Given an integer array arr[], print kth distinct element in this array. The given array may contain duplicates and the output should print the k-th element among all unique elements. If k is more than the number of distinct elements, print -1.
Examples:
Input: arr[] = {1, 2, 1, 3, 4, 2}, k = 2
Output: 4
Explanation: The First non-repeating element is 3 and the Second non-repeating element is 4
Input: {2, 2, 2, 2}, k = 2
Output: -1
Explanation: There is no distinct (non-repeating) element in the array.Input: arr[] = {1, 2, 50, 10, 20, 2}, k = 3
Output: TheFirst non-repeating element is 1 and the Second non-repeating element is 10.
Try It Yourself
Table of Content
[Naive Approach] Nested Loop – O(n^2) Time and O(1) Space
The idea is to use two nested loops where,
- The outer loop picks elements from left to right
- Inner loop checks if the picked element is present somewhere else. If not present, then increment the count of distinct elements. If the count becomes k, return the current element.
#include <bits/stdc++.h>
using namespace std;
int kthDistinct(vector<int> arr,
int k)
{
int dist_count = 0, n = arr.size();
for (int i = 0; i < n; i++)
{
// Check if current element is
// present somewhere else.
int j;
for (j = 0; j < n; j++)
if (i != j && arr[j] == arr[i])
break;
// If element is unique
if (j == n)
dist_count++;
if (dist_count == k)
return arr[i];
}
return -1;
}
int main ()
{
vector<int> arr = {1, 2, 1, 3, 4, 2};
int k = 2;
cout << kthDistinct(arr, k);
return 0;
}
class GfG
{
// Returns k-th distinct element in arr.
static int kthDistinct(int arr[],
int k)
{
int n = arr.length;
int dist_count = 0;
for (int i = 0; i < n; i++)
{
// Check if current element is
// present somewhere else.
int j;
for (j = 0; j < n; j++)
if (i != j && arr[j] == arr[i])
break;
// If element is unique
if (j == n)
dist_count++;
if (dist_count == k)
return arr[i];
}
return -1;
}
public static void main (String[] args)
{
int arr[] = {1, 2, 1, 3, 4, 2};
int k = 2;
System.out.print(kthDistinct(arr, k));
}
}
def kthDistinct(arr, k):
dist_count = 0
n = len(arr)
for i in range(n):
# Check if current element is
# present somewhere else.
j = 0
while j < n:
if (i != j and arr[j] == arr[i]):
break
j += 1
# If element is unique
if (j == n):
dist_count += 1
if (dist_count == k):
return arr[i]
return -1
arr = [1, 2, 1, 3, 4, 2]
k = 2
print(kthDistinct(arr, k))
using System;
class GfG
{
static int kthDistinct(int []arr, int k)
{
int n = arr.Length;
int dist_count = 0;
for (int i = 0; i < n; i++)
{
// Check if current element is
// present somewhere else.
int j;
for (j = 0; j < n; j++)
if (i != j && arr[j] == arr[i])
break;
// If element is unique
if (j == n)
dist_count++;
if (dist_count == k)
return arr[i];
}
return -1;
}
public static void Main ()
{
int []arr = {1, 2, 1, 3, 4, 2};
int k = 2;
Console.Write(kthDistinct(arr, k));
}
}
function kthDistinct(arr, k)
{
var dist_count = 0;
var n = arr.length;
for (var i = 0; i < n; i++)
{
// Check if current element is
// present somewhere else.
var j;
for (j = 0; j < n; j++)
if (i != j && arr[j] == arr[i])
break;
// If element is unique
if (j == n)
dist_count++;
if (dist_count == k)
return arr[i];
}
return -1;
}
var arr = [1, 2, 1, 3, 4, 2];
var k = 2;
console.log( kthDistinct(arr, k));
Output
4
[Expected Approach] Hashing – O(n) Time and O(n) Space
The idea is to traverse the array and use a hash table to store the the elements and their count of occurrence..
Traverse the array again to find and count elements with occurrence count equal to 1 and if count becomes k then return the current element.
#include <bits/stdc++.h>
using namespace std;
int kthDistinct(vector<int> arr, int k)
{
// Traverse input array and
// store counts if individual
// elements.
int n = arr.size();
unordered_map<int, int> h;
for (int i = 0; i < n; i++)
h[arr[i]]++;
// If size of hash is
// less than k.
if (h.size() < k)
return -1;
// Traverse array again and
// find k-th element with
// count as 1.
int dist_count = 0;
for (int i = 0; i < n; i++)
{
if (h[arr[i]] == 1)
dist_count++;
if (dist_count == k)
return arr[i];
}
return -1;
}
// Driver Code
int main ()
{
vector<int>arr = {1, 2, 1, 3, 4, 2};
cout << kthDistinct(arr, 2);
return 0;
}
import java.util.*;
class GfG {
static int kthDistinct(int arr[], int k)
{
// int dist_count = 0;
int n = arr.length;
Map<Integer, Integer> h
= new HashMap<Integer, Integer>();
for (int i = 0; i < n; i++) {
if (h.containsKey(arr[i]))
h.put(arr[i], h.get(arr[i]) + 1);
else
h.put(arr[i], 1);
}
// If size of hash is
// less than k.
if (h.size() < k)
return -1;
// Traverse array again and
// find k-th element with
// count as 1.
int dist_count = 0;
for (int i = 0; i < n; i++) {
if (h.get(arr[i]) == 1)
dist_count++;
if (dist_count == k)
return arr[i];
}
return -1;
}
public static void main(String[] args)
{
int arr[] = { 1, 2, 1, 3, 4, 2 };
System.out.println(kthDistinct(arr, 2));
}
}
def kthDistinct(arr, KthIndex):
size = len(arr)
dict = {}
vect = []
for i in range(size):
if(arr[i] in dict):
dict[arr[i]] = dict[arr[i]] + 1
else:
dict[arr[i]] = 1
for i in range(size):
if(dict[arr[i]] > 1):
continue
else:
KthIndex = KthIndex - 1
if(KthIndex == 0):
return arr[i]
return -1
arr = [1, 2, 1, 3, 4, 2]
print(kthDistinct(arr, 2))
using System;
using System.Collections.Generic;
class GfG
{
// Returns k-th distinct
// element in arr.
static int kthDistinct(int []arr, int k)
{
int n = arr.Length;
Dictionary<int, int> h = new Dictionary<int, int>();
for (int i = 0; i < n; i++)
{
if(h.ContainsKey(arr[i]))
{
var val = h[arr[i]];
h.Remove(arr[i]);
h.Add(arr[i], val + 1);
}
else
h.Add(arr[i], 1);
}
// If size of hash is
// less than k.
if (h.Count < k)
return -1;
// Traverse array again and
// find k-th element with
// count as 1.
int dist_count = 0;
for (int i = 0; i < n; i++)
{
if (h[arr[i]] == 1)
dist_count++;
if (dist_count == k)
return arr[i];
}
return -1;
}
// Driver Code
public static void Main (String[] args)
{
int []arr = {1, 2, 1, 3, 4, 2};
Console.WriteLine(kthDistinct(arr, 2));
}
}
function kthDistinct(arr,k)
{
// int dist_count = 0;
let n = arr.length;
let h = new Map();
for (let i = 0; i < n; i++)
{
if(h.has(arr[i]))
h.set(arr[i], h.get(arr[i]) + 1);
else
h.set(arr[i], 1);
}
// If size of hash is
// less than k.
if (h.length < k)
return -1;
// Traverse array again and
// find k-th element with
// count as 1.
let dist_count = 0;
for (let i = 0; i < n; i++)
{
if (h.get(arr[i]) == 1)
dist_count++;
if (dist_count == k)
return arr[i];
}
return -1;
}
// Driver Code
let arr=[1, 2, 1, 3, 4, 2];
console.log(kthDistinct(arr, 2));
Output
4
