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.
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