# Java Program for Least frequent element in an array

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

 `// 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.*/`

Output:Â

`3`

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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

 `//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 count =  ` `                   ``new` `HashMap(); ` `                    `  `        ``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 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. `

Output:Â

`3`

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Time Complexity : O(n)Â
Auxiliary Space : O(n)
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