给定一个N个整数的数组arr [] 。任务是创建给定数组arr []的频率数组freq []并找到频率数组的最大元素。如果数组freq []中两个元素的频率相同，则返回值较小的元素。
例子：
Input: arr[] = {1, 1, 1, 2, 3, 2, 2, 3, 5, 5, 5, 5, 4, 4, 4, 4, 4};
Output: 3
Explanation:
frequency of elements is given by:
val -> freq[]
1 -> 3
2 -> 3
3 -> 2
4 -> 5
5 -> 4
Element 3 has the maximum frequency in the frequency array.
Input: arr[] = { 3, 5, 15, 51, 15, 14, 14, 14, 14, 4};
Output: 1
Explanation:
frequency of elements is given by:
val -> freq[]
3 -> 1
4 -> 1
5 -> 1
14 -> 4
15 -> 2
51 -> 1
Element 1 has the maximum frequency in the frequency array.
方法：
- 将arr []的元素的频率存储在一个映射图map1中，其中arr []的元素作为键，其频率作为值。
- 现在，将map1元素的频率存储在另一个映射中，例如map2 。
- 遍历map2以获取最高元素。
- 如果存在多个最高元素，则打印出具有较低值的元素。
下面是上述方法的实现：
C++14
// C++14 program for the above approach
#include
using namespace std;
// Function to get the highest
// frequency of frequency array
int findElement(int a[], int n)
{
// To find the maximum frequency
// initialize it with INT_MIN
int mx = INT_MIN;
int ans = 0;
// Initialize maps to store the
// count of element in array
map map1, map2;
// Store the frequency of
// element of array in map1
for (int i = 0; i < n; i++) {
map1[a[i]]++;
}
// Storing the frequency i.e.,
// (x.second) which is count of
// element in array
for (auto x : map1) {
map2[x.second]++;
}
for (auto x : map2) {
// Check if the fequency of
// element is greater than mx
if (x.second > mx) {
mx = x.second;
// Store the value to check
// when frequency is same
ans = x.first;
}
// If frequency of 2 element is
// same than storing minimum value
else if (x.second == mx) {
ans = min(ans, x.first);
}
}
// Return the highest frequency
return ans;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 1, 1, 1, 2, 3, 2, 2, 3, 5,
5, 5, 5, 4, 4, 4, 4, 4 };
// Size of the array
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << findElement(arr, n) << endl;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to get the highest
// frequency of frequency array
static int findElement(int a[], int n)
{
// To find the maximum frequency
// initialize it with INT_MIN
int mx = Integer.MIN_VALUE;
int ans = 0;
// Initialize maps to store the
// count of element in array
Map map1 = new HashMap<>(),
map2 = new HashMap<>();
// Store the frequency of
// element of array in map1
for(int i = 0; i < n; i++)
{
map1.put(a[i], map1.getOrDefault(a[i], 0) + 1);
}
// Storing the frequency i.e.,
// (x.second) which is count of
// element in array
for(Integer x : map1.values())
{
map2.put(x, map2.getOrDefault(x, 0) + 1);
}
for(Map.Entry x : map2.entrySet())
{
// Check if the fequency of
// element is greater than mx
if (x.getValue() > mx)
{
mx = x.getValue();
// Store the value to check
// when frequency is same
ans = x.getKey();
}
// If frequency of 2 element is
// same than storing minimum value
else if (x.getValue() == mx)
{
ans = Math.min(ans, x.getKey());
}
}
// Return the highest frequency
return ans;
}
// Driver code
public static void main (String[] args)
{
// Given array arr[]
int arr[] = { 1, 1, 1, 2, 3, 2, 2, 3, 5,
5, 5, 5, 4, 4, 4, 4, 4 };
// Size of the array
int n = arr.length;
// Function call
System.out.println(findElement(arr, n));
}
}
// This code is contributed by offbeat Python3
# Python3 program for the above approach
import sys
# Function to get the highest
# frequency of frequency array
def findElement(a, n):
# To find the maximum frequency
# initialize it with INT_MIN
mx = -sys.maxsize - 1
ans = 0
# Initialize maps to store the
# count of element in array
map1 = {}
map2 = {}
# Store the frequency of
# element of array in map1
for i in a:
map1[i] = map1.get(i, 0) + 1
# Storing the frequency i.e.,
# (x.second) which is count of
# element in array
for x in map1:
map2[map1[x]] = map2.get(map1[x], 0) + 1
for x in map2:
# Check if the fequency of
# element is greater than mx
if (map2[x] > mx):
mx = map2[x]
# Store the value to check
# when frequency is same
ans = x
# If frequency of 2 element is
# same than storing minimum value
elif (map2[x] == mx):
ans = min(ans, x)
# Return the highest frequency
return ans
# Driver Code
if __name__ == '__main__':
# Given array arr[]
arr = [ 1, 1, 1, 2, 3, 2, 2, 3,
5, 5, 5, 5, 4, 4, 4, 4, 4]
# Size of the array
n = len(arr)
# Function call
print(findElement(arr, n))
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to get the highest
// frequency of frequency array
static int findElement(int[] a, int n)
{
// To find the maximum frequency
// initialize it with INT_MIN
int mx = Int32.MinValue;
int ans = 0;
// Initialize maps to store the
// count of element in array
Dictionary map1 = new Dictionary(),
map2 = new Dictionary();
// Store the frequency of
// element of array in map1
for(int i = 0; i < n; i++)
{
if (map1.ContainsKey(a[i]))
map1[a[i]] = map1[a[i]] + 1;
else
map1[a[i]] = 1;
}
// Storing the frequency i.e.,
// (x.second) which is count of
// element in array
foreach(KeyValuePair xx in map1)
{
int x = xx.Value;
if (map2.ContainsKey(x))
map2[x] = map2[x] + 1;
else
map2[x] = 1;
}
foreach(KeyValuePair x in map2)
{
// Check if the fequency of
// element is greater than mx
if (x.Value > mx)
{
mx = x.Value;
// Store the value to check
// when frequency is same
ans = x.Key;
}
// If frequency of 2 element is
// same than storing minimum value
else if (x.Value == mx)
{
ans = Math.Min(ans, x.Key);
}
}
// Return the highest frequency
return ans;
}
// Driver code
static public void Main ()
{
// Given array arr[]
int[] arr = { 1, 1, 1, 2, 3, 2, 2, 3, 5,
5, 5, 5, 4, 4, 4, 4, 4 };
// Size of the array
int n = arr.Length;
// Function call
Console.WriteLine(findElement(arr, n));
}
}
// This code is contributed by offbeat 输出：
3
时间复杂度： O(N)
辅助空间： O(N)