Absolute distinct count in a sorted array

Given a sorted array of integers, return the number of distinct absolute values among the elements of the array. The input can contain duplicates values.
Examples:

Input: [-3, -2, 0, 3, 4, 5]
Output: 5
There are 5 distinct absolute values
among the elements of this array, i.e.
0, 2, 3, 4 and 5)

Input:  [-1, -1, -1, -1, 0, 1, 1, 1, 1]
Output: 2

Input:  [-1, -1, -1, -1, 0]
Output: 2

Input:  [0, 0, 0]
Output: 1

The solution should do only one scan of the input array and should not use any extra space. i.e. expected time complexity is O(n) and auxiliary space is O(1).

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

One simple solution is to use set. For each element of the input array, we insert its absolute value in the set. As set doesn’t support duplicate elements, the element’s absolute value will be inserted only once. Therefore, the required count is size of the set.

Below is the implementation of the idea.

 // C++ program to find absolute distinct // count of an array in O(n) time. #include using namespace std;    // The function returns number of // distinct absolute values among // the elements of the array int distinctCount(int arr[], int n) {     unordered_set s;        // Note that set keeps only one     // copy even if we try to insert     // multiple values     for (int i = 0 ; i < n; i++)         s.insert(abs(arr[i]));        return s.size(); }    // Driver code int main() {     int arr[] = {-2, -1, 0, 1, 1};     int n = sizeof(arr)/sizeof(arr);        cout << "Count of absolute distinct values : "          << distinctCount(arr, n);        return 0; }

 // java code to find absolute distinct // count of an array in O(n) time. import java.util.*;    class GFG  {     // The function returns number of     // distinct absolute values among     // the elements of the array     static int distinctCount(int arr[], int n)     {         Set s = new HashSet ();            // Note that set keeps only one         // copy even if we try to insert         // multiple values         for (int i = 0 ; i < n; i++)         s.add(Math.abs(arr[i]));                    return s.size();                }            // Driver code     public static void main(String[] args)     {         int arr[] = {-2, -1, 0, 1, 1};         int n = arr.length;            System.out.println("Count of absolute distinct values : "                            + distinctCount(arr, n));                } }    // This code is contributed by prerna saini

 # Python3 code to find absolute distinct # count of an array in O(n) time.    # This function returns number of # distinct absolute values among # the elements of the array def distinctCount(arr, n):     s = set()            # set keeps all unique elements     for i in range(n):         s.add(abs(arr[i]))     return len(s)    # Driver Code arr = [-2, -1, 0, 1, 1] n = len(arr) print("Count of absolute distinct values:",                      distinctCount(arr, n))    # This code is contributed # by Adarsh_Verma

 // C# code to find absolute distinct // count of an array in O(n) time. using System; using System.Collections.Generic;    class GFG  {     // The function returns number of     // distinct absolute values among     // the elements of the array     static int distinctCount(int []arr, int n)     {         HashSet s = new HashSet();            // Note that set keeps only one         // copy even if we try to insert         // multiple values         for (int i = 0 ; i < n; i++)         s.Add(Math.Abs(arr[i]));                    return s.Count;                }            // Driver code     public static void Main()     {         int []arr = {-2, -1, 0, 1, 1};         int n = arr.Length;            Console.Write("Count of absolute distinct values : "                         + distinctCount(arr, n));                } }    // This code is contributed by PrinciRaj1992

Output :
Count of absolute distinct values : 3

Time Complexity : O(n)
Auxiliary Space : O(n)

The above implementation takes O(n) extra space, how to do in O(1) extra space?

The idea is to take advantage of the fact that the array is already Sorted. We initialize the count of distinct elements to number of elements in the array. We start with two index variables from two corners of the array and check for pair in the input array with sum as 0. If pair with 0 sum is found or duplicates are encountered, we decrement the count of distinct elements.Finally we return the updated count.

Below is the implementation of above approach.

 // C++ program to find absolute distinct // count of an array using O(1) space. #include using namespace std;    // The function returns return number // of distinct absolute values // among the elements of the array int distinctCount(int arr[], int n) {     // initialize count as number of elements     int count = n;     int i = 0, j = n - 1, sum = 0;        while (i < j)     {         // Remove duplicate elements from the         // left of the current window (i, j)         // and also decrease the count         while (i != j && arr[i] == arr[i + 1])             count--, i++;            // Remove duplicate elements from the         // right of the current window (i, j)         // and also decrease the count         while (i != j && arr[j] == arr[j - 1])             count--, j--;            // break if only one element is left         if (i == j)             break;            // Now look for the zero sum pair         // in current window (i, j)         sum = arr[i] + arr[j];            if (sum == 0)         {             // decrease the count if (positive,             // negative) pair is encountered             count--;             i++, j--;         }         else if(sum < 0)             i++;         else             j--;     }        return count; }    // Driver code int main() {     int arr[] = {-2, -1, 0, 1, 1};     int n = sizeof(arr)/sizeof(arr);        cout << "Count of absolute distinct values : "          << distinctCount(arr, n);        return 0; }

 // Java program to find absolute distinct  // count of an array using O(1) space.     import java.io.*;    class GFG {               // The function returns return number  // of distinct absolute values  // among the elements of the array  static int distinctCount(int arr[], int n)  {      // initialize count as number of elements      int count = n;      int i = 0, j = n - 1, sum = 0;         while (i < j)      {          // Remove duplicate elements from the          // left of the current window (i, j)          // and also decrease the count          while (i != j && arr[i] == arr[i + 1])          {             count--;             i++;          }         // Remove duplicate elements from the          // right of the current window (i, j)          // and also decrease the count          while (i != j && arr[j] == arr[j - 1])          {             count--;             j--;          }         // break if only one element is left          if (i == j)              break;             // Now look for the zero sum pair          // in current window (i, j)          sum = arr[i] + arr[j];             if (sum == 0)          {              // decrease the count if (positive,              // negative) pair is encountered              count--;              i++;             j--;          }          else if(sum < 0)              i++;          else             j--;      }         return count;  }     // Driver code             public static void main (String[] args) {            int arr[] = {-2, -1, 0, 1, 1};      int n = arr.length;         System.out.println ("Count of absolute distinct values : "+              distinctCount(arr, n));                           } }

 # Python3 program to find absolute distinct # count of an array using O(1) space.    # The function returns return number # of distinct absolute values # among the elements of the array def distinctCount(arr, n):        # initialize count as number of elements     count = n;     i = 0; j = n - 1; sum = 0;        while (i < j):                    # Remove duplicate elements from the         # left of the current window (i, j)         # and also decrease the count         while (i != j and arr[i] == arr[i + 1]):             count = count - 1;              i = i + 1;            # Remove duplicate elements from the         # right of the current window (i, j)         # and also decrease the count         while (i != j and arr[j] == arr[j - 1]):             count = count - 1;             j = j - 1;            # break if only one element is left         if (i == j):             break;            # Now look for the zero sum pair         # in current window (i, j)         sum = arr[i] + arr[j];            if (sum == 0):                        # decrease the count if (positive,             # negative) pair is encountered             count = count - 1;             i = i + 1;             j = j - 1;                        elif(sum < 0):             i = i + 1;         else:             j = j - 1;            return count;       # Driver code arr = [-2, -1, 0, 1, 1]; n = len(arr);    print("Count of absolute distinct values : ",                        distinctCount(arr, n));    # This code is contributed # by Akanksha Rai

 //C# program to find absolute distinct  // count of an array using O(1) space.  using System;    class GFG {               // The function returns return number  // of distinct absolute values  // among the elements of the array  static int distinctCount(int []arr, int n)  {      // initialize count as number of elements      int count = n;      int i = 0, j = n - 1, sum = 0;         while (i < j)      {          // Remove duplicate elements from the          // left of the current window (i, j)          // and also decrease the count          while (i != j && arr[i] == arr[i + 1])          {             count--;             i++;          }         // Remove duplicate elements from the          // right of the current window (i, j)          // and also decrease the count          while (i != j && arr[j] == arr[j - 1])          {             count--;             j--;          }         // break if only one element is left          if (i == j)              break;             // Now look for the zero sum pair          // in current window (i, j)          sum = arr[i] + arr[j];             if (sum == 0)          {              // decrease the count if (positive,              // negative) pair is encountered              count--;              i++;             j--;          }          else if(sum < 0)              i++;          else             j--;      }         return count;  }     // Driver code             public static void Main () {            int []arr = {-2, -1, 0, 1, 1};      int n = arr.Length;         Console.WriteLine("Count of absolute distinct values : "+             distinctCount(arr, n));                // This code is contributed by inder_verma         } }



Output :
Count of absolute distinct values : 3

Time Complexity: O(n)
Auxiliary Space: O(1)

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Article Tags :
Practice Tags :