Smallest element repeated exactly ‘k’ times (not limited to small range)
Last Updated :
19 Jul, 2022
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.
- Create a hash map that stores elements and their frequencies.
- Traverse given array. For every element being traversed, increment its frequency.
- Traverse hash map and print the smallest element with frequency k.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
int smallestKFreq( int a[], int n, int k)
{
unordered_map< int , int > m;
for ( int i = 0; i < n; i++)
m[a[i]]++;
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;
}
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
import java.util.*;
class GFG {
public static int smallestKFreq( int a[], int n, int k)
{
HashMap<Integer, Integer> m = new HashMap<Integer, Integer>();
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 );
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 ;
}
public static void main(String[] args)
{
int arr[] = { 2 , 2 , 1 , 3 , 1 };
int k = 2 ;
System.out.println(smallestKFreq(arr, arr.length, k));
}
}
|
Python3
from collections import defaultdict
import sys
def smallestKFreq(arr, n, k):
mp = defaultdict( lambda : 0 )
for i in range (n):
mp[arr[i]] + = 1
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
arr = [ 2 , 2 , 1 , 3 , 1 ]
k = 2
n = len (arr)
print (smallestKFreq(arr, n, k))
|
C#
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 >();
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);
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;
}
public static void Main(String[] args)
{
int []arr = { 2, 2, 1, 3, 1 };
int k = 2;
Console.WriteLine(smallestKFreq(arr, arr.Length, k));
}
}
|
Javascript
<script>
function smallestKFreq(a, n, k) {
let m = new Map();
for (let i = 0; i < n; i++)
if (m.has(a[i]))
m.set(a[i], m.get(a[i]) + 1);
else m.set(a[i], 1);
let res = Number.MAX_SAFE_INTEGER;
let s = m.keys();
for (let temp of s)
if (m.get(temp) == k)
res = Math.min(res, temp);
return (res != Number.MAX_SAFE_INTEGER) ? res : -1;
}
let arr = [2, 2, 1, 3, 1];
let k = 2;
document.write(smallestKFreq(arr, arr.length, k));
</script>
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Time Complexity : O(n)
Auxiliary Space : O(n)
Related Article:
Smallest number repeating k times
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