# 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 :

```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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

 `// CPP program to find the least frequent element ` `// in an array. ` `#include ` `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); ` `    ``cout << leastFrequent(arr, n); ` `    ``return` `0; ` `} `

 `// 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.*/`

 `# 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# 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 `

 ` `

Output:
```3
```

Time Complexity : O(n Log n)
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.

 `// CPP program to find the least frequent element ` `// in an array. ` `#include ` `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); ` `    ``cout << leastFrequent(arr, n); ` `    ``return` `0; ` `} `

 `//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. `

 `# 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# 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) `

Output:
```3
```

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

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