Smallest element repeated exactly ‘k’ times (not limited to small range)

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++

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// 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;
}

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Java

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// 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.

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Python3

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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 

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C#

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// 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

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Output:

1

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

Related Article:
Smallest number repeating k times

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Improved By : shrikanth13, Rajput-Ji



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