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Copy pathKthPermutationSquenceUsingMathFormula.java
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KthPermutationSquenceUsingMathFormula.java
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/*
The set [1,2,3,...,n] contains a total of n! unique permutations.
By listing and labeling all of the permutations in order, we get the following sequence for n = 3:
"123"
"132"
"213"
"231"
"312"
"321"
Given n and k, return the kth permutation sequence.
Note:
Given n will be between 1 and 9 inclusive.
Given k will be between 1 and n! inclusive.
Example 1:
Input: n = 3, k = 3
Output: "213"
Example 2:
Input: n = 4, k = 9
Output: "2314"
*/
class Solution {
int[] fact = new int[10];
ArrayList<Integer> listOfNumbers = null;
public String getPermutation(int n, int K) {
//Fill factorial table from 0 -> 9
fillFact();
//Fill list of eligible numbers
listOfNumbers = new ArrayList<>();
fillListOfNumbers(n);
return KthPermSequence(K-1); //just to be aligned with List Index that why K-1;
}
private String KthPermSequence(int K){
if(listOfNumbers.size() == 0){
return "";
}
//Find the index of List which element to be picked
//find the bucket in which K will land -> K / (n-1)!
int index = K / fact[listOfNumbers.size() - 1];
//Take the digit and append it to the list
StringBuilder sb = new StringBuilder();
sb.append(listOfNumbers.get(index));
//Update K -> after landing into a particular bucket update the K for Subbucket
K = K % fact[listOfNumbers.size() - 1];
//Remove the element which is already added from List
listOfNumbers.remove(index);
return (sb.toString() + KthPermSequence(K));
}
private void fillFact(){
fact[0] = 1;
for(int i = 1; i <= 9; i++){
fact[i] = i * fact[i-1];
}
}
private void fillListOfNumbers(int n){
for(int i=1; i<=n; i++)
{
listOfNumbers.add(i);
}
}
}