# Given an array of integers, find the first integer that is unique

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## Answers

1. #### Emerson

• 2015/10/27

Inserting to a map is `O(log n)` not `O(n log n)` so inserting `n` keys will be `n log n`. also its better to use `set`.

2. #### Riley

• 2016/12/2

put integer (number as key, its index as value) to it one by one (O(n^2 lgn)) , if have duplicate, remove the entry from the map (O(lg n)) ,

3. #### Ira

• 2015/7/30

Given an array of integers, find the first integer that is unique. my solution: use std::map. put integer (number as key, its index as value) to it one by one (O(n^2 lgn)), if have duplicate, remove the entry from the map (O(lg n)), after putting all numbers into the map, iterate the map and find the key with smallest index O(n).

4. #### Smith

• 2018/9/27

Given an array of integers, find the first integer that is unique. my solution: use std::map. put integer (number as key, its index as value) to it one by

5. #### Scott

• 2021/6/14

Although it's O(n^2), the following has small coefficients, isn't too bad on the cache, and uses `memmem()` which is fast.

``` for(int x=0;x<len-1;x++)
if(memmem(&array[x+1], sizeof(int)*(len-(x+1)), array[x], sizeof(int))==NULL &&
memmem(&array[x+1], sizeof(int)*(x-1), array[x], sizeof(int))==NULL)
return array[x];
```
6. #### Baker

• 2018/5/18

Given an array of integers, find the first integer that is unique. my solution: use std::map. put integer (number as key, its index as value) to it one by one (O (n^2 lgn)), if have duplicate, remove the entry from the map (O (lg n)), after putting all numbers into the map, iterate the map and find the key with smallest index O (n).

7. #### Adriel

• 2017/10/31

1) Copy the given array to an auxiliary array temp[]. 2) Sort the temp array using a O(nLogn) time sorting algorithm. 3) Scan the input array

8. #### Elijah

• 2020/6/1

By returning an array you can still get the first unique by applying [0] to the result and indicate no unique entry with an empty array. It also does not change your complexity: avg. n+k with k being the count for different values, but still 2n tops.

9. #### Madden

• 2018/5/28

All this for no reason. Just using 2 `for-loops` & a variable would give you a simple `O(n^2)` algo.

If you are taking all the trouble of using a hash map, then it might as well be what @Micheal Goldshteyn suggests

UPDATE: I know this question is 1 year old. But was looking through the questions I answered and came across this. Thought there is a better solution than using a hashtable.

When we say unique, we will have a pattern. Eg: [5, 5, 66, 66, 7, 1, 1, 77]. In this lets have moving window of 3. first consider (5,5,66). we can easily estab. that there is duplicate here. So move the window by 1 element so we get (5,66,66). Same here. move to next (66,66,7). Again dups here. next (66,7,1). No dups here! take the middle element as this has to be the first unique in the set. The left element belongs to the dup so could 1. Hence 7 is the first unique element.

space: O(1) time: O(n) * O(m^2) = O(n) * 9 ≈ O(n)

10. #### Kannon

• 2021/3/17

Given an array of integers. All numbers occur twice except one number which occurs once. Find the number in O(n) time & constant extra

11. #### Nasir

• 2017/10/18

1.Take xor of all elements,call it xor. 2.Set the rightmost bit of xor and store it in B. 3.Do the following: number_1=0,number_2=0; for (i = 0 to n-1) { if (array [i] & B) number_1 = number_1^array [i]; else number_2 = number_2^array [i]; } The number_1 and number_2 are the required numbers. Share.

12. #### Marlon

• 2021/7/18

Given an array, find the first integer, which is unique in the array. Unique means the number does not repeat and appears only once in the whole array.

13. #### Ruiz

• 2015/3/27
```public static string firstUnique(int[] input)
{
int size = input.Length;
bool[] dupIndex = new bool[size];
for (int i = 0; i < size; ++i)
{
if (dupIndex[i])
{
continue;
}
else if (i == size - 1)
{
return input[i].ToString();
}
for (int j = i + 1; j < size; ++j)
{
if (input[i]==input[j])
{
dupIndex[j] = true;
break;
}
else if (j == size - 1)
{
return input[i].ToString();
}
}
}
return "No unique element";
}
```
14. #### Bernard

• 2019/6/24

1) Copy the given array to an auxiliary array temp[]. 2) Sort the temp array using a O(nLogn) time sorting algorithm. 3) Scan the input array from left to right. For every element, count its occurrences in temp[] using binary search. As soon as we find an element that occurs more than once, we return the element. This step can be done in O(nLogn) time.

15. #### Gideon

• 2019/2/26

Write a program that, Given an array of n integers and given a number K, Ans: No, the pair should consist of two different array elements.

16. #### Martinelli

• 2017/4/25

/** * * @param inputArray * @return returns unique Element in the array. * -1 if no unique element is available in the array. */ public static int xorApproach(int[] inputArray) { int result = 0; for(int i=0;i<inputArray.length;i++) { result ^= inputArray[i]; } return (result>0 ? result : -1); }

17. #### Richard

• 2019/7/14

I believe that the following would be the optimal solution, at least based on time / space complexity:

Step 1: Store the integers in a hash map, which holds the integer as a key and the count of the number of times it appears as the value. This is generally an O(n) operation and the insertion / updating of elements in the hash table should be constant time, on the average. If an integer is found to appear more than twice, you really don't have to increment the usage count further (if you don't want to).

Step 2: Perform a second pass over the integers. Look each up in the hash map and the first one with an appearance count of one is the one you were looking for (i.e., the first single appearing integer). This is also O(n), making the entire process O(n).

Some possible optimizations for special cases:

Optimization A: It may be possible to use a simple array instead of a hash table. This guarantees O(1) even in the worst case for counting the number of occurrences of a particular integer as well as the lookup of its appearance count. Also, this enhances real time performance, since the hash algorithm does not need to be executed. There may be a hit due to potentially poorer locality of reference (i.e., a larger sparse table vs. the hash table implementation with a reasonable load factor). However, this would be for very special cases of integer orderings and may be mitigated by the hash table's hash function producing pseudorandom bucket placements based on the incoming integers (i.e., poor locality of reference to begin with).

Each byte in the array would represent the count (up to 255) for the integer represented by the index of that byte. This would only be possible if the difference between the lowest integer and the highest (i.e., the cardinality of the domain of valid integers) was small enough such that this array would fit into memory. The index in the array of a particular integer would be its value minus the smallest integer present in the data set.

For example on modern hardware with a 64-bit OS, it is quite conceivable that a 4GB array can be allocated which can handle the entire domain of 32-bit integers. Even larger arrays are conceivable with sufficient memory.

The smallest and largest integers would have to be known before processing, or another linear pass through the data using the minmax algorithm to find out this information would be required.

Optimization B: You could optimize Optimization A further, by using at most 2 bits per integer (One bit indicates presence and the other indicates multiplicity). This would allow for the representation of four integers per byte, extending the array implementation to handle a larger domain of integers for a given amount of available memory. More bit games could be played here to compress the representation further, but they would only support special cases of data coming in and therefore cannot be recommended for the still mostly general case.

18. #### Kaden

• 2015/12/6

The first integer (corresponding to arr[0]) is your pivot element, p. Given an array of integers, find and print the minimum absolute difference between

19. #### Conti

• 2019/3/26

public class ArrayDemo { public static void main(String[] args) { int[] anArray; // declare an array of integers anArray = new int[10]; // create an array

20. #### Noah

• 2018/2/21

Use the first for loop to fix one array element. Use the second loop to look for duplicate elements in the remaining elements. Increment the count variable if

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Error processing SSI file
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