Absolute distinct count in a sorted array
Last Updated :
23 Jul, 2025
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).
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.
Algorithm:
Step 1: Define a function named distinctCount which takes two parameters, an array of integers arr[] and an integer n.
Step 2: Declare an empty HashSet named s.
Step 3: Traverse the array arr[] using a for loop from i=0 to i<n.
Step 4: Insert the absolute value of arr[i] in the set s using the add() function.
Step 5: After the for loop, return the size of set s using the size() function.
Below is the implementation of the idea.
C++
// C++ program to find absolute distinct
// count of an array in O(n) time.
#include <bits/stdc++.h>
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<int> 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[0]);
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<Integer> s = new HashSet<Integer> ();
// 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<int> s = new HashSet<int>();
// 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
<script>
// Javascript code to find absolute distinct
// count of an array in O(n) time.
// The function returns number of
// distinct absolute values among
// the elements of the array
function distinctCount(arr, n)
{
let s = new Set();
// Note that set keeps only one
// copy even if we try to insert
// multiple values
for (let i = 0 ; i < n; i++)
s.add(Math.abs(arr[i]));
return s.size;
}
let arr = [-2, -1, 0, 1, 1];
let n = arr.length;
document.write("Count of absolute distinct values : "
+ distinctCount(arr, n));
</script>
OutputCount 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++
// C++ program to find absolute distinct
// count of an array using O(1) space.
#include <bits/stdc++.h>
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[0]);
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
}
}
<?php
// PHP 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
function 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 && $arr[$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
$arr = array(-2, -1, 0, 1, 1);
$n = sizeof($arr);
echo "Count of absolute distinct values : " .
distinctCount($arr, $n);
// This code is contributed
// by Akanksha Rai
?>
<script>
// Javascript 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
function distinctCount(arr, n)
{
// initialize count as number of elements
let count = n;
let 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;
}
let arr = [-2, -1, 0, 1, 1];
let n = arr.length;
document.write(
"Count of absolute distinct values : " + distinctCount(arr, n));
</script>
OutputCount of absolute distinct values : 3
Time Complexity: O(n)
Auxiliary Space: O(1)
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