Count all elements in the array which appears at least K times after their first occurrence
Last Updated :
11 Jul, 2025
Given an array arr[] of N integer elements and an integer K. The task is to count all distinct arr[i] such that arr[i] appears at least K times in the index range i + 1 to n - 1.
Examples:
Input: arr[] = {1, 2, 1, 3}, K = 1
Output: 1
arr[0] = 1 is the only element that appears at least once in the index range [1, 3] i.e. arr[2]
Input: arr[] = {1, 2, 3, 2, 1, 3, 1, 2, 1}, K = 2
Output: 2
[Naive Approach] Using Nested for loops – O(n^2) Time and O(n) Space
To count valid elements efficiently, we iterate from i = 0 to n-1 and track the occurrences of arr[i] in the range [i+1, n-1]. If an element appears K or more times, we increment the result. While iterating, if the current element has been checked before, skip it to prevent redundant processing.
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of
// all distinct valid elements
int countOccurrence(int n, vector<int> arr, int k)
{
int cnt, ans = 0;
// Traverse the complete array
for (int i = 0; i < n; i++) {
cnt = 0;
bool check = false;
// To check current element is previously checked
for(int j=0;j<i;j++){
if(arr[i]==arr[j]){
check = true;
break;
}
}
// If current element is previously checked
// don't check it again
if(check==1) continue;
// Count occurrence of arr[i] in range [i + 1, n - 1]
for (int j = i + 1; j < n; j++) {
if (arr[j] == arr[i])
cnt++;
// If count becomes equal to K
// break the loop
if (cnt >= k)
break;
}
// If cnt >= K
// increment ans by 1
if (cnt >= k)
ans++;
}
return ans;
}
// Driver code
int main()
{
vector<int> arr = { 1, 2, 1, 3 };
int n = arr.size();
int k = 1;
cout << countOccurrence(n, arr, k);
return 0;
}
// Java implementation of the approach
class GFG
{
// Function to return the count of
// all distinct valid elements
public static int countOccurrence(int n, int[] arr, int k)
{
int cnt, ans = 0;
// Traverse the complete array
for (int i = 0; i < n; i++)
{
cnt = 0;
boolean check = false;
// To check current element is previously checked
for(int j=0;j<i;j++){
if(arr[i]==arr[j]){
check = true;
break;
}
}
// If current element is previously checked
// don't check it again
if(check==true)
continue;
// Count occurrence of arr[i] in range [i + 1, n - 1]
for (int j = i + 1; j < n; j++)
{
if (arr[j] == arr[i])
cnt++;
// If count becomes equal to K
// break the loop
if (cnt >= k)
break;
}
// If cnt >= K
// increment ans by 1
if (cnt >= k)
ans++;
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
int[] arr = {1, 2, 1, 3};
int n = arr.length;
int k = 1;
System.out.println(countOccurrence(n, arr, k));
}
}
// This code is contributed by
// sanjeev2552
# Python3 implementation of the approach
# Function to return the count of
# all distinct valid elements
def countOccurrence(n, arr, k):
cnt, ans = 0, 0
# Traverse the complete array
for i in range(n):
cnt = 0
check = False;
# To check current element is previously checked
for j in range(0,i):
if(arr[i]==arr[j]):
check = True;
break;
# If current element is previously checked
# don't check it again
if(check == True):
continue;
# Count occurrence of arr[i] in
# range [i + 1, n - 1]
for j in range(i + 1, n):
if (arr[j] == arr[i]):
cnt += 1
# If count becomes equal to K
# break the loop
if (cnt >= k):
break
# If cnt >= K
# increment ans by 1
if (cnt >= k):
ans += 1
return ans
# Driver code
arr = [1, 2, 1, 3]
n = len(arr)
k = 1
print(countOccurrence(n, arr, k))
# This code is contributed
# by mohit kumar
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the count of
// all distinct valid elements
public static int countOccurrence(int n,
int[] arr, int k)
{
int cnt, ans = 0;
// Traverse the complete array
for (int i = 0; i < n; i++)
{
cnt = 0;
// To check current element is previously checked
bool check = false;
for(int j=0;j<i;j++){
if(arr[i]==arr[j]){
check = true;
break;
}
}
// If current element is previously checked
// don't check it again
if(check == true)continue;
// Count occurrence of arr[i]
// in range [i + 1, n - 1]
for (int j = i + 1; j < n; j++)
{
if (arr[j] == arr[i])
cnt++;
// If count becomes equal to K
// break the loop
if (cnt >= k)
break;
}
// If cnt >= K
// increment ans by 1
if (cnt >= k)
ans++;
}
return ans;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = {1, 2, 1, 3};
int n = arr.Length;
int k = 1;
Console.WriteLine(countOccurrence(n, arr, k));
}
}
// This code is contributed by Rajput-Ji
// JavaScript implementation of the approach
// Function to return the count of
// all distinct valid elements
function countOccurrence(n, arr, k)
{
let cnt, ans = 0;
// To avoid duplicates
// Traverse the complete array
for (let i = 0; i < n; i++)
{
cnt = 0;
// To check current element is previously checked
let check = false;
for(let j=0;j<i;j++){
if(arr[i]==arr[j]){
check = true;
break;
}
}
// If current element is previously checked
// don't check it again
if(check == true)continue;
// Count occurrence of arr[i]
// in range [i + 1, n - 1]
for (let j = i + 1; j < n; j++)
{
if (arr[j] == arr[i])
cnt++;
// If count becomes equal to K
// break the loop
if (cnt >= k)
break;
}
// If cnt >= K
// increment ans by 1
if (cnt >= k)
ans++;
}
return ans;
}
// Driver code
let arr = [1, 2, 1, 3];
let n = arr.length;
let k = 1;
console.log(countOccurrence(n, arr, k));
[Better Approach] Using Binary Search – O(n*log(n)) Time and O(1) Space
To efficiently count valid elements, we note that if an element must appear at least K times after its first occurrence, its total frequency should be at least K+1. We first sort the array, enabling binary search for quick frequency lookup. The frequency is determined as (last occurrence index - first occurrence index) + 1. If it meets or exceeds K+1, we increment the count. Finally, we return the total count.
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
//Function to find count of all elements that occurs
//atleast k times in the array after first occurrence
int countOccurrence(int arr[], int n, int k)
{ int count = 0;
sort(arr,arr+n);//sort array for binary search
for(int i = 0 ; i < n ;i++)
{
//index of first and last occ of arr[i]
int first_index = lower_bound(arr,arr+n,arr[i])- arr;
int last_index = upper_bound(arr,arr+n,arr[i])- arr-1;
i = last_index; // assign i to last_index to avoid counting
// same element multiple time
int fre = last_index-first_index+1;//finding frequency
if(fre >= k+1)
{ // if frequency >= k+1 ,means elements occur atleast k times
//then increase the count by 1
count += 1;
}
}
return count;//return final answer
}
// Drive code
int main()
{
int arr[] = { 1, 2, 1, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 1;
// Function call
cout << countOccurrence( arr, n, k);
return 0;
}
// This Approach is contributed by nikhilsainiofficial546
import java.util.Arrays;
public class Main {
// Function to find count of all elements that occurs
// atleast k times in the array after first occurrence
static int countOccurrence(int[] arr, int n, int k) {
int count = 0;
Arrays.sort(arr); // sort array for binary search
for (int i = 0; i < n; i++) {
// index of first and last occ of arr[i]
int first_index = Arrays.binarySearch(arr, arr[i]);
if (first_index < 0) { // element not found
continue;
}
int last_index = first_index;
while (last_index + 1 < n && arr[last_index + 1] == arr[i]) {
last_index++;
}
i = last_index; // assign i to last_index to avoid counting
// same element multiple time
int fre = last_index - first_index + 1; // finding frequency
if (fre >= k + 1) {
// if frequency >= k+1, means elements occur atleast k times
// then increase the count by 1
count += 1;
}
}
return ++count; // return final answer
}
// Drive code
public static void main(String[] args) {
int[] arr = { 1, 2, 1, 3 };
int n = arr.length;
int k = 1;
// Function call
System.out.println(countOccurrence(arr, n, k));
}
}
# Function to find count of all elements that occur
# at least k times in the array after first occurrence
def countOccurrence(arr, n, k):
count = 0
arr.sort() # sort array for binary search
i = 0
while i < n:
# index of first and last occurrence of arr[i]
first_index = arr.index(arr[i])
last_index = n - arr[::-1].index(arr[i]) - 1
i = last_index # assign i to last_index to avoid counting
# the same element multiple times
fre = last_index - first_index + 1 # finding frequency
# if frequency >= k+1, means elements occur at least k times
# then increase the count by 1
if fre >= k+1:
count += 1
i+=1
return count # return final answer
# Drive code
arr = [1, 2, 1, 3]
n = len(arr)
k = 1
# Function call
print(countOccurrence(arr, n, k))
using System;
class MainClass {
//Function to find count of all elements that occurs
//atleast k times in the array after first occurrence
static int countOccurrence(int[] arr, int n, int k) {
int count = 0;
Array.Sort(arr); //sort array for binary search
for (int i = 0; i < n; i++) {
//index of first and last occ of arr[i]
int first_index = Array.BinarySearch(arr, arr[i]);
if (first_index < 0) { // element not found
continue;
}
int last_index = first_index;
while (last_index + 1 < n && arr[last_index + 1] == arr[i]) {
last_index++;
}
i = last_index; // assign i to last_index to avoid counting
// same element multiple time
int fre = last_index - first_index + 1; //finding frequency
if (fre >= k + 1) {
// if frequency >= k+1, means elements occur atleast k times
// then increase the count by 1
count += 1;
}
}
return ++count; //return final answer
}
//Drive code
public static void Main() {
int[] arr = { 1, 2, 1, 3 };
int n = arr.Length;
int k = 1;
//Function call
Console.WriteLine(countOccurrence(arr, n, k));
}
}
// js equivalent
//Function to find count of all elements that occurs
//atleast k times in the array after first occurrence
function countOccurrence(arr, n, k) {
let count = 0;
//sort array for binary search
arr.sort((a, b) => a - b);
for (let i = 0; i < n; i++) {
//index of first and last occ of arr[i]
let first_index = arr.indexOf(arr[i]);
let last_index = arr.lastIndexOf(arr[i]);
i = last_index; // assign i to last_index to avoid counting
// same element multiple time
let fre = last_index - first_index + 1; //finding frequency
if (fre >= k + 1) {
// if frequency >= k+1 ,means elements occur atleast k times
//then increase the count by 1
count += 1;
}
}
return count; //return final answer
}
// Drive code
let arr = [1, 2, 1, 3];
let n = arr.length;
let k = 1;
// Function call
console.log(countOccurrence(arr, n, k));
[Expected Approach] Using Hash Map – O(n) Time and O(n) Space
To efficiently count valid elements, we use a hash map to store the occurrence of each element while traversing the array from right to left (n-1 to 0). If an element's occurrence reaches K, we increment the count. Otherwise, we update its occurrence in the hash map.
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of
// all distinct valid elements
int countOccurrence(int n, int arr[], int k)
{
int cnt, ans = 0;
// To avoid duplicates
unordered_map<int, bool> hash;
// To store the count of arr[i]
// in range [i + 1, n - 1]
unordered_map<int, int> occurrence;
for (int i = n - 1; i >= 0; i--) {
// To avoid duplicates
if (hash[arr[i]] == true)
continue;
// If occurrence in range i+1 to n becomes
// equal to K then increment ans by 1
if (occurrence[arr[i]] >= k) {
ans++;
hash[arr[i]] = true;
}
// Otherwise increase occurrence of arr[i] by 1
else
occurrence[arr[i]]++;
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 1, 2, 1, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 1;
cout << countOccurrence(n, arr, k);
return 0;
}
// Java implementation of the approach
import java.util.HashMap;
class GFG
{
// Function to return the count of
// all distinct valid elements
public static int countOccurrence(int n, int[] arr, int k)
{
int ans = 0;
// To avoid duplicates
HashMap<Integer, Boolean> hash = new HashMap<>();
// To store the count of arr[i]
// in range [i + 1, n - 1]
HashMap<Integer, Integer> occurrence = new HashMap<>();
for (int i = n-1; i>=0; i--)
{
// To avoid duplicates
if (hash.get(arr[i]) != null &&
hash.get(arr[i]) == true)
continue;
// If occurrence in range i+1 to n becomes
// equal to K then increment ans by 1
if (occurrence.get(arr[i]) != null &&
occurrence.get(arr[i]) >= k)
{
ans++;
hash.put(arr[i], true);
}
// Otherwise increase occurrence of arr[i] by 1
else
{
if (occurrence.get(arr[i]) == null)
occurrence.put(arr[i], 1);
else
{
int temp = occurrence.get(arr[i]);
occurrence.put(arr[i], ++temp);
}
}
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
int[] arr = {1, 2, 1, 3};
int n = arr.length;
int k = 1;
System.out.println(countOccurrence(n, arr, k));
}
}
// This code is contributed by
// sanjeev2552
# Python3 implementation of the approach
# Function to return the count of
# all distinct valid elements
def countOccurrence(n, arr, k) :
ans = 0;
# To avoid duplicates
hash = dict.fromkeys(arr,0);
# To store the count of arr[i]
# in range [i + 1, n - 1]
occurrence = dict.fromkeys(arr, 0);
for i in range(n - 1, -1, -1) :
# To avoid duplicates
if (hash[arr[i]] == True) :
continue;
# If occurrence in range i+1 to n
# becomes equal to K then increment
# ans by 1
if (occurrence[arr[i]] >= k) :
ans += 1;
hash[arr[i]] = True;
# Otherwise increase occurrence
# of arr[i] by 1
else :
occurrence[arr[i]] += 1;
return ans;
# Driver code
if __name__ == "__main__" :
arr = [ 1, 2, 1, 3 ];
n = len(arr) ;
k = 1;
print(countOccurrence(n, arr, k));
# This code is contributed by Ryuga
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the count of
// all distinct valid elements
public static int countOccurrence(int n,
int[] arr,
int k)
{
int ans = 0;
// To avoid duplicates
Dictionary<int,
bool> hash = new Dictionary<int,
bool>();
// To store the count of arr[i]
// in range [i + 1, n - 1]
Dictionary<int,
int> occurrence = new Dictionary<int,
int>();
for (int i = n - 1; i >= 0; i--)
{
// To avoid duplicates
if (hash.ContainsKey(arr[i]) &&
hash[arr[i]] == true)
continue;
// If occurrence in range i+1 to n becomes
// equal to K then increment ans by 1
if (occurrence.ContainsKey(arr[i]) &&
occurrence[arr[i]] >= k)
{
ans++;
hash.Add(arr[i], true);
}
// Otherwise increase occurrence of arr[i] by 1
else
{
if (!occurrence.ContainsKey(arr[i]))
occurrence.Add(arr[i], 1);
else
{
int temp = occurrence[arr[i]];
occurrence.Add(arr[i], ++temp);
}
}
}
return ans;
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = {1, 2, 1, 3};
int n = arr.Length;
int k = 1;
Console.WriteLine(countOccurrence(n, arr, k));
}
}
// This code is contributed by 29AjayKumar
// Javascript implementation of the approach
// Function to return the count of
// all distinct valid elements
function countOccurrence(n, arr, k)
{
let ans = 0;
// To avoid duplicates
let hash = new Map();
// To store the count of arr[i]
// in range [i + 1, n - 1]
let occurrence = new Map();
for(let i = n - 1; i >= 0; i--)
{
// To avoid duplicates
if (hash.get(arr[i]) != null &&
hash.get(arr[i]) == true)
continue;
// If occurrence in range i+1 to n becomes
// equal to K then increment ans by 1
if (occurrence.get(arr[i]) != null &&
occurrence.get(arr[i]) >= k)
{
ans++;
hash.set(arr[i], true);
}
// Otherwise increase occurrence of arr[i] by 1
else
{
if (occurrence.get(arr[i]) == null)
occurrence.set(arr[i], 1);
else
{
let temp = occurrence.get(arr[i]);
occurrence.set(arr[i], ++temp);
}
}
}
return ans;
}
// Driver Code
let arr = [ 1, 2, 1, 3 ];
let n = arr.length;
let k = 1;
console.log(countOccurrence(n, arr, k))
// This code is contributed by unknown2108
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