Given an array, find the least frequent element in it. If there are multiple elements that appear least number of times, print any one of them.
Examples :Â
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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
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A simple solution is to run two loops. The outer loop picks all elements one by one. The inner loop finds frequency of the picked element and compares with the minimum so far. Time complexity of this solution is O(n2)
A better solution is to do sorting. We first sort the array, then linearly traverse the array.
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Java
import java.io.*;
import java.util.*;
class GFG {
static int leastFrequent( int arr[], int n)
{
Arrays.sort(arr);
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 (curr_count < min_count)
{
min_count = curr_count;
res = arr[n - 1 ];
}
return res;
}
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));
}
}
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Time Complexity : O(n Log n)Â
Auxiliary Space : O(1)
An efficient solution is to use hashing. We create a hash table and store elements and their frequency counts as key value pairs. Finally we traverse the hash table and print the key with minimum value.
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Java
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
class GFG {
static int leastFrequent( int arr[], int n)
{
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 );
}
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;
}
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));
}
}
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Time Complexity : O(n)Â
Auxiliary Space : O(n)
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Please refer complete article on Least frequent element in an array for more details!