查找数组中元素的索引，该数组将其之前的大多数元素相除

给定一个数组arr ，任务是找到一个数组中元素的索引，该数组将它之前的大多数元素相除
例子：

Input: arr = {5, 2, 1, 4, 5, 8, 2}
Output: 6
Explanation
arr[6] = 2 
it divides 2, 4, and 8 (3 elements)

Input: arr = {8, 1, 28, 4, 1, 6, 7}
Output: 4

方法：

• 维护一张地图。
• 对于每个 arr[i]，通过查看 ar[i] 的映射来更新计数变量，并将 ar[i] 的所有除数插入到映射中。
• 如果 cnt > maxx 则更新 maxElement。
• 最后用 maxElement 返回索引。

以下是上述方法的实现：

CPP

// C++ program find the index of the element
// in an array which divides
// most elements before it
 
#include 
using namespace std;
 
// Function to get the max element
// divisible by arr[i]
int maxElement(int arr[], int n)
{
 
    map mp;
    int maxx = -1, maxElement = -1;
 
    for (int i = 0; i < n; i++) {
        int num = arr[i];
        int cnt = 0;
 
        // Update count for A[i]
        if (mp.find(num) != mp.end()) {
            cnt += mp[num];
        }
 
        // Generate Divisor For A[i]
        for (int j = 1; j * j <= num; j++) {
            if (num % j == 0) {
                mp[j]++;
                if (j != num / j)
                    mp[num / j]++;
            }
        }
 
        // Update Max Element
        if (cnt > maxx) {
            maxElement = i;
            maxx = cnt;
        }
    }
 
    return maxElement;
}
 
// Driver code
int main()
{
 
    int arr[] = { 5, 2, 1, 4, 5, 8, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << maxElement(arr, n) << '\n';
 
    return 0;
}

Java

// Java program find the index of the element
// in an array which divides
// most elements before it
import java.util.*;
 
class GFG
{
 
    // Function to get the max element
    // divisible by arr[i]
    static int maxElement(int arr[], int n)
    {
 
        HashMap mp = new HashMap();
        int maxx = -1, maxElement = -1;
 
        for (int i = 0; i < n; i++)
        {
            int num = arr[i];
            int cnt = 0;
 
            // Update count for A[i]
            if (mp.containsKey(num))
            {
                cnt += mp.get(num);
            }
 
            // Generate Divisor For A[i]
            for (int j = 1; j * j <= num; j++)
            {
                if (num % j == 0)
                {
                    if (mp.containsKey(j))
                        mp.put(j, mp.get(j) + 1);
                    else
                        mp.put(j, 1);
                    if (j != num / j)
                        if (mp.containsKey(num / j))
                            mp.put(num / j, mp.get(num / j) + 1);
                        else
                            mp.put(num / j, 1);
                }
            }
 
            // Update Max Element
            if (cnt > maxx)
            {
                maxElement = i;
                maxx = cnt;
            }
        }
 
        return maxElement;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        int arr[] = { 5, 2, 1, 4, 5, 8, 2 };
        int n = arr.length;
 
        System.out.print(maxElement(arr, n));
    }
}
 
// This code is contributed by 29AjayKumar

Python

# Python3 program find the index of the element
# in an array which divides
# most elements before it
 
# Function to get the max element
# divisible by arr[i]
def maxElement(arr, n):
 
    mp = dict()
    maxx = -1
    maxElement = -1
 
    for i in range(n):
        num = arr[i]
        cnt = 0
 
        # Update count for A[i]
        if (num in mp):
            cnt += mp[num]
 
        # Generate Divisor For A[i]
        j = 1
 
        while j * j <= num:
            if (num % j == 0):
                mp[j] = mp.get(j, 0) + 1
                if (j != num // j):
                    mp[num // j] = mp.get(num//j, 0) + 1
            j += 1
 
        # Update Max Element
        if (cnt > maxx):
            maxElement = i
            maxx = cnt
 
    return maxElement
 
# Driver code
arr = [5, 2, 1, 4, 5, 8, 2]
n = len(arr)
 
print(maxElement(arr, n))
 
# This code is contributed by mohit kumar 29

C#

// C# program find the index of the element
// in an array which divides
// most elements before it
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to get the max element
    // divisible by arr[i]
    static int maxElement(int []arr, int n)
    {
 
        Dictionary mp = new Dictionary();
        int maxx = -1, maxElement = -1;
 
        for (int i = 0; i < n; i++)
        {
            int num = arr[i];
            int cnt = 0;
 
            // Update count for A[i]
            if (mp.ContainsKey(num))
            {
                cnt += mp[num];
            }
 
            // Generate Divisor For A[i]
            for (int j = 1; j * j <= num; j++)
            {
                if (num % j == 0)
                {
                    if (mp.ContainsKey(j))
                        mp[j] = mp[j] + 1;
                    else
                        mp.Add(j, 1);
                    if (j != num / j)
                        if (mp.ContainsKey(num / j))
                            mp[num / j] = mp[num / j] + 1;
                        else
                            mp.Add(num / j, 1);
                }
            }
 
            // Update Max Element
            if (cnt > maxx)
            {
                maxElement = i;
                maxx = cnt;
            }
        }
 
        return maxElement;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
 
        int []arr = { 5, 2, 1, 4, 5, 8, 2 };
        int n = arr.Length;
 
        Console.Write(maxElement(arr, n));
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

      // JavaScript program find the index of the element
      // in an array which divides
      // most elements before it
      // Function to get the max element
      // divisible by arr[i]
      function maxElement(arr, n) {
        var mp = {};
        var maxx = -1,
          maxElement = -1;
 
        for (var i = 0; i < n; i++) {
          var num = arr[i];
          var cnt = 0;
 
          // Update count for A[i]
          if (mp.hasOwnProperty(num)) {
            cnt += mp[num];
          }
 
          // Generate Divisor For A[i]
          for (var j = 1; j * j <= num; j++) {
            if (num % j === 0) {
              if (mp.hasOwnProperty(j)) mp[j] = mp[j] + 1;
              else mp[j] = 1;
              if (j !== num / j)
                if (mp.hasOwnProperty(num / j)) mp[num / j] = mp[num / j] + 1;
                else mp[num / j] = 1;
            }
          }
 
          // Update Max Element
          if (cnt > maxx) {
            maxElement = i;
            maxx = cnt;
          }
        }
 
        return maxElement;
      }
 
      // Driver code
      var arr = [5, 2, 1, 4, 5, 8, 2];
      var n = arr.length;
 
      document.write(maxElement(arr, n));
       
      // This code is contributed by rdtank.
    

输出：

6

时间复杂度：O(N√max(Arr))