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Head of Career Skills Development & Coaching

*Based on past data of successful IK students

Find a duplicated number in a loose permutation of numbers. A permutation is an array that is size `n`

, and also has positive numbers from 1 to `n`

. A loose permutation is a permutation where some numbers are missing and some are duplicated, but the total number is still `n`

.

```
{
"arr": [1, 2, 7, 3, 4, 4, 5]
}
```

Output:

```
4
```

We can see that 4 is a duplicate number here as it is present 2 times in the array.

```
{
"arr": [1, 2, 3]
}
```

Output:

```
-1
```

- You can only use a constant amount of extra memory.
- If no duplicate number is present, return -1.
- If multiple duplicates exist, return any.

Constraints:

- 1 <=
`n`

<= 10^{6} - 1 <= any element of the input list <=
`n`

We provided one solution.

Throughout this editorial, we will refer to the length of the input array as `n`

.

- We are given an input array of integers.
- We start iterating over this array and for every integer encountered, say we encounter integer 5, we negate the value at index
`abs(integer) -1`

, here`abs(5) - 1 = 4`

, in the input array. - So if the array was [5, 1, 2, 3, 3], after iterating through the first element, that is 5, it would become [5, 1, 2, 3, -3].
- Note that we are using
`abs(integer)`

because over the subsequent iterations, we may start getting negative numbers at certain array positions, hence we use the absolute values of array integers. - Now, if we try to negate a number at a certain position, which is already negated, we can be certain that this number is a duplicate because that position which corresponds to the number has already been negated.
- If the number was a unique one, then it would not had that position negated.
- To illustrate let's come back to our array [5, 1, 2, 3, 3], after iterating through the first 4 elements and updating the array as we discussed, we get the following array [-5, -1, -2, 3, -3].
- Now when we encounter the last element, that is -3, we try to negate the value at index
`abs(-3) - 1`

, which is value at index 2, and that is already negated. So we identify 3 as a duplicate.

O(n).

We iterate all the elements of the array only once.

O(1).

O(n).

Space used for input: O(n).

Auxiliary space used: O(1).

Space used for output: O(1).

So, total space complexity: O(n).

```
/*
Asymptotic complexity in terms of the length of the input array \`n\`:
* Time: O(n).
* Auxiliary space: O(1).
* Total space: O(n).
*/
int find_duplicate(vector<int> &arr) {
for (int i = 0; i < arr.size(); i++) {
int value = abs(arr[i]) - 1;
if (arr[value] < 0) {
return abs(arr[i]);
}
else {
arr[value] = -arr[value];
}
}
// If no duplicate is present, we return -1.
return -1;
}
```

We hope that these solutions to finding duplicate in a loose permutation problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart’s FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as:

- Back-end Engineering Interview Course
- Front-end Engineering Interview Course
- Full Stack Developer Interview Course

To learn more, register for the FREE webinar.

Note: Input and Output will already be taken care of.

Find a duplicated number in a loose permutation of numbers. A permutation is an array that is size `n`

, and also has positive numbers from 1 to `n`

. A loose permutation is a permutation where some numbers are missing and some are duplicated, but the total number is still `n`

.

```
{
"arr": [1, 2, 7, 3, 4, 4, 5]
}
```

Output:

```
4
```

We can see that 4 is a duplicate number here as it is present 2 times in the array.

```
{
"arr": [1, 2, 3]
}
```

Output:

```
-1
```

- You can only use a constant amount of extra memory.
- If no duplicate number is present, return -1.
- If multiple duplicates exist, return any.

Constraints:

- 1 <=
`n`

<= 10^{6} - 1 <= any element of the input list <=
`n`

We provided one solution.

Throughout this editorial, we will refer to the length of the input array as `n`

.

- We are given an input array of integers.
- We start iterating over this array and for every integer encountered, say we encounter integer 5, we negate the value at index
`abs(integer) -1`

, here`abs(5) - 1 = 4`

, in the input array. - So if the array was [5, 1, 2, 3, 3], after iterating through the first element, that is 5, it would become [5, 1, 2, 3, -3].
- Note that we are using
`abs(integer)`

because over the subsequent iterations, we may start getting negative numbers at certain array positions, hence we use the absolute values of array integers. - Now, if we try to negate a number at a certain position, which is already negated, we can be certain that this number is a duplicate because that position which corresponds to the number has already been negated.
- If the number was a unique one, then it would not had that position negated.
- To illustrate let's come back to our array [5, 1, 2, 3, 3], after iterating through the first 4 elements and updating the array as we discussed, we get the following array [-5, -1, -2, 3, -3].
- Now when we encounter the last element, that is -3, we try to negate the value at index
`abs(-3) - 1`

, which is value at index 2, and that is already negated. So we identify 3 as a duplicate.

O(n).

We iterate all the elements of the array only once.

O(1).

O(n).

Space used for input: O(n).

Auxiliary space used: O(1).

Space used for output: O(1).

So, total space complexity: O(n).

```
/*
Asymptotic complexity in terms of the length of the input array \`n\`:
* Time: O(n).
* Auxiliary space: O(1).
* Total space: O(n).
*/
int find_duplicate(vector<int> &arr) {
for (int i = 0; i < arr.size(); i++) {
int value = abs(arr[i]) - 1;
if (arr[value] < 0) {
return abs(arr[i]);
}
else {
arr[value] = -arr[value];
}
}
// If no duplicate is present, we return -1.
return -1;
}
```

We hope that these solutions to finding duplicate in a loose permutation problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

If you are preparing for a tech interview at FAANG or any other Tier-1 tech company, register for Interview Kickstart’s FREE webinar to understand the best way to prepare.

Interview Kickstart offers interview preparation courses taught by FAANG+ tech leads and seasoned hiring managers. Our programs include a comprehensive curriculum, unmatched teaching methods, and career coaching to help you nail your next tech interview.

We offer 18 interview preparation courses, each tailored to a specific engineering domain or role, including the most in-demand and highest-paying domains and roles, such as:

- Back-end Engineering Interview Course
- Front-end Engineering Interview Course
- Full Stack Developer Interview Course

To learn more, register for the FREE webinar.

**Attend our free webinar to amp up your career and get the salary you deserve.**

Hosted By

Ryan Valles

Founder, Interview Kickstart