El elemento más pequeño de una array que se repite exactamente ‘k’ veces.

Dada una array de tamaño n, el objetivo es encontrar el número más pequeño que se repite exactamente ‘k’ veces donde k > 0? 
Suponga que la array solo tiene números enteros positivos y 1 <= arr[i] < 1000 para cada i = 0 a n -1.

Ejemplos: 

Input : arr[] = {2 2 1 3 1}
        k = 2
Output: 1
Explanation:
Here in array,
2 is repeated 2 times
1 is repeated 2 times 
3 is repeated 1 time
Hence 2 and 1 both are repeated 'k' times
i.e 2 and min(2, 1) is 1

Input : arr[] = {3 5 3 2}
        k = 1
Output : 2
Explanation:
Both 2 and 5 are repeating 1 time but
min(5, 2) is 2

Enfoque simple: un enfoque simple es usar dos bucles anidados. El bucle exterior elige un elemento uno por uno comenzando desde el elemento más a la izquierda. El bucle interno verifica si el mismo elemento está presente en el lado derecho. Si está presente, aumente el conteo y haga negativo el número que obtuvimos en el lado derecho para evitar que cuente nuevamente. 

Implementación:

// C++ program to find smallest number
// in array that is repeated exactly
// 'k' times.
#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 1000;
 
int findDuplicate(int arr[], int n, int k)
{
    // Since arr[] has numbers in range from
    // 1 to MAX
    int res = MAX + 1;
    for (int i = 0; i < n; i++) {
        if (arr[i] > 0) {
 
            // set count to 1 as number is present
            // once
            int count = 1;
            for (int j = i + 1; j < n; j++)
                if (arr[i] == arr[j])
                    count += 1;
 
            // If frequency of number is equal to 'k'
            if (count == k)
                res = min(res, arr[i]);
        }
    }
    return res;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 2, 1, 3, 1 };
    int k = 2;
    int n = sizeof(arr) / (sizeof(arr[0]));
    cout << findDuplicate(arr, n, k);
    return 0;
}

// Java program to find smallest number
// in array that is repeated exactly
// 'k' times.
public class GFG {
    static final int MAX = 1000;
 
    // finds the smallest number in arr[]
    // that is repeated k times
    static int findDuplicate(int arr[], int n, int k)
    {
        // Since arr[] has numbers in range from
        // 1 to MAX
        int res = MAX + 1;
        for (int i = 0; i < n; i++) {
            if (arr[i] > 0) {
 
                // set count to 1 as number is
                // present once
                int count = 1;
                for (int j = i + 1; j < n; j++)
                    if (arr[i] == arr[j])
                        count += 1;
 
                // If frequency of number is equal
                // to 'k'
                if (count == k)
                    res = Math.min(res, arr[i]);
            }
        }
        return res;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 2, 2, 1, 3, 1 };
        int k = 2;
        int n = arr.length;
        System.out.println(findDuplicate(arr, n, k));
    }
}
// This article is contributed by Sumit Ghosh

# Python 3 program to find smallest
# number in array that is repeated
# exactly 'k' times.
MAX = 1000
 
def findDuplicate(arr, n, k):
 
    # Since arr[] has numbers in
    # range from 1 to MAX
    res = MAX + 1
     
    for i in range(0, n):
        if (arr[i] > 0):
 
            # set count to 1 as number
            # is present once
            count = 1
            for j in range(i + 1, n):
                if (arr[i] == arr[j]):
                    count += 1
 
            # If frequency of number is equal to 'k'
            if (count == k):
                res = min(res, arr[i])
     
    return res
 
# Driver code
arr = [2, 2, 1, 3, 1]
k = 2
n = len(arr)
print(findDuplicate(arr, n, k))
 
# This code is contributed by Smitha Dinesh Semwal.

// C# program to find smallest number
// in array that is repeated exactly
// 'k' times.
using System;
 
public class GFG {
     
    static int MAX = 1000;
 
    // finds the smallest number in arr[]
    // that is repeated k times
    static int findDuplicate(int[] arr,
                              int n, int k)
    {
         
        // Since arr[] has numbers in range
        // from 1 to MAX
        int res = MAX + 1;
         
        for (int i = 0; i < n; i++)
        {
            if (arr[i] > 0)
            {
 
                // set count to 1 as number
                // is present once
                int count = 1;
                for (int j = i + 1; j < n; j++)
                    if (arr[i] == arr[j])
                        count += 1;
 
                // If frequency of number is
                // equal to 'k'
                if (count == k)
                    res = Math.Min(res, arr[i]);
            }
        }
         
        return res;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 2, 2, 1, 3, 1 };
        int k = 2;
        int n = arr.Length;
         
        Console.WriteLine(
                      findDuplicate(arr, n, k));
    }
}
 
// This article is contributed by vt_m.

<?php
// PHP program to find smallest number
// in array that is repeated exactly
// 'k' times.
function findDuplicate($arr, $n, $k)
{
    // Since arr[] has numbers in
    // range from 1 to MAX
    $MAX = 1000;
    $res = $MAX + 1;
    for ($i = 0; $i < $n; $i++)
    {
        if ($arr[$i] > 0)
        {
 
            // set count to 1 as number is
            // present once
            $count = 1;
            for ($j = $i + 1; $j < $n; $j++)
                if ($arr[$i] == $arr[$j])
                    $count += 1;
 
            // If frequency of number is
            // equal to 'k'
            if ($count == $k)
                $res = min($res, $arr[$i]);
        }
    }
    return $res;
}
 
// Driver code
$arr = array(2, 2, 1, 3, 1);
$k = 2;
$n = count($arr);
echo findDuplicate($arr, $n, $k);
 
// This code is contributed by Rajput-Ji
?>

<script>
 
// Javascript program to find smallest number
// in array that is repeated exactly
// 'k' times
 
    let MAX = 1000;
   
    // finds the smallest number in arr[]
    // that is repeated k times
    function findDuplicate(arr, n, k)
    {
        // Computing frequencies of all elements
        let freq = new Array(MAX).fill(0);
   
        for (let i = 0; i < n; i++) {
            if (arr[i] < 1 && arr[i] > MAX) {
                document.write("Out of range");
                return -1;
            }
            freq[arr[i]] += 1;
        }
   
        // Finding the smallest element with
        // frequency as k
        for (let i = 0; i < MAX; i++) {
   
            // If frequency of any of the number
            // is equal to k starting from 0
            // then return the number
            if (freq[i] == k)
                return i;
        }
   
        return -1;
    }
     
// driver program
     
        let arr = [ 2, 2, 1, 3, 1 ];
        let k = 2;
        let n = arr.length;
        document.write(findDuplicate(arr, n, k));
 
// This code is contributed by code_hunt.
</script>

1

Complejidad de Tiempo : O(n ² ) 
Espacio Auxiliar : O(1) 

Esta solución no requiere que los elementos de la array estén en un rango limitado.

Mejor solución : ordene la array de entrada y encuentre el primer elemento con exactamente k recuento de apariciones. 

Implementación:

// C++ program to find smallest number
// in array that is repeated exactly
// 'k' times.
#include <bits/stdc++.h>
using namespace std;
 
int findDuplicate(int arr[], int n, int k)
{
    // Sort the array
    sort(arr, arr + n);
 
    // Find the first element with exactly
    // k occurrences.
    int i = 0;
    while (i < n) {
        int j, count = 1;
        for (j = i + 1; j < n && arr[j] == arr[i]; j++)
            count++;
 
        if (count == k)
            return arr[i];
 
        i = j;
    }
 
    return -1;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 2, 1, 3, 1 };
    int k = 2;
    int n = sizeof(arr) / (sizeof(arr[0]));
    cout << findDuplicate(arr, n, k);
    return 0;
}

// Java program to find smallest number
// in array that is repeated exactly
// 'k' times.
import java.util.Arrays;
public class GFG {
 
    // finds the smallest number in arr[]
    // that is repeated k times
    static int findDuplicate(int arr[], int n, int k)
    {
        // Sort the array
        Arrays.sort(arr);
 
        // Find the first element with exactly
        // k occurrences.
        int i = 0;
        while (i < n) {
            int j, count = 1;
            for (j = i + 1; j < n && arr[j] == arr[i]; j++)
                count++;
 
            if (count == k)
                return arr[i];
 
            i = j;
        }
 
        return -1;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 2, 2, 1, 3, 1 };
        int k = 2;
        int n = arr.length;
        System.out.println(findDuplicate(arr, n, k));
    }
}
// This article is contributed by Sumit Ghosh

# Python program to find smallest number
# in array that is repeated exactly
# 'k' times.
 
def findDuplicate(arr, n, k):
     
    # Sort the array
    arr.sort()
  
    # Find the first element with exactly
    # k occurrences.
    i = 0
    while (i < n):
        j, count = i + 1, 1
        while (j < n and arr[j] == arr[i]):
            count += 1
            j += 1
  
        if (count == k):
            return arr[i]
  
        i = j
         
    return -1
  
# Driver code
arr = [ 2, 2, 1, 3, 1 ];
k = 2
n = len(arr)
print(findDuplicate(arr, n, k))
 
# This code is contributed by Sachin Bisht

// C# program to find smallest number
// in array that is repeated exactly
// 'k' times.
using System;
 
public class GFG {
 
    // finds the smallest number in arr[]
    // that is repeated k times
    static int findDuplicate(int[] arr,
                             int n, int k)
    {
         
        // Sort the array
        Array.Sort(arr);
 
        // Find the first element with
        // exactly k occurrences.
        int i = 0;
        while (i < n) {
            int j, count = 1;
            for (j = i + 1; j < n &&
                     arr[j] == arr[i]; j++)
                count++;
 
            if (count == k)
                return arr[i];
 
            i = j;
        }
 
        return -1;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 2, 2, 1, 3, 1 };
        int k = 2;
        int n = arr.Length;
         
        Console.WriteLine(
               findDuplicate(arr, n, k));
    }
}
 
// This article is contributed by vt_m.

<?php
// PHP program to find smallest number
// in array that is repeated exactly
// 'k' times.
 
// finds the smallest number in arr[]
// that is repeated k times
function findDuplicate($arr, $n, $k)
{
    // Sort the array
    sort($arr);
 
    // Find the first element with
    // exactly k occurrences.
    $i = 0;
    while ($i < $n)
    {
        $j; $count = 1;
        for ($j = $i + 1; $j < $n &&
             $arr[$j] == $arr[$i]; $j++)
            $count++;
 
        if ($count == $k)
            return $arr[$i];
 
        $i = $j;
    }
 
    return -1;
}
 
// Driver code
$arr = array( 2, 2, 1, 3, 1 );
$k = 2;
$n = sizeof($arr);
echo(findDuplicate($arr, $n, $k));
 
// This code is contributed
// by Code_Mech.
?>

<script>
 
    // JavaScript program to find smallest number
    // in array that is repeated exactly
    // 'k' times.
     
    // finds the smallest number in arr[]
    // that is repeated k times
    function findDuplicate(arr, n, k)
    {
          
        // Sort the array
        arr.sort();
  
        // Find the first element with
        // exactly k occurrences.
        let i = 0;
        while (i < n) {
            let j, count = 1;
            for (j = i + 1; j < n &&
                     arr[j] == arr[i]; j++)
                count++;
  
            if (count == k)
                return arr[i];
  
            i = j;
        }
  
        return -1;
    }
     
    let arr = [ 2, 2, 1, 3, 1 ];
    let k = 2;
    let n = arr.length;
 
    document.write(findDuplicate(arr, n, k));
     
</script>

1

Complejidad de Tiempo : O(n Log n) 
Espacio Auxiliar : O(1)

Enfoque eficiente: el enfoque eficiente se basa en el hecho de que la array tiene números en un rango pequeño (1 a 1000). Resolvemos este problema usando una array de frecuencias de tamaño máximo y almacenamos la frecuencia de cada número en esa array. 

Implementación:

// C++ program to find smallest number
// in array that is repeated exactly
// 'k' times.
#include <bits/stdc++.h>
using namespace std;
 
const int MAX = 1000;
 
int findDuplicate(int arr[], int n, int k)
{
    // Computing frequencies of all elements
    int freq[MAX];
    memset(freq, 0, sizeof(freq));
    for (int i = 0; i < n; i++) {
        if (arr[i] < 1 && arr[i] > MAX) {
            cout << "Out of range";
            return -1;
        }
        freq[arr[i]] += 1;
    }
 
    // Finding the smallest element with
    // frequency as k
    for (int i = 0; i < MAX; i++) {
 
        // If frequency of any of the number
        // is equal to k starting from 0
        // then return the number
        if (freq[i] == k)
            return i;
    }
 
    return -1;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 2, 1, 3, 1 };
    int k = 2;
    int n = sizeof(arr) / (sizeof(arr[0]));
    cout << findDuplicate(arr, n, k);
    return 0;
}

// Java program to find smallest number
// in array that is repeated exactly
// 'k' times.
public class GFG {
 
    static final int MAX = 1000;
 
    // finds the smallest number in arr[]
    // that is repeated k times
    static int findDuplicate(int arr[], int n, int k)
    {
        // Computing frequencies of all elements
        int[] freq = new int[MAX];
 
        for (int i = 0; i < n; i++) {
            if (arr[i] < 1 && arr[i] > MAX) {
                System.out.println("Out of range");
                return -1;
            }
            freq[arr[i]] += 1;
        }
 
        // Finding the smallest element with
        // frequency as k
        for (int i = 0; i < MAX; i++) {
 
            // If frequency of any of the number
            // is equal to k starting from 0
            // then return the number
            if (freq[i] == k)
                return i;
        }
 
        return -1;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 2, 2, 1, 3, 1 };
        int k = 2;
        int n = arr.length;
        System.out.println(findDuplicate(arr, n, k));
    }
}
// This article is contributed by Sumit Ghosh

# Python program to find smallest number
# in array that is repeated exactly
# 'k' times.
 
MAX = 1000
  
def findDuplicate(arr, n, k):
     
    # Computing frequencies of all elements
    freq = [0 for i in range(MAX)]
     
    for i in range(n):
        if (arr[i] < 1 and arr[i] > MAX):
            print("Out of range")
            return -1
        freq[arr[i]] += 1
  
    # Finding the smallest element with
    # frequency as k
    for i in range(MAX):
     
        # If frequency of any of the number
        # is equal to k starting from 0
        # then return the number
        if (freq[i] == k):
            return i
     
    return -1
     
# Driver code
arr = [ 2, 2, 1, 3, 1 ]
k = 2
n = len(arr)
print(findDuplicate(arr, n, k))
 
# This code is contributed by Sachin Bisht

// C# program to find smallest number
// in array that is repeated exactly
// 'k' times.
using System;
 
public class GFG {
 
    static int MAX = 1000;
 
    // finds the smallest number in arr[]
    // that is repeated k times
    static int findDuplicate(int[] arr,
                            int n, int k)
    {
         
        // Computing frequencies of all
        // elements
        int[] freq = new int[MAX];
 
        for (int i = 0; i < n; i++)
        {
            if (arr[i] < 1 && arr[i] > MAX)
            {
                Console.WriteLine("Out of range");
                return -1;
            }
             
            freq[arr[i]] += 1;
        }
 
        // Finding the smallest element with
        // frequency as k
        for (int i = 0; i < MAX; i++) {
 
            // If frequency of any of the
            // number is equal to k starting
            // from 0 then return the number
            if (freq[i] == k)
                return i;
        }
 
        return -1;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 2, 2, 1, 3, 1 };
        int k = 2;
        int n = arr.Length;
         
        Console.WriteLine(
               findDuplicate(arr, n, k));
    }
}
 
// This article is contributed by vt_m.

<?php
// PHP program to find smallest number
// in array that is repeated exactly
// 'k' times.
 
$MAX = 1000;
 
function findDuplicate($arr, $n, $k)
{
    global $MAX;
     
    // Computing frequencies of all elements
    $freq=array_fill(0, $MAX, 0);
 
    for ($i = 0; $i < $n; $i++)
    {
        if ($arr[$i] < 1 && $arr[$i] > $MAX)
        {
            echo "Out of range";
            return -1;
        }
        $freq[$arr[$i]] += 1;
    }
 
    // Finding the smallest element with
    // frequency as k
    for ($i = 0; $i < $MAX; $i++)
    {
 
        // If frequency of any of the number
        // is equal to k starting from 0
        // then return the number
        if ($freq[$i] == $k)
            return $i;
    }
 
    return -1;
}
 
    // Driver code
    $arr = array( 2, 2, 1, 3, 1 );
    $k = 2;
    $n = count($arr);
    echo findDuplicate($arr, $n, $k);
 
// This code is contributed by mits
?>

<script>
 
// javascript program to find smallest number
// in array that is repeated exactly
// 'k' times.
 
var MAX = 1000;
 
// finds the smallest number in arr
// that is repeated k times
function findDuplicate(arr , n , k)
{
    // Computing frequencies of all elements
    var freq = Array.from({length: MAX}, (_, i) => 0);
 
    for (var i = 0; i < n; i++) {
        if (arr[i] < 1 && arr[i] > MAX) {
            document.write("Out of range");
            return -1;
        }
        freq[arr[i]] += 1;
    }
 
    // Finding the smallest element with
    // frequency as k
    for (var i = 0; i < MAX; i++) {
 
        // If frequency of any of the number
        // is equal to k starting from 0
        // then return the number
        if (freq[i] == k)
            return i;
    }
 
    return -1;
}
 
// Driver code
var arr = [ 2, 2, 1, 3, 1 ];
var k = 2;
var n = arr.length;
document.write(findDuplicate(arr, n, k));
 
// This code is contributed by 29AjayKumar
</script>

1

Complejidad temporal: O(MAX + n) 
Espacio auxiliar: O(MAX)

¿Podemos resolverlo en tiempo O(n) si el rango no está limitado?  
Consulte el elemento más pequeño repetido exactamente ‘k’ veces (no limitado a un rango pequeño)

Este artículo es una contribución de Abhijit Shankhdhar . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.