357. Count Numbers with Unique Digits - cocoder39/coco39_LC GitHub Wiki

357. Count Numbers with Unique Digits

Following the hint. Let f(n) = count of number with unique digits of length n.

f(1) = 10. (0, 1, 2, 3, ...., 9)

f(2) = 9 * 9. Because for each number i from 1, ..., 9, we can pick j to form a 2-digit number ij and there are 9 numbers that are different from i for j to choose from.

f(3) = f(2) * 8 = 9 * 9 * 8. Because for each number with unique digits of length 2, say ij, we can pick k to form a 3 digit number ijk and there are 8 numbers that are different from i and j for k to choose from.

Similarly f(4) = f(3) * 7 = 9 * 9 * 8 * 7....

...

f(10) = 9 * 9 * 8 * 7 * 6 * ... * 1

f(11) = 0 = f(12) = f(13)....

//dp[i] is count of numbers with unique digits with length k
    //dp[1] = 10, dp[2] = 9 * 9, dp[3] = 9 * 9 * 8 ...
    int countNumbersWithUniqueDigits(int n) {
        if (n == 0) {
            return 1;
        }
        
        int res = 10; //n == 1
        int dp = 9;
        for (int len = 1; len < n; len++) {
            dp *= (10 - len);
            res += dp;
        }
        return res;
    }