Shortest subarray to be removed to make all Array elements unique
Given an array arr[] containing N elements, the task is to remove a subarray of minimum possible length from the given array such that all remaining elements are distinct. Print the minimum possible length of the subarray.
Examples:
Input: N = 5, arr[] = {1, 2, 1, 2, 3}
Output: 2
Explanation:
Remove the sub array {2, 1} to make the elements distinct.
Input: N = 5, arr[] = {1, 2, 3, 4, 5}
Output: 0
Explanation:
Elements are already distinct.
Naive Approach: The naive approach for this problem is to simply check for all the possible subarrays and find the length of the smallest subarray after removal of which all the elements in the array become distinct.
Time complexity: O(N3)
Efficient Approach:
- Let ans be the length of the minimum subarray that on removing from the given array, makes the elements of the array unique.
- We can easily observe that if all array elements become distinct after removing a subarray of length ans, then this condition is also true for all values greater than ans.
- This means that the solution for this problem is a monotonically increasing function and we can apply binary search on the answer.
- Now, for a particular length K of subarray, we can check if elements of prefix and suffix of all sub arrays of length K are distinct or not.
- We can do this by using a sliding window technique.
- Use a hash map to store the frequencies of elements in prefix and suffix, on moving the window forward, increment frequency of the last element of prefix and decrement frequency of the first element of suffix.
Below is the implementation of the above approach:
C++
// C++ program to make array elements // distinct by removing at most // one subarray of minimum length #include <bits/stdc++.h> using namespace std; // Function to check if elements of // Prefix and suffix of each sub array // of size K are distinct or not bool check( int a[], int n, int k) { // Hash map to store frequencies of // elements of prefix and suffix map< int , int > m; // Variable to store number of // occurrences of an element other // than one int extra = 0; // Adding frequency of elements of suffix // to hash for subarray starting from first // index // There is no prefix for this sub array for ( int i = k; i < n; i++) m[a[i]]++; // Counting extra elements in current Hash // map for ( auto x : m) extra += x.second - 1; // If there are no extra elements return // true if (extra == 0) return true ; // Check for remaining sub arrays for ( int i = 1; i + k - 1 < n; i++) { // First element of suffix is now // part of subarray which is being // removed so, check for extra elements if (m[a[i + k - 1]] > 1) extra--; // Decrement frequency of first // element of the suffix m[a[i + k - 1]]--; // Increment frequency of last // element of the prefix m[a[i - 1]]++; // Check for extra elements if (m[a[i - 1]] > 1) extra++; // If there are no extra elements // return true if (extra == 0) return true ; } return false ; } // Function for calculating minimum // length of the subarray, which on // removing make all elements distinct int minlength( int a[], int n) { // Possible range of length of subarray int lo = 0, hi = n + 1; int ans = 0; // Binary search to find minimum ans while (lo < hi) { int mid = (lo + hi) / 2; if (check(a, n, mid)) { ans = mid; hi = mid; } else lo = mid + 1; } return ans; } // Driver code int main() { int a[5] = { 1, 2, 1, 2, 3 }; int n = sizeof (a) / sizeof ( int ); cout << minlength(a, n); } |
Java
// Java program to make array elements // pairwise distinct by removing at most // one subarray of minimum length import java.util.*; import java.lang.*; class GFG{ // Function to check if elements of // Prefix and suffix of each sub array // of size K are pairwise distinct or not static boolean check( int a[], int n, int k) { // Hash map to store frequencies of // elements of prefix and suffix Map<Integer, Integer> m = new HashMap<>(); // Variable to store number of // occurrences of an element other // than one int extra = 0 ; // Adding frequency of elements of suffix // to hash for subarray starting from first // index // There is no prefix for this sub array for ( int i = k; i < n; i++) m.put(a[i], m.getOrDefault(a[i], 0 ) + 1 ); // Counting extra elements in current Hash // map for (Integer x : m.values()) extra += x - 1 ; // If there are no extra elements return // true if (extra == 0 ) return true ; // Check for remaining sub arrays for ( int i = 1 ; i + k - 1 < n; i++) { // First element of suffix is now // part of subarray which is being // removed so, check for extra elements if (m.get(a[i + k - 1 ]) > 1 ) extra--; // Decrement frequency of first // element of the suffix m.put(a[i + k - 1 ], m.get(a[i + k - 1 ]) - 1 ); // Increment frequency of last // element of the prefix m.put(a[i - 1 ], m.get(a[i - 1 ]) + 1 ); // Check for extra elements if (m.get(a[i - 1 ]) > 1 ) extra++; // If there are no extra elements // return true if (extra == 0 ) return true ; } return false ; } // Function for calculating minimum // length of the subarray, which on // removing make all elements pairwise // distinct static int minlength( int a[], int n) { // Possible range of length of subarray int lo = 0 , hi = n + 1 ; int ans = 0 ; // Binary search to find minimum ans while (lo < hi) { int mid = (lo + hi) / 2 ; if (check(a, n, mid)) { ans = mid; hi = mid; } else lo = mid + 1 ; } return ans; } // Driver Code public static void main (String[] args) { int a[] = { 1 , 2 , 1 , 2 , 3 }; int n = a.length; System.out.println(minlength(a, n)); } } // This code is contributed by offbeat |
Python3
# Python3 program to make array elements # pairwise distinct by removing at most # one subarray of minimum length from collections import defaultdict # Function to check if elements of # Prefix and suffix of each sub array # of size K are pairwise distinct or not def check(a, n, k): # Hash map to store frequencies of # elements of prefix and suffix m = defaultdict( int ) # Variable to store number of # occurrences of an element other # than one extra = 0 # Adding frequency of elements of suffix # to hash for subarray starting from first # index # There is no prefix for this sub array for i in range (k, n): m[a[i]] + = 1 # Counting extra elements in current Hash # map for x in m: extra + = m[x] - 1 # If there are no extra elements return # true if (extra = = 0 ): return True # Check for remaining sub arrays for i in range ( 1 , i + k - 1 < n): # First element of suffix is now # part of subarray which is being # removed so, check for extra elements if (m[a[i + k - 1 ]] > 1 ): extra - = 1 # Decrement frequency of first # element of the suffix m[a[i + k - 1 ]] - = 1 # Increment frequency of last # element of the prefix m[a[i - 1 ]] + = 1 # Check for extra elements if (m[a[i - 1 ]] > 1 ): extra + = 1 # If there are no extra elements # return true if (extra = = 0 ): return True return False # Function for calculating minimum # length of the subarray, which on # removing make all elements pairwise # distinct def minlength(a, n): # Possible range of length of subarray lo = 0 hi = n + 1 ans = 0 # Binary search to find minimum ans while (lo < hi): mid = (lo + hi) / / 2 if (check(a, n, mid)): ans = mid hi = mid else : lo = mid + 1 return ans # Driver code if __name__ = = "__main__" : a = [ 1 , 2 , 1 , 2 , 3 ] n = len (a) print (minlength(a, n)) # This code is contributed by chitranayal |
C#
// C# program to make array elements // pairwise distinct by removing at most // one subarray of minimum length using System; using System.Collections.Generic; class GFG{ // Function to check if elements of // Prefix and suffix of each sub array // of size K are pairwise distinct or not static bool check( int []a, int n, int k) { // Hash map to store frequencies of // elements of prefix and suffix Dictionary< int , int > m = new Dictionary< int , int >(); // Variable to store number of // occurrences of an element other // than one int extra = 0; // Adding frequency of elements of suffix // to hash for subarray starting from first // index // There is no prefix for this sub array for ( int i = k; i < n; i++) if (m.ContainsKey(a[i])) m[a[i]] = m[a[i]] + 1; else m.Add(a[i], 1); // Counting extra elements in current Hash // map foreach ( int x in m.Keys) extra += m[x] - 1; // If there are no extra elements return // true if (extra == 0) return true ; // Check for remaining sub arrays for ( int i = 1; i + k - 1 < n; i++) { // First element of suffix is now // part of subarray which is being // removed so, check for extra elements if (m[a[i + k - 1]] > 1) extra--; // Decrement frequency of first // element of the suffix m[a[i + k - 1]] = m[a[i + k - 1]] - 1; // Increment frequency of last // element of the prefix m[a[i - 1]] = m[a[i - 1]] + 1; // Check for extra elements if (m[a[i - 1]] > 1) extra++; // If there are no extra elements // return true if (extra == 0) return true ; } return false ; } // Function for calculating minimum // length of the subarray, which on // removing make all elements pairwise // distinct static int minlength( int []a, int n) { // Possible range of length of subarray int lo = 0, hi = n + 1; int ans = 0; // Binary search to find minimum ans while (lo < hi) { int mid = (lo + hi) / 2; if (check(a, n, mid)) { ans = mid; hi = mid; } else lo = mid + 1; } return ans; } // Driver Code public static void Main(String[] args) { int []a = { 1, 2, 1, 2, 3 }; int n = a.Length; Console.WriteLine(minlength(a, n)); } } // This code is contributed by Amit Katiyar |
Javascript
<script> // Javascript program to make array elements // pairwise distinct by removing at most // one subarray of minimum length // Function to check if elements of // Prefix and suffix of each sub array // of size K are pairwise distinct or not function check(a, n, k) { // Hash map to store frequencies of // elements of prefix and suffix let m = new Map(); // Variable to store number of // occurrences of an element other // than one let extra = 0; // Adding frequency of elements of suffix // to hash for subarray starting from first // index // There is no prefix for this sub array for (let i = k; i < n; i++) m.set(a[i], m.get(a[i])== null ? 1 :m.get(a[i])+ 1); // Counting extra elements in current Hash // map for (let x of m.values()) extra += x - 1; // If there are no extra elements return // true if (extra == 0) return true ; // Check for remaining sub arrays for (let i = 1; i + k - 1 < n; i++) { // First element of suffix is now // part of subarray which is being // removed so, check for extra elements if (m.get(a[i + k - 1]) > 1) extra--; // Decrement frequency of first // element of the suffix m.set(a[i + k - 1], m.get(a[i + k - 1]) - 1); // Increment frequency of last // element of the prefix m.set(a[i - 1], m.get(a[i - 1]) + 1); // Check for extra elements if (m.get(a[i - 1]) > 1) extra++; // If there are no extra elements // return true if (extra == 0) return true ; } return false ; } // Function for calculating minimum // length of the subarray, which on // removing make all elements pairwise // distinct function minlength(a,n) { // Possible range of length of subarray let lo = 0, hi = n + 1; let ans = 0; // Binary search to find minimum ans while (lo < hi) { let mid = Math.floor((lo + hi) / 2); if (check(a, n, mid)) { ans = mid; hi = mid; } else lo = mid + 1; } return ans; } // Driver Code let a = [1, 2, 1, 2, 3 ]; let n = a.length; document.write(minlength(a, n)); // This code is contributed by avanitrachhadiya2155 </script> |
2
Time Complexity: O(N * log(N)), where N is the size of the array, This is because the function “check” runs in O(n) time, and the function “minlength” performs a binary search on the length of the subarray, which takes O(log n) time.
Auxiliary Space: O(N), as it uses a hash map to store the frequencies of elements in each subarray, which can have up to n elements.
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