Given an array of size n, the goal is to find out the smallest number that is repeated exactly ‘k’ times where k > 0?

And

Examples:

Input : a[] = {2, 1, 3, 1, 2, 2} k = 3 Output : 2 Input : a[] = {3, 4, 3, 2, 1, 5, 5} k = 2 Output : 3 Explanation: As 3 is smaller than 5. So 3 should be printed.

We have discussed different solutions of this problem in below post.

Smallest element in an array that is repeated exactly ‘k’ times

The solutions discussed above are either limited to small range work in more than linear time. In this post a hashing based solution is discussed that works in O(n) time and is applicable to any range. Below are abstract steps.

1) Create a hash map that stores elements and their frequencies.

2) Traverse given array. For every element being traversed, increment its frequency.

3) Traverse hash map and print the smallest element with frequency k.

## C++

`// C++ program to find the smallest element` `// with frequency exactly k.` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `int` `smallestKFreq(` `int` `a[], ` `int` `n, ` `int` `k)` `{` ` ` `unordered_map<` `int` `, ` `int` `> m;` ` ` ` ` `// Map is used to store the count of` ` ` `// elements present in the array` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `m[a[i]]++;` ` ` ` ` `// Traverse the map and find minimum` ` ` `// element with frequency k.` ` ` `int` `res = INT_MAX;` ` ` `for` `(` `auto` `it = m.begin(); it != m.end(); ++it)` ` ` `if` `(it->second == k)` ` ` `res = min(res, it->first);` ` ` ` ` `return` `(res != INT_MAX)? res : -1;` `}` ` ` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 2, 2, 1, 3, 1 };` ` ` `int` `k = 2;` ` ` `int` `n = ` `sizeof` `(arr) / (` `sizeof` `(arr[0]));` ` ` `cout << smallestKFreq(arr, n, k);` ` ` `return` `0;` `}` |

## Java

`// Java program to find the smallest element` `// with frequency exactly k.` `import` `java.util.*;` ` ` `class` `GFG {` ` ` ` ` `public` `static` `int` `smallestKFreq(` `int` `a[], ` `int` `n, ` `int` `k)` ` ` `{` ` ` `HashMap<Integer, Integer> m = ` `new` `HashMap<Integer, Integer>();` ` ` ` ` `// Map is used to store the count of` ` ` `// elements present in the array` ` ` `for` `(` `int` `i = ` `0` `; i < n; i ++)` ` ` ` ` `if` `(m.containsKey(a[i]))` ` ` `m.put(a[i], m.get(a[i]) + ` `1` `);` ` ` ` ` `else` `m.put(a[i], ` `1` `);` ` ` ` ` `// Traverse the map and find minimum` ` ` `// element with frequency k.` ` ` `int` `res = Integer.MAX_VALUE;` ` ` `Set<Integer> s = m.keySet();` ` ` ` ` `for` `(` `int` `temp : s)` ` ` `if` `(m.get(temp) == k)` ` ` `res = Math.min(res, temp);` ` ` ` ` `return` `(res != Integer.MAX_VALUE)? res : -` `1` `;` ` ` `}` ` ` ` ` `/* Driver program to test above function */` ` ` `public` `static` `void` `main(String[] args) ` ` ` `{` ` ` `int` `arr[] = { ` `2` `, ` `2` `, ` `1` `, ` `3` `, ` `1` `};` ` ` `int` `k = ` `2` `;` ` ` ` ` `System.out.println(smallestKFreq(arr, arr.length, k));` ` ` ` ` `}` ` ` `}` `// This code is contributed by Arnav Kr. Mandal.` |

## Python3

`from` `collections ` `import` `defaultdict` `import` `sys` ` ` `# Python program to find the smallest element ` `# with frequency exactly k. ` `def` `smallestKFreq(arr, n, k):` ` ` `mp ` `=` `defaultdict(` `lambda` `: ` `0` `)` ` ` ` ` ` ` `# Map is used to store the count of ` ` ` `# elements present in the array ` ` ` `for` `i ` `in` `range` `(n):` ` ` `mp[arr[i]] ` `+` `=` `1` ` ` ` ` `# Traverse the map and find minimum ` ` ` `# element with frequency k. ` ` ` `res ` `=` `sys.maxsize` ` ` `res1 ` `=` `sys.maxsize` ` ` ` ` `for` `key,values ` `in` `mp.items():` ` ` ` ` `if` `values ` `=` `=` `k:` ` ` `res ` `=` `min` `(res, key)` ` ` `return` `res ` `if` `res !` `=` `res1 ` `else` `-` `1` ` ` `# Driver code` `arr ` `=` `[` `2` `, ` `2` `, ` `1` `, ` `3` `, ` `1` `]` `k ` `=` `2` `n ` `=` `len` `(arr)` `print` `(smallestKFreq(arr, n, k))` ` ` `# This code is contributed by Shrikant13 ` |

## C#

`// C# program to find the smallest element` `// with frequency exactly k.` `using` `System;` `using` `System.Linq;` `using` `System.Collections.Generic; ` ` ` `class` `GFG ` `{` ` ` ` ` `public` `static` `int` `smallestKFreq(` `int` `[]a, ` `int` `n, ` `int` `k)` ` ` `{` ` ` `Dictionary<` `int` `,` `int` `> m = ` `new` `Dictionary<` `int` `,` `int` `>();` ` ` ` ` `// Map is used to store the count of` ` ` `// elements present in the array` ` ` `for` `(` `int` `i = 0; i < n; i ++)` ` ` ` ` `if` `(m.ContainsKey(a[i]))` ` ` `{` ` ` `var` `v = m[a[i]];` ` ` `m.Remove(a[i]);` ` ` `m.Add(a[i],v + 1);` ` ` `}` ` ` `else` `m.Add(a[i], 1);` ` ` ` ` `// Traverse the map and find minimum` ` ` `// element with frequency k.` ` ` `int` `res = ` `int` `.MaxValue;` ` ` `HashSet<` `int` `> s = ` `new` `HashSet<` `int` `>(m.Keys.ToArray());` ` ` ` ` `foreach` `(` `int` `temp ` `in` `s)` ` ` `if` `(m[temp] == k)` ` ` `res = Math.Min(res, temp);` ` ` ` ` `return` `(res != ` `int` `.MaxValue)? res : -1;` ` ` `}` ` ` ` ` `/* Driver code */` ` ` `public` `static` `void` `Main(String[] args) ` ` ` `{` ` ` `int` `[]arr = { 2, 2, 1, 3, 1 };` ` ` `int` `k = 2;` ` ` ` ` `Console.WriteLine(smallestKFreq(arr, arr.Length, k));` ` ` ` ` `}` `}` ` ` `// This code is contributed by Rajput-Ji` |

**Output:**

1

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

**Related Article:**

Smallest number repeating k times

This article is contributed by **Gitanjali Sharma**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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