Given an unsorted array that may contain duplicates. Also given a number k which is smaller than size of array. Write a function that returns true if array contains duplicates within k distance.
Examples:
Input: k = 3, arr[] = {1, 2, 3, 4, 1, 2, 3, 4} Output: false All duplicates are more than k distance away. Input: k = 3, arr[] = {1, 2, 3, 1, 4, 5} Output: true 1 is repeated at distance 3. Input: k = 3, arr[] = {1, 2, 3, 4, 5} Output: false Input: k = 3, arr[] = {1, 2, 3, 4, 4} Output: true
A Simple Solution is to run two loops. The outer loop picks every element ‘arr[i]’ as a starting element, the inner loop compares all elements which are within k distance of ‘arr[i]’. The time complexity of this solution is O(kn).
We can solve this problem in Θ(n) time using Hashing. The idea is to one by add elements to hash. We also remove elements which are at more than k distance from current element. Following is detailed algorithm.
1) Create an empty hashtable.
2) Traverse all elements from left from right. Let the current element be ‘arr[i]’
….a) If current element ‘arr[i]’ is present in hashtable, then return true.
….b) Else add arr[i] to hash and remove arr[i-k] from hash if i is greater than or equal to k
C/C++
#include<bits/stdc++.h> using namespace std; /* C++ program to Check if a given array contains duplicate elements within k distance from each other */ bool checkDuplicatesWithinK( int arr[], int n, int k) { // Creates an empty hashset set< int > myset; // Traverse the input array for ( int i = 0; i < n; i++) { // If already present n hash, then we found // a duplicate within k distance if (myset.find(arr[i]) != myset.end()) return true ; // Add this item to hashset myset.insert(arr[i]); // Remove the k+1 distant item if (i >= k) myset.erase(arr[i-k]); } return false ; } // Driver method to test above method int main () { int arr[] = {10, 5, 3, 4, 3, 5, 6}; int n = sizeof (arr) / sizeof (arr[0]); if (checkDuplicatesWithinK(arr, n, 3)) cout << "Yes" ; else cout << "No" ; } //This article is contributed by Chhavi |
Java
/* Java program to Check if a given array contains duplicate elements within k distance from each other */ import java.util.*; class Main { static boolean checkDuplicatesWithinK( int arr[], int k) { // Creates an empty hashset HashSet<Integer> set = new HashSet<>(); // Traverse the input array for ( int i= 0 ; i<arr.length; i++) { // If already present n hash, then we found // a duplicate within k distance if (set.contains(arr[i])) return true ; // Add this item to hashset set.add(arr[i]); // Remove the k+1 distant item if (i >= k) set.remove(arr[i-k]); } return false ; } // Driver method to test above method public static void main (String[] args) { int arr[] = { 10 , 5 , 3 , 4 , 3 , 5 , 6 }; if (checkDuplicatesWithinK(arr, 3 )) System.out.println( "Yes" ); else System.out.println( "No" ); } } |
Python 3
# Python 3 program to Check if a given array # contains duplicate elements within k distance # from each other def checkDuplicatesWithinK(arr, n, k): # Creates an empty list myset = [] # Traverse the input array for i in range (n): # If already present n hash, then we # found a duplicate within k distance if arr[i] in myset: return True # Add this item to hashset myset.append(arr[i]) # Remove the k+1 distant item if (i > = k): myset.remove(arr[i - k]) return False # Driver Code if __name__ = = "__main__" : arr = [ 10 , 5 , 3 , 4 , 3 , 5 , 6 ] n = len (arr) if (checkDuplicatesWithinK(arr, n, 3 )): print ( "Yes" ) else : print ( "No" ) # This code is contributed by ita_c |
C#
/* C# program to Check if a given array contains duplicate elements within k distance from each other */ using System; using System.Collections.Generic; class GFG { static bool checkDuplicatesWithinK( int []arr, int k) { // Creates an empty hashset HashSet< int > set = new HashSet< int >(); // Traverse the input array for ( int i = 0; i < arr.Length; i++) { // If already present n hash, then we found // a duplicate within k distance if ( set .Contains(arr[i])) return true ; // Add this item to hashset set .Add(arr[i]); // Remove the k+1 distant item if (i >= k) set .Remove(arr[i - k]); } return false ; } // Driver code public static void Main (String[] args) { int []arr = {10, 5, 3, 4, 3, 5, 6}; if (checkDuplicatesWithinK(arr, 3)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code has been contributed // by 29AjayKumar |
Output:
Yes
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