Find unique number in Integer array/list using python

Question

Given an integer array/list(ARR) of size N. Where N is equal to [2M + 1]. Now, in the given array/list, 'M' numbers are present twice and one number is present only once. Find and return that number that is unique in the array/list.

Note: Unique element is always present in the array/list according to the given condition.

I have done this question using WHILE loop, but when I am trying to do this using FOR loop it is not working

and, I have seen some codes for this type of problem, but as I am a learner and have done only some basic python so and don't know yet about the COMPLEXITY(O^N).

please help and thanks in advance

WHILE loop code:


def find_unique(li,n):
    for i in range(n):
        j = 0
        while(j < n):
            if i != j:
                if li[i] == li[j]:
                    break
            j = j + 1
        if j == n:
            return li[i]
n = 7
li = [2,3,1,6,3,6,2]
ele = find_unique(li,n)
print(ele)

FOR loop code:

def find_unique(li,n):
    for i in range(n):
        for j in range(0,n):
            if i == j:
                continue
            if(li[i] == li[j] and j ==n):
                    return li[i]
        

n = 7
li = [2,3,1,6,3,6,2]
ele = find_unique(li,n)
print(ele)

convert to numpy.ndarray or a pandas dataframe . They both have a function for it. numpy.org/doc/stable/reference/generated/numpy.unique.html and pandas.pydata.org/docs/reference/api/pandas.unique.html . You can look at some examples in the documentation. Python is slow in for loops, so, whenever you can, convert to numpy array and use the functions of that as they're quite fast. It's called vectorization if you want to read more about it. — Sina Meftah
– Sina Meftah, Commented May 23, 2021 at 10:12
@SinaMeftah it is not relevant to the question as the author wants to find the issue in code. — DeGo
– DeGo, Commented May 23, 2021 at 10:14
@DeepakGouda yes, i know j is not reaching n . but, how to correct it. — Nipun
– Nipun, Commented May 23, 2021 at 10:15

Aditya Vikram · Accepted Answer · 2021-05-23 10:15:21Z

3

def unique(arr,N):
    ans=0
    for i in range(N):
        ans=ans^arr[i]
    return ans

answered May 23, 2021 at 10:15

Aditya Vikram

511 bronze badge

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4 Comments

ChimeraObscura Over a year ago

I like this answer - I knew there had to be a mathematical way to do this. (BTW you can simplify this by iterating over the list directly. "for n in arr: ans = ans ^ n")

pppig Over a year ago

I have to say this idea is great, but it doesn't seem to be all that satisfying. For example, [2, 3, 6, 3, 6, 2, 3, 11].

Aditya Vikram Over a year ago

@pppig your test case is fine but in the question it is already provided every numbers are present in pairs and there is only one unique value, but in your case 3 is present 3 times which is violating the problem statement . Hope I was able to answer your query.

pppig Over a year ago

@AdityaVikram Oh, you are right, after reading your answer, I can not help but want to understand its thinking, so that I forget the description of the question, this idea is really great.

pppig · Accepted Answer · 2021-05-23 10:18:02Z

1

from collections import Counter

li = [2, 3, 1, 6, 3, 6, 2]
print([key for key, val in Counter(li).items() if val == 1])

# [1]

answered May 23, 2021 at 10:18

pppig

1,2951 gold badge9 silver badges12 bronze badges

Comments

user15801675 · Accepted Answer · 2021-05-23 10:14:42Z

0

I just replaced some code

def find_unique(li,n):
    for i in range(n):
        for j in range(0,n):
            if i != j:
                if li[i] == li[j]:
                    break
            j = j + 1
        if j == n:
            return li[i]
n = 7
li = [2,3,1,6,3,6,2]
ele = find_unique(li,n)
print(ele)

answered May 23, 2021 at 10:14

user15801675

1 Comment

Nipun Over a year ago

thanks, for the answer. i just want to know why there is j = j + 1, if that is being done by the for loop

Naaman · Accepted Answer · 2021-05-23 10:20:32Z

0

from collections import Counter
def find_unique(li,n):
    a = Counter(li).most_common()
    return a[len(a)-1][0]    
n = 7
li = [2,3,1,6,3,6,2]
ele = find_unique(li,n)
print(ele)

answered May 23, 2021 at 10:20

Naaman

723 bronze badges

Comments

Anand Narayanan · Accepted Answer · 2021-05-23 10:43:21Z

0

Since this is a very specific case - with all items repeating twice but one. You can pull off a simple one-liner using this property

find_unique = lambda li, n: 2*sum(set(li)) - sum(li)

Once you have defined the function - Now use it

n = 7
li = [2,3,1,6,3,6,2]
ele = find_unique(li, n)
print(ele)

answered May 23, 2021 at 10:43

Anand Narayanan

831 silver badge6 bronze badges

Comments

Tarun Tanmay · Accepted Answer · 2022-08-31 19:59:26Z

0

JAVA CODE:

public class Solution{

public static int findUnique(int[] arr){
    
    int n = arr.length;
    int similar = arr[0]; //first element of array is stored as similar
    //now compare similar with every element of array
    for(int i=1; i<=n-1; i++)
    {
        similar = similar ^ arr[i];
        //XOR operation 1^1 || 0 ^0 = 0
        //XOR operation 1^0 || 0^1= 1
        
        //therefore, similar ^ similar = 0
        //similar ^ unique  = unique
        //similar takes the value of unique after XOR
    }
    
    return similar; //unique element is returned using similar named variable
}

}

answered Aug 31, 2022 at 19:59

Tarun Tanmay

793 bronze badges

Comments

Juan Saavedra · Accepted Answer · 2023-01-28 17:53:26Z

0

Check this code:

def find_unique(li,n):
    unique_numbers = list(set(li))
    for i in unique_numbers:
        if li.count(i) == 1:
            return i
    return "No number once"

It return the number that is once in the array. Size N is not necessary. Set and Count are functions from list variables. Set return unique elements in the array and count return the number of times an element is in the array.

answered Jan 28, 2023 at 17:53

Juan Saavedra

614 bronze badges

Comments

Rabiul Islam · Accepted Answer · 2023-11-28 14:36:50Z

0

From an array, to find unique values with its count; just use np.unique function as follows:

unique_values = np.unique(any_2d_array.flatten(), return_counts=True)

For example after counting an image (of mask), I got the result. That means the array contains 3 unique values 1, 2, 3 and these appears 22938, 198766 and 18296 times, respectively.

To get count into a separate variable, you can use like this:

unique_values, count = np.unique(any_2d_array.flatten(), return_counts=True)

answered Nov 28, 2023 at 14:36

Rabiul Islam

1311 silver badge4 bronze badges

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