Leetcode Count Numbers with Unique Digits

Title Description given a non-negative integer n, count all numbers with unique digits, X,where 0≤x <10n. Example:given N=2,return  the. (The answer should is the total numbersinchThe range of0≤x < -, excluding [ One, A, -, -, -, the, the, the, About]) Hint:1. A Direct isTo use the backtracking approach.2. Backtracking should contains three states which is (the current number,
Number of steps toGetThat number and a bitmask which represent which number isMarked asVisited so farinchThe current number).
Start with State (0,0,0) and count all valid number till we reach number of steps equals to 10n.3. This problem can also is solvedusingADynamicprogramming approach and some knowledge of combinatorics.4. Let f (k) =count of numbers with a unique digits with length equals K.5. F (1) =Ten,..., f (k) =9*9*8* ... (9-K +2) [The first factor is 9Because a number cannot start with0].

That is, given an integer n, calculates the number of all numbers that do not contain duplicates from [0^n]

1. Backtracking:

If you enter N, there is a total of n digits regardless of the 10^n. Use Cur[n] to represent the current number. Use a flag[10] to represent 0~9 digits, each time a number is placed, the flag corresponds to a number of 1,

The code is as follows:

classsolution (object):defcountnumberswithuniquedigits (self, n):""": Type N:int:rtype:int"""        ifn = =0:return1Flag= [0]*10cur= [0]*N Step=0; Count=[0]; Self.helper (0, flag, cur, count, N)returnCOUNT[0]defHelper (self, step, flag, cur, COUNT, N):ifStep >=N:count[0]+ = 1return         forIinchRange (0, 10):            ifFlag[i]! = 1or(i==0 andStep!=n andSUM (cur[step:]) = =0): #解决诸如 [0,0,0], [0, 1,0] Such numbers do not count the question Cur[step]=I flag[i]= 1Self.helper (Step+1, flag, cur, count, N) flag[i]=0Else:                Continue

The above backtracking method timed out in time t_t

2. Dynamic Planning:

According to Tip 4, f (k) = 9 * 9 * ... (9-k+2)

classsolution (object):defcountnumberswithuniquedigits (self, n):""": Type N:int:rtype:int"""        ifn = =0:return1ifn = = 1:            return10Res=0 forIinchRange (1, n+1): Res+=Self.count (i)returnResdefcount (self, num):ifnum = = 0:return1ifnum = = 1:return10Res= 1 forIinchRange (9, 11-num-1, 1): Res*=IreturnRes*9

Leetcode Count Numbers with Unique Digits