Given a function rand7 which generates a uniform random integer in the range 1 to 7, write a function rand10 which generates a uniform random integer in the range 1 to 10.
Do NOT use system’s Math.random().
Example 1:
Input: 1Output: [7]
Example 2:
Input: 2Output: [8,4]
Example 3:
Input: 3Output: [8,1,10]
Note:
rand7 is predefined.
Each testcase has one argument: n, the number of times that rand10 is called.
Follow up:
What is the expected value for the number of calls to rand7() function?
Could you minimize the number of calls to rand7()?
Input: A = [1,1,2,2,2,2], target = 5Output: 12Explanation:
A[i] = 1, A[j] = A[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.
We have a sorted set of digits D, a non-empty subset of {'1','2','3','4','5','6','7','8','9'}. (Note that '0' is not included.)
Now, we write numbers using these digits, using each digit as many times as we want. For example, if D = {'1','3','5'}, we may write numbers such as '13', '551', '1351315'.
Return the number of positive integers that can be written (using the digits of D) that are less than or equal to N.
Example 1:
Input: D = ["1","3","5","7"], N = 100Output: 20Explanation:
The 20 numbers that can be written are:
1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.
Example 2:
Input: D = ["1","4","9"], N = 1000000000Output: 29523Explanation:
We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers,
81 four digit numbers, 243 five digit numbers, 729 six digit numbers,
2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers.
In total, this is 29523 integers that can be written using the digits of D.
Note:
D is a subset of digits '1'-'9' in sorted order.
1 <= N <= 10^9
Solution -1: DFS (TLE)
Time complexity: O(|D|^log10(N))
Space complexity: O(n)
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// Author: Huahua
classSolution{
public:
intatMostNGivenDigitSet(vector<string>& D, int N) {
int ans = 0;
dfs(D,0,N,ans);
returnans;
}
private:
voiddfs(constvector<string>& D, int cur, int N, int& ans) {
if (cur > N) return;
if(cur>0&& cur <= N) ++ans;
for(conststring& d : D)
dfs(D, cur * 10 + d[0] - '0', N, ans);
}
};
Solution 1: Math
Time complexity: O(log10(N))
Space complexity: O(1)
Suppose N has n digits.
less than n digits
We can use all the numbers from D to construct numbers of with length 1,2,…,n-1 which are guaranteed to be less than N.
e.g. n = 52125, D = [1, 2, 5]
format X: e.g. 1, 2, 5 counts = |D| ^ 1
format XX: e.g. 11,12,15,21,22,25,51,52,55, counts = |D|^2
format XXX: counts = |D|^3
format XXXX: counts = |D|^4
exact n digits
if all numbers in D != N[0], counts = |d < N[0] | d in D| * |D|^(n-1), and we are done.