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 (ie, for n = 3):

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and k, return the k^th permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

public String getPermutation(int n, int k) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<Integer> nums = new ArrayList<Integer>();
        for(int i=1;i<=n;i++){
            nums.add(i);
        }
        k--;//k originally starts from 1
        int index=0;
        while(k!=0){
            int frac = frac(n-index-1);
            int targetNumIndex = k/frac+index;
            int targetNum = nums.remove(targetNumIndex);
            nums.add(index,targetNum);
            index++;
            k=k%frac;
        }
        String output = "";
        for(Integer num:nums){
            output+=num;
        }
        return output;
    }
    
    public int frac(int n){
        int sum = 1;
        for(int i=1;i<=n;i++){
            sum*=i;
        }
        return sum;
    }

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