Time complexity for my solution to all permutations of integer array containing distinct numbers
0
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What's the time complexity of my code? I ran this through www.leetcode.com and it's optimal. I think its O(n*n!). First I thought it was O(n^2*n!) : The extra n since we make n recursive calls. However, only the first call to permute() is dominant, and kind of dwarfs the rest since n! is >>> (n-1)!
Thanks upfront!
class Solution {
public List<List<Integer>> permute(int nums) {
return permute(nums, nums.length - 1);
}
private List<List<Integer>> permute(int nums, int n) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(n < 0) {
List<Integer> permutation = new ArrayList<Integer>();
result.add(permutation);
return result;
}
// below returns (n-1)! results of size n-1 each
List<List<Integer>> prefixes = permute(nums, n-1);
for(List<Integer> prefix : prefixes) {
List<List<Integer>> permutations = insert(nums[n], prefix);
result.addAll(permutations);
}
return result;
}
// O(n^2) worst case when size of list is n-1
private List<List<Integer>> insert(int num, List<Integer> list) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
for(int i = 0; i <= list.size(); i++) {
List<Integer> clone = new ArrayList<Integer>();
clone.addAll(list);
clone.add(i, num);
result.add(clone);
}
return result;
}
}
java algorithm array recursion combinatorics
asked 8 mins ago
AranAran
1
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0
$begingroup$
What's the time complexity of my code? I ran this through www.leetcode.com and it's optimal. I think its O(n*n!). First I thought it was O(n^2*n!) : The extra n since we make n recursive calls. However, only the first call to permute() is dominant, and kind of dwarfs the rest since n! is >>> (n-1)!
Thanks upfront!
class Solution {
public List<List<Integer>> permute(int nums) {
return permute(nums, nums.length - 1);
}
private List<List<Integer>> permute(int nums, int n) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(n < 0) {
List<Integer> permutation = new ArrayList<Integer>();
result.add(permutation);
return result;
}
// below returns (n-1)! results of size n-1 each
List<List<Integer>> prefixes = permute(nums, n-1);
for(List<Integer> prefix : prefixes) {
List<List<Integer>> permutations = insert(nums[n], prefix);
result.addAll(permutations);
}
return result;
}
// O(n^2) worst case when size of list is n-1
private List<List<Integer>> insert(int num, List<Integer> list) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
for(int i = 0; i <= list.size(); i++) {
List<Integer> clone = new ArrayList<Integer>();
clone.addAll(list);
clone.add(i, num);
result.add(clone);
}
return result;
}
}
java algorithm array recursion combinatorics
asked 8 mins ago
AranAran
1
New contributor
Aran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
0
0
0
$begingroup$
What's the time complexity of my code? I ran this through www.leetcode.com and it's optimal. I think its O(n*n!). First I thought it was O(n^2*n!) : The extra n since we make n recursive calls. However, only the first call to permute() is dominant, and kind of dwarfs the rest since n! is >>> (n-1)!
Thanks upfront!
class Solution {
public List<List<Integer>> permute(int nums) {
return permute(nums, nums.length - 1);
}
private List<List<Integer>> permute(int nums, int n) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(n < 0) {
List<Integer> permutation = new ArrayList<Integer>();
result.add(permutation);
return result;
}
// below returns (n-1)! results of size n-1 each
List<List<Integer>> prefixes = permute(nums, n-1);
for(List<Integer> prefix : prefixes) {
List<List<Integer>> permutations = insert(nums[n], prefix);
result.addAll(permutations);
}
return result;
}
// O(n^2) worst case when size of list is n-1
private List<List<Integer>> insert(int num, List<Integer> list) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
for(int i = 0; i <= list.size(); i++) {
List<Integer> clone = new ArrayList<Integer>();
clone.addAll(list);
clone.add(i, num);
result.add(clone);
}
return result;
}
}
java algorithm array recursion combinatorics
asked 8 mins ago
AranAran
1
New contributor
Aran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
What's the time complexity of my code? I ran this through www.leetcode.com and it's optimal. I think its O(n*n!). First I thought it was O(n^2*n!) : The extra n since we make n recursive calls. However, only the first call to permute() is dominant, and kind of dwarfs the rest since n! is >>> (n-1)!
Thanks upfront!
class Solution {
public List<List<Integer>> permute(int nums) {
return permute(nums, nums.length - 1);
}
private List<List<Integer>> permute(int nums, int n) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(n < 0) {
List<Integer> permutation = new ArrayList<Integer>();
result.add(permutation);
return result;
}
// below returns (n-1)! results of size n-1 each
List<List<Integer>> prefixes = permute(nums, n-1);
for(List<Integer> prefix : prefixes) {
List<List<Integer>> permutations = insert(nums[n], prefix);
result.addAll(permutations);
}
return result;
}
// O(n^2) worst case when size of list is n-1
private List<List<Integer>> insert(int num, List<Integer> list) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
for(int i = 0; i <= list.size(); i++) {
List<Integer> clone = new ArrayList<Integer>();
clone.addAll(list);
clone.add(i, num);
result.add(clone);
}
return result;
}
}
java algorithm array recursion combinatorics
java algorithm array recursion combinatorics
asked 8 mins ago
AranAran
1
New contributor
Aran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 mins ago
AranAran
1
New contributor
Aran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 mins ago
AranAran
1
New contributor
Aran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 mins ago
AranAran
1
asked 8 mins ago
AranAran
1
1
New contributor
Aran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Aran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Aran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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