Maximum number of unique values in the array after performing given operations

18 Apr, 2022
Given an array arr[] of size N and every element in the array arr[] is in the range [1, N] and the array may contain duplicates. The task is to find the maximum number of unique values that can be obtained such that the value at any index i can either be:  
Increased by 1.
Decreased by 1.
Left as it is.
Note: The operation can be performed only once with each index and have to be performed for all the indices in the array arr[].Examples:  
Input: arr[] = {1, 2, 4, 4} Output: 4 Explanation: One way is to perform the following operations for the index: 1: Leave the value at the first index(1) as it is. 2: Leave the value at the second index(2) as it is. 3: Leave the value at the third index(4) as it is. 4: Increment the value at the fourth index(4) by 1. Then the array becomes {1, 2, 4, 5} and there are 4 unique values.Input: arr[]={3, 3, 3, 3} Output: 3 Explanation: One way is to perform the following operations for the index: 1: Leave the value at the first index(3) as it is. 2: Decrement the value at the second index(3) by 1. 3: Leave the value at the third index(3) as it is. 4: Increment the value at the fourth index(3) by 1. Then the array becomes {3, 2, 3, 4} and there are 3 unique values.  
 
Approach:  
For some arbitrary element X present in the array at index i, we decide what operation to perform on it by taking the following things into consideration: We decrement the value X by 1 if the value (X – 1) is not present in the array and there are one or more other X’s present in the array at different indices.We don’t change the value X if the value X is present only once on the array.We increment the value X by 1 if the value (X + 1) is not present in the array and there are one or more other X’s present in the array at different indices.
We decrement the value X by 1 if the value (X – 1) is not present in the array and there are one or more other X’s present in the array at different indices.We don’t change the value X if the value X is present only once on the array.We increment the value X by 1 if the value (X + 1) is not present in the array and there are one or more other X’s present in the array at different indices.
We decrement the value X by 1 if the value (X – 1) is not present in the array and there are one or more other X’s present in the array at different indices.
We don’t change the value X if the value X is present only once on the array.
We increment the value X by 1 if the value (X + 1) is not present in the array and there are one or more other X’s present in the array at different indices.
By taking the above decisions for every element, we can be sure that the final count of unique elements which we get is the maximum.
However, to perform the above steps for every index and count the occurrences of the element X and continuously update the array arr[], the time taken would be quadratic which is not feasible for large-sized arrays.
One alternative to reduce the time complexity is to initially sort the array. By sorting, all the elements in the array are grouped and all the repeated values come together.
After sorting the array, since the range of the numbers is already given and it is fixed, a hash map can be used where the keys of the hash are the numbers in the range [1, N] and the value for each key is boolean which is used to determine if the key is present in the array or not.
In this problem, since the indices themselves are the keys of the hash, an array freq[] of size (N + 2) is used to implement the hash.
Below is the implementation of the above approach:  
C++
C
Java
Python3
C#
Javascript
// C++ program to find the maximum number of// unique values in the array #include <bits/stdc++.h>using namespace std; // Function to find the maximum number of// unique values in the vectorint uniqueNumbers(vector<int> arr, int n){    // Sorting the given vector    sort(arr.begin(),arr.end());     // This array will store the frequency    // of each number in the array    // after performing the given operation      // initialise the array with 0    vector<int> freq(n+2,0) ;     // Loop to apply operation on    // each element of the array    for (int x = 0; x < n; x++) {         // Incrementing the value at index x        // if the value  arr[x] - 1 is        // not present in the array        if (freq[arr[x] - 1] == 0) {            freq[arr[x] - 1]++;        }         // If arr[x] itself is not present, then it        // is left as it is        else if (freq[arr[x]] == 0) {            freq[arr[x]]++;        }         // If both arr[x] - 1 and arr[x] are present        // then the value is incremented by 1        else {            freq[arr[x] + 1]++;        }    }     // Variable to store the number of unique values    int unique = 0;     // Finding the unique values    for (int x = 0; x <= n + 1; x++) {        if (freq[x]) {            unique++;        }    }     // Returning the number of unique values    return unique;} // Driver Codeint main(){    vector<int> arr= { 3, 3, 3, 3 };     // Size of the vector    int n = arr.size();     int ans = uniqueNumbers(arr, n);    cout << ans;    return 0;}
// C program to find the maximum number of// unique values in the array#include <stdio.h>#include <stdlib.h> // comparison function for qsortint cmpfunc (const void * a, const void * b) {   return ( *(int*)a - *(int*)b );} // Function to find the maximum number of// unique values in the arrayint uniqueNumbers(int* arr, int n){    // Sorting the given array       qsort(arr, n, sizeof(int), cmpfunc);     // This array will store the frequency    // of each number in the array    // after performing the given operation    int freq[n+2];      // initialise the array with 0      for(int i = 0 ; i < n + 2 ; i++)      freq[i] = 0;     // Loop to apply operation on    // each element of the array    for (int x = 0; x < n; x++) {         // Incrementing the value at index x        // if the value  arr[x] - 1 is        // not present in the array        if (freq[arr[x] - 1] == 0) {            freq[arr[x] - 1]++;        }         // If arr[x] itself is not present, then it        // is left as it is        else if (freq[arr[x]] == 0) {            freq[arr[x]]++;        }         // If both arr[x] - 1 and arr[x] are present        // then the value is incremented by 1        else {            freq[arr[x] + 1]++;        }    }     // Variable to store the number of unique values    int unique = 0;     // Finding the unique values    for (int x = 0; x <= n + 1; x++) {        if (freq[x]) {            unique++;        }    }     // Returning the number of unique values    return unique;} // Driver Codeint main(){    int arr[] = { 3, 3, 3, 3 };     // Size of the array    int n = sizeof(arr)/sizeof(arr[0]);     int ans = uniqueNumbers(arr, n);    printf("%d", ans);    return 0;} // This code is contributed by phalashi.
// Java program to find the maximum number of// unique values in the arrayimport java.util.*; class GFG {         // Function to find the maximum number of    // unique values in the array    static int uniqueNumbers(int arr[], int n)    {        // Sorting the given array        Arrays.sort(arr);             // This array will store the frequency        // of each number in the array        // after performing the given operation        int freq[] = new int[n + 2];             // Initialising the array with all zeroes        for(int i = 0; i < n + 2; i++)            freq[i] = 0;         // Loop to apply operation on        // each element of the array        for (int x = 0; x < n; x++) {                 // Incrementing the value at index x            // if the value arr[x] - 1 is            // not present in the array            if (freq[arr[x] - 1] == 0) {                freq[arr[x] - 1]++;            }                 // If arr[x] itself is not present, then it            // is left as it is            else if (freq[arr[x]] == 0) {                freq[arr[x]]++;            }                 // If both arr[x] - 1 and arr[x] are present            // then the value is incremented by 1            else {                freq[arr[x] + 1]++;            }        }             // Variable to store the number of unique values        int unique = 0;             // Finding the unique values        for (int x = 0; x <= n + 1; x++) {            if (freq[x] != 0) {                unique++;            }        }             // Returning the number of unique values        return unique;    }         // Driver Code    public static void main (String[] args)    {        int []arr = { 3, 3, 3, 3 };             // Size of the array        int n = 4;             int ans = uniqueNumbers(arr, n);        System.out.println(ans);    }} // This code is contributed by Yash_R
# Python program to find the maximum number of# unique values in the array # Function to find the maximum number of# unique values in the arraydef uniqueNumbers(arr, n):     # Sorting the given array    arr.sort()     # This array will store the frequency    # of each number in the array    # after performing the given operation    freq =[0]*(n + 2)     # Loop to apply the operation on    # each element of the array    for val in arr:         # Incrementing the value at index x        # if the value  arr[x] - 1 is        # not present in the array        if(freq[val-1]== 0):            freq[val-1]+= 1         # If arr[x] itself is not present, then it        # is left as it is        elif(freq[val]== 0):            freq[val]+= 1         # If both arr[x] - 1 and arr[x] are present        # then the value is incremented by 1        else:            freq[val + 1]+= 1         # Variable to store the    # number of unique values    unique = 0     # Finding the number of unique values    for val in freq:        if(val>0):            unique+= 1         return unique # Driver codeif __name__ == "__main__":    arr =[3, 3, 3, 3]    n = 4    print(uniqueNumbers(arr, n))
// C# program to find the maximum number of// unique values in the arrayusing System; class GFG {         // Function to find the maximum number of    // unique values in the array    static int uniqueNumbers(int []arr, int n)    {        // Sorting the given array        Array.Sort(arr);             // This array will store the frequency        // of each number in the array        // after performing the given operation        int []freq = new int[n + 2];             // Initialising the array with all zeroes        for(int i = 0; i < n + 2; i++)            freq[i] = 0;         // Loop to apply operation on        // each element of the array        for (int x = 0; x < n; x++) {                 // Incrementing the value at index x            // if the value arr[x] - 1 is            // not present in the array            if (freq[arr[x] - 1] == 0) {                freq[arr[x] - 1]++;            }                 // If arr[x] itself is not present, then it            // is left as it is            else if (freq[arr[x]] == 0) {                freq[arr[x]]++;            }                 // If both arr[x] - 1 and arr[x] are present            // then the value is incremented by 1            else {                freq[arr[x] + 1]++;            }        }             // Variable to store the number of unique values        int unique = 0;             // Finding the unique values        for (int x = 0; x <= n + 1; x++) {            if (freq[x] != 0) {                unique++;            }        }             // Returning the number of unique values        return unique;    }         // Driver Code    public static void Main (string[] args)    {        int []arr = { 3, 3, 3, 3 };             // Size of the array        int n = 4;             int ans = uniqueNumbers(arr, n);        Console.WriteLine(ans);    }} // This code is contributed by Yash_R
<script>// javascript program to find the maximum number of// unique values in the array     // Function to find the maximum number of    // unique values in the array    function uniqueNumbers(arr , n) {        // Sorting the given array        arr.sort((a,b)=>a-b);         // This array will store the frequency        // of each number in the array        // after performing the given operation        var freq = Array(n + 2).fill(0);         // Initialising the array with all zeroes        for (i = 0; i < n + 2; i++)            freq[i] = 0;         // Loop to apply operation on        // each element of the array        for (x = 0; x < n; x++) {             // Incrementing the value at index x            // if the value arr[x] - 1 is            // not present in the array            if (freq[arr[x] - 1] == 0) {                freq[arr[x] - 1]++;            }             // If arr[x] itself is not present, then it            // is left as it is            else if (freq[arr[x]] == 0) {                freq[arr[x]]++;            }             // If both arr[x] - 1 and arr[x] are present            // then the value is incremented by 1            else {                freq[arr[x] + 1]++;            }        }         // Variable to store the number of unique values        var unique = 0;         // Finding the unique values        for (x = 0; x <= n + 1; x++) {            if (freq[x] != 0) {                unique++;            }        }         // Returning the number of unique values        return unique;    }     // Driver Code             var arr = [ 3, 3, 3, 3 ];         // Size of the array        var n = 4;         var ans = uniqueNumbers(arr, n);        document.write(ans); // This code contributed by Rajput-Ji</script>
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Time Complexity Analysis:
The time taken to sort the given array is O(N * log(N)) where N is the size of the array.
The time taken to run a loop over the sorted array to perform the operations is O(N).
The time taken to run a loop over the hash to count the unique values is O(N).
Therefore, the overall time complexity is O(N * log(N)) + O(N) + O(N). Since N * log(N) is greater, the final time complexity of the above approach is O(N * log(N)).
Auxiliary Space: O(N)
The extra space taken by the freq array is O(N). Therefore the auxiliary space is O(N).
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