数组中最少的元素
原文： https://www.geeksforgeeks.org/least-frequent-element-array/
给定一个数组，在其中找到最不频繁的元素。 如果有多个元素出现的次数最少，请打印其中的任何一个。
示例：
Input : arr[] = {1, 3, 2, 1, 2, 2, 3, 1}
Output : 3
3 appears minimum number of times in given
array.
Input : arr[] = {10, 20, 30}
Output : 10 or 20 or 30
一个简单的解决方案是运行两个循环。 外循环一一挑选所有元素。 内循环找到所拾取元素的频率，并与到目前为止的最小值进行比较。 该解决方案的时间复杂度为O(n^2)。
更好的解决方案是进行排序。 我们首先对数组进行排序，然后线性遍历数组。
C++
// CPP program to find the least frequent element
// in an array.
#include <bits/stdc++.h>
using namespace std;
int leastFrequent(int arr[], int n)
{
// Sort the array
sort(arr, arr + n);
// find the min frequency using linear traversal
int min_count = n+1, res = -1, curr_count = 1;
for (int i = 1; i < n; i++) {
if (arr[i] == arr[i - 1])
curr_count++;
else {
if (curr_count < min_count) {
min_count = curr_count;
res = arr[i - 1];
}
curr_count = 1;
}
}
// If last element is least frequent
if (curr_count < min_count)
{
min_count = curr_count;
res = arr[n - 1];
}
return res;
}
// driver program
int main()
{
int arr[] = {1, 3, 2, 1, 2, 2, 3, 1};
int n = sizeof(arr) / sizeof(arr[0]);
cout << leastFrequent(arr, n);
return 0;
}
Java
// Java program to find the least frequent element
// in an array.
import java.io.*;
import java.util.*;
class GFG {
static int leastFrequent(int arr[], int n)
{
// Sort the array
Arrays.sort(arr);
// find the min frequency using
// linear traversal
int min_count = n+1, res = -1;
int curr_count = 1;
for (int i = 1; i < n; i++) {
if (arr[i] == arr[i - 1])
curr_count++;
else {
if (curr_count < min_count) {
min_count = curr_count;
res = arr[i - 1];
}
curr_count = 1;
}
}
// If last element is least frequent
if (curr_count < min_count)
{
min_count = curr_count;
res = arr[n - 1];
}
return res;
}
// driver program
public static void main(String args[])
{
int arr[] = {1, 3, 2, 1, 2, 2, 3, 1};
int n = arr.length;
System.out.print(leastFrequent(arr, n));
}
}
/*This code is contributed by Nikita Tiwari.*/
Python3
# Python 3 program to find the least
# frequent element in an array.
def leastFrequent(arr, n) :
# Sort the array
arr.sort()
# find the min frequency using
# linear traversal
min_count = n + 1
res = -1
curr_count = 1
for i in range(1, n) :
if (arr[i] == arr[i - 1]) :
curr_count = curr_count + 1
else :
if (curr_count < min_count) :
min_count = curr_count
res = arr[i - 1]
curr_count = 1
# If last element is least frequent
if (curr_count < min_count) :
min_count = curr_count
res = arr[n - 1]
return res
# Driver program
arr = [1, 3, 2, 1, 2, 2, 3, 1]
n = len(arr)
print(leastFrequent(arr, n))
# This code is contributed
# by Nikita Tiwari.
C#
// C# program to find the least
// frequent element in an array.
using System;
class GFG {
static int leastFrequent(int[] arr, int n)
{
// Sort the array
Array.Sort(arr);
// find the min frequency
// using linear traversal
int min_count = n + 1, res = -1;
int curr_count = 1;
for (int i = 1; i < n; i++)
{
if (arr[i] == arr[i - 1])
curr_count++;
else
{
if (curr_count < min_count)
{
min_count = curr_count;
res = arr[i - 1];
}
curr_count = 1;
}
}
// If last element is least frequent
if (curr_count < min_count)
{
min_count = curr_count;
res = arr[n - 1];
}
return res;
}
// Driver code
static public void Main ()
{
int[] arr = {1, 3, 2, 1, 2, 2, 3, 1};
int n = arr.Length;
// Function calling
Console.Write(leastFrequent(arr, n));
}
}
// This code is contributed by Shrikant13
PHP
<?php
// PHP program to find the
// least frequent element
// in an array.
function leastFrequent($arr, $n)
{
// Sort the array
sort($arr);
sort($arr , $n);
// find the min frequency
// using linear traversal
$min_count = $n + 1;
$res = -1;
$curr_count = 1;
for($i = 1; $i < $n; $i++)
{
if ($arr[$i] == $arr[$i - 1])
$curr_count++;
else
{
if ($curr_count < $min_count)
{
$min_count = $curr_count;
$res = $arr[$i - 1];
}
$curr_count = 1;
}
}
// If last element is
// least frequent
if ($curr_count < $min_count)
{
$min_count = $curr_count;
$res = $arr[$n - 1];
}
return $res;
}
// Driver Code
{
$arr = array(1, 3, 2, 1, 2, 2, 3, 1);
$n = sizeof($arr) / sizeof($arr[0]);
echo leastFrequent($arr, $n);
return 0;
}
// This code is contributed by nitin mittal
?>
输出：
3
时间复杂度：O(N log N)。
辅助空间：O(1)。
有效解决方案是使用哈希。 我们创建一个哈希表，并将元素及其频率计数存储为键值对。 最后，我们遍历哈希表并打印具有最小值的键。
C++
// CPP program to find the least frequent element
// in an array.
#include <bits/stdc++.h>
using namespace std;
int leastFrequent(int arr[], int n)
{
// Insert all elements in hash.
unordered_map<int, int> hash;
for (int i = 0; i < n; i++)
hash[arr[i]]++;
// find the min frequency
int min_count = n+1, res = -1;
for (auto i : hash) {
if (min_count >= i.second) {
res = i.first;
min_count = i.second;
}
}
return res;
}
// driver program
int main()
{
int arr[] = {1, 3, 2, 1, 2, 2, 3, 1};
int n = sizeof(arr) / sizeof(arr[0]);
cout << leastFrequent(arr, n);
return 0;
}
Java
//Java program to find the least frequent element
//in an array
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
class GFG {
static int leastFrequent(int arr[],int n)
{
// Insert all elements in hash.
Map<Integer,Integer> count =
new HashMap<Integer,Integer>();
for(int i = 0; i < n; i++)
{
int key = arr[i];
if(count.containsKey(key))
{
int freq = count.get(key);
freq++;
count.put(key,freq);
}
else
count.put(key,1);
}
// find min frequency.
int min_count = n+1, res = -1;
for(Entry<Integer,Integer> val : count.entrySet())
{
if (min_count >= val.getValue())
{
res = val.getKey();
min_count = val.getValue();
}
}
return res;
}
// driver program
public static void main (String[] args) {
int arr[] = {1, 3, 2, 1, 2, 2, 3, 1};
int n = arr.length;
System.out.println(leastFrequent(arr,n));
}
}
// This code is contributed by Akash Singh.
Python3
# Python3 program to find the most
# frequent element in an array.
import math as mt
def leastFrequent(arr, n):
# Insert all elements in Hash.
Hash = dict()
for i in range(n):
if arr[i] in Hash.keys():
Hash[arr[i]] += 1
else:
Hash[arr[i]] = 1
# find the max frequency
min_count = n + 1
res = -1
for i in Hash:
if (min_count >= Hash[i]):
res = i
min_count = Hash[i]
return res
# Driver Code
arr = [1, 3, 2, 1, 2, 2, 3, 1]
n = len(arr)
print(leastFrequent(arr, n))
# This code is contributed by
# mohit kumar 29
C
// C# program to find the
// least frequent element
// in an array.
using System;
using System.Collections.Generic;
class GFG
{
static int leastFrequent(int []arr,
int n)
{
// Insert all elements in hash.
Dictionary<int, int> count =
new Dictionary<int,
int>();
for (int i = 0; i < n; i++)
{
int key = arr[i];
if(count.ContainsKey(key))
{
int freq = count[key];
freq++;
count[key] = freq;
}
else
count.Add(key, 1);
}
// find the min frequency
int min_count = n + 1, res = -1;
foreach (KeyValuePair<int,
int> pair in count)
{
if (min_count >= pair.Value)
{
res = pair.Key;
min_count = pair.Value;
}
}
return res;
}
// Driver Code
static void Main()
{
int []arr = new int[]{1, 3, 2, 1,
2, 2, 3, 1};
int n = arr.Length;
Console.Write(leastFrequent(arr, n));
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)
输出：
3
时间复杂度：O(n)。
辅助空间：O(n)。
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