# Maximum Unique Element in every subarray of size K

Given an array and an integer K. We need to find the maximum of every segment of length K which has no duplicates in that segment.

Examples:

Input : a[] = {1, 2, 2, 3, 3},
K = 3.
Output : 1 3 2
For segment (1, 2, 2), Maximum = 1.
For segment (2, 2, 3), Maximum = 3.
For segment (2, 3, 3), Maximum = 2.

Input : a[] = {3, 3, 3, 4, 4, 2},
K = 4.
Output : 4 Nothing 3

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to run two loops. For every subarray find all distinct elements and print maximum unique element.

An efficient solution is to use sliding window technique. We maintain two structures in every window.
1) A hash table to store counts of all elements in current window.
2) A self balancing BST (implemented using set in C++ STL and TreeSet in Java). The idea is to quickly find maximum element and update maximum element.

We process first K-1 elements and store their counts in hash table. We also store unique elements in set. Now we one by one process last element of every window. If current element is unique, we add it to set. We also increase its count. After processing last element, we print maximum from set. Before starting next iteration, we remove first element of previous window.

## C++

 // C++ code to calculate maximum unique // element of every segment of array #include using namespace std;    void find_max(int A[], int N, int K) {     // Storing counts of first K-1 elements     // Also storing distinct elements.     map Count;     for (int i = 0; i < K - 1; i++)         Count[A[i]]++;     set Myset;     for (auto x : Count)         if (x.second == 1)             Myset.insert(x.first);        // Before every iteration of this loop,     // we maintain that K-1 elements of current     // window are processed.     for (int i = K - 1; i < N; i++) {            // Process K-th element of current window         Count[A[i]]++;         if (Count[A[i]] == 1)             Myset.insert(A[i]);         else             Myset.erase(A[i]);            // If there are no distinct         // elements in current window         if (Myset.size() == 0)             printf("Nothing\n");            // Set is ordered and last element         // of set gives us maximum element.         else             printf("%d\n", *Myset.rbegin());            // Remove first element of current         // window before next iteration.         int x = A[i - K + 1];         Count[x]--;         if (Count[x] == 1)             Myset.insert(x);         if (Count[x] == 0)             Myset.erase(x);     } }    // Driver code int main() {     int a[] = { 1, 2, 2, 3, 3 };     int n = sizeof(a) / sizeof(a[0]);     int k = 3;     find_max(a, n, k);     return 0; }

## Java

 // Java code to calculate maximum unique // element of every segment of array import java.io.*; import java.util.*; class GFG {        static void find_max(int[] A, int N, int K)     {         // Storing counts of first K-1 elements         // Also storing distinct elements.         HashMap Count = new HashMap<>();         for (int i = 0; i < K - 1; i++)             if (Count.containsKey(A[i]))                 Count.put(A[i], 1 + Count.get(A[i]));             else                 Count.put(A[i], 1);            TreeSet Myset = new TreeSet();         for (Map.Entry x : Count.entrySet()) {             if (Integer.parseInt(String.valueOf(x.getValue())) == 1)                 Myset.add(Integer.parseInt(String.valueOf(x.getKey())));         }            // Before every iteration of this loop,         // we maintain that K-1 elements of current         // window are processed.         for (int i = K - 1; i < N; i++) {                // Process K-th element of current window             if (Count.containsKey(A[i]))                 Count.put(A[i], 1 + Count.get(A[i]));             else                 Count.put(A[i], 1);                if (Integer.parseInt(String.valueOf(Count.get(A[i]))) == 1)                 Myset.add(A[i]);             else                 Myset.remove(A[i]);                // If there are no distinct             // elements in current window             if (Myset.size() == 0)                 System.out.println("Nothing");                // Set is ordered and last element             // of set gives us maximum element.             else                 System.out.println(Myset.last());                // Remove first element of current             // window before next iteration.             int x = A[i - K + 1];             Count.put(x, Count.get(x) - 1);                if (Integer.parseInt(String.valueOf(Count.get(x))) == 1)                 Myset.add(x);             if (Integer.parseInt(String.valueOf(Count.get(x))) == 0)                 Myset.remove(x);         }     }        // Driver code     public static void main(String args[])     {         int[] a = { 1, 2, 2, 3, 3 };         int n = a.length;         int k = 3;         find_max(a, n, k);     } }    // This code is contributed by rachana soma

## Python3

 # Python3 code to calculate maximum unique # element of every segment of array def find_max(A, N, K):            # Storing counts of first K-1 elements     # Also storing distinct elements.     Count = dict()     for i in range(K - 1):         Count[A[i]] = Count.get(A[i], 0) + 1        Myset = dict()     for x in Count:         if (Count[x] == 1):             Myset[x] = 1        # Before every iteration of this loop,     # we maintain that K-1 elements of current     # window are processed.     for i in range(K - 1, N):            # Process K-th element of current window         Count[A[i]] = Count.get(A[i], 0) + 1            if (Count[A[i]] == 1):             Myset[A[i]] = 1         else:             del Myset[A[i]]            # If there are no distinct         # elements in current window         if (len(Myset) == 0):             print("Nothing")            # Set is ordered and last element         # of set gives us maximum element.         else:             maxm = -10**9             for i in Myset:                 maxm = max(i, maxm)             print(maxm)            # Remove first element of current         # window before next iteration.         x = A[i - K + 1]         if x in Count.keys():             Count[x] -= 1             if (Count[x] == 1):                 Myset[x] = 1             if (Count[x] == 0):                 del Myset[x]    # Driver code a = [1, 2, 2, 3, 3 ] n = len(a) k = 3 find_max(a, n, k)    # This code is contributed  # by mohit kumar

Output:

1
3
2

Time Complexity : O(N Log K)

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