the maximum element
    return result
 
 
# Driver program to test above function
arr = [2, 3, 3, 5, 3, 4, 1, 7]
n = len(arr)
k = 8
print("The maximum repeating number is",maxRepeating(arr, n, k))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

//C# program to find the maximum repeating
// number
using System;
 
class GFG {
     
    // Returns maximum repeating element
    // in arr[0..n-1].
    // The array elements are in range
    // from 0 to k-1
    static int maxRepeating(int []arr,
                             int n, int k)
    {
        // Iterate though input array, for
        // every element arr[i], increment
        // arr[arr[i]%k] by k
        for (int i = 0; i< n; i++)
            arr[(arr[i]%k)] += k;
 
        // Find index of the maximum
        // repeating element
        int max = arr[0], result = 0;
         
        for (int i = 1; i < n; i++)
        {
            if (arr[i] > max)
            {
                max = arr[i];
                result = i;
            }
        }
 
        // Return index of the
        // maximum element
        return result;
    }
 
    /*Driver function to check for
     above function*/
    public static void Main ()
    {
 
        int []arr = {2, 3, 3, 5, 3, 4, 1, 7};
        int n = arr.Length;
        int k=8;
         
        Console.Write("Maximum repeating "
                          + "element is: "
                  + maxRepeating(arr,n,k));
    }
}
 
// This code is contributed by Sam007.

PHP

 $max)
        {
            $max = $arr[$i];
            $result = $i;
        }
    }
 
    /* Uncomment this code to
       get the original array back
       for (int i = 0; i< n; i++)
       arr[i] = arr[i] % k; */
 
    // Return index of the
    // maximum element
    return $result;
}
 
    // Driver Code
    $arr = array(2, 3, 3, 5, 3, 4, 1, 7);
    $n = sizeof($arr);
    $k = 8;
 
    echo "The maximum repeating number is ",
        maxRepeating($arr, $n, $k);
 
// This Code is contributed by Ajit
?>

Javascript

 
// JavaScript program to find the maximum repeating number
 
// Returns maximum repeating element in arr[0..n-1].
// The array elements are in range from 0 to k-1
function maxRepeating(arr, n, k)
{
    // Iterate though input array, for every element
    // arr[i], increment arr[arr[i]%k] by k
    for (let i = 0; i< n; i++)
        arr[arr[i]%k] += k;
 
    // Find index of the maximum repeating element
    let max = arr[0], result = 0;
    for (let i = 1; i < n; i++)
    {
        if (arr[i] > max)
        {
            max = arr[i];
            result = i;
        }
    }
 
    /* Uncomment this code to get the original array back
    for (int i = 0; i< n; i++)
        arr[i] = arr[i]%k; */
 
    // Return index of the maximum element
    return result;
}
 
// Driver program to test above function
 
    let arr = [2, 3, 3, 5, 3, 4, 1, 7];
    let n = arr.length;
    let k = 8;
 
    document.write("The maximum repeating number is " +
        maxRepeating(arr, n, k) + "<br>");
 
// This code is contributed by Surbhi Tyagi.
 

输出：

The maximum repeating number is 3