Shortest subarray to be removed to make all Array elements unique

• Difficulty Level : Hard
• Last Updated : 11 Nov, 2021

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 pairwise distinct. Print the minimum possible length of the subarray.
Examples:

Input: N = 5, arr[] = {1, 2, 1, 2, 3}
Output:
Explanation:
Remove the sub array {2, 1} to make the elements distinct.
Input: N = 5, arr[] = {1, 2, 3, 4, 5}
Output:
Explanation:

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 pairwise 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 pairwise 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// pairwise distinct by removing at most// one subarray of minimum length #include using namespace std; // Function to check if elements of// Prefix and suffix of each sub array// of size K are pairwise distinct or notbool check(int a[], int n, int k){    // Hash map to store frequencies of    // elements of prefix and suffix    map 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 pairwise// distinctint 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 codeint main(){    int a = { 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 lengthimport 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 notstatic boolean check(int a[], int n, int k){         // Hash map to store frequencies of    // elements of prefix and suffix    Map 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// distinctstatic 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 Codepublic 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 lengthfrom collections import defaultdict # Function to check if elements of# Prefix and suffix of each sub array# of size K are pairwise distinct or notdef 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# distinctdef 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 codeif __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 lengthusing 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 notstatic bool check(int []a, int n, int k){         // Hash map to store frequencies of    // elements of prefix and suffix    Dictionary m = new Dictionary();         // 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// distinctstatic 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 Codepublic 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


Output:
2

Time Complexity: O(N * log(N)), where N is the size of the array.

Auxiliary Space: O(N)

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