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Duplicate In A Loose Permutation Problem

Duplicate In A Loose Permutation Problem Statement

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.

Example One

{
"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.

Example Two

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

Output:

-1

Notes

  • 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 <= 106
  • 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.

Duplicate In A Loose Permutation Solution: Optimal

  • 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.

Time Complexity

O(n).

We iterate all the elements of the array only once.

Auxiliary Space Used

O(1).

Space Complexity

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).

Code For Duplicate In A Loose Permutation Solution: Optimal

/*
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:

‍To learn more, register for the FREE webinar.

Try yourself in the Editor

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

Duplicate In A Loose Permutation Problem

Duplicate In A Loose Permutation Problem Statement

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.

Example One

{
"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.

Example Two

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

Output:

-1

Notes

  • 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 <= 106
  • 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.

Duplicate In A Loose Permutation Solution: Optimal

  • 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.

Time Complexity

O(n).

We iterate all the elements of the array only once.

Auxiliary Space Used

O(1).

Space Complexity

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).

Code For Duplicate In A Loose Permutation Solution: Optimal

/*
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:

‍To learn more, register for the FREE webinar.

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