233. Number of Digit One 详细解答

Approach #1 Brute force [Time Limit Exceeded]

Intuition

Do as directed in question.

Algorithm

• Iterate over $i$ from $1$ to $n$:
• Convert $i$ to string and count $\text{’1’}$ in each integer string
• Add count of $\text{’1’}$ in each string to the sum, say $countr$

C++

int countDigitOne(int n){    int countr = 0;    for (int i = 1; i <= n; i++) {        string str = to_string(i);        countr += count(str.begin(), str.end(), '1');    }    return countr;}

Complexity Analysis

• Time complexity: O(n*log_{10}(n))O(n∗log​10​​(n)).
• We iterate from $1$ to $n$

• In each iteration, we convert integer to string and count '1' in string which takes linear time in number of digits in $i$, which is log_{10}(n)log​10​​(n).

• Space complexity: O(log_{10}(n))O(log​10​​(n)) Extra space for the countr and the converted string $\text{str}$.

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Approach #2 Solve it mathematically [Accepted]

Intuition

In Approach #1, we manually calculated the number of all the '1'​′​​1​′​​s in the digits, but this is very slow. Hence, we need a way to find a pattern in the way '1'​′​​1​′​​s (or for that matter any digit) appears in the numbers. We could then use the pattern to formulate the answer.

Consider the $1$s in $\text{ones}$ place , $\text{tens}$ place, $\text{hundreds}$ place and so on... An analysis has been performed in the following figure.

From the figure, we can see that from digit '1' at $\text{ones}$ place repeat in group of 1 after interval of $10$. Similarly, '1' at $\text{tens}$ place repeat in group of 10 after interval of $100$. This can be formulated as $(n/(i\ast 10))\ast i$.

Also, notice that if the digit at $\text{tens}$ place is $\text{’1’}$, then the number of terms with $\text{’1’s}$ is increased by $x+1$, if the number is say $\text{"ab1x"}$. As if digits at $\text{tens}$place is greater than $1$, then all the $10$ occurances of numbers with '1'​′​​1​′​​ at $\text{tens}$ place have taken place, hence, we add $10$. This is formluated as $\min (\max ((\text{n mod (i*10)})-i+1,0),i)$.

Lets take an example, say $n=1234$.

No of $\text{’1’}$ in $\text{ones}$ place = $1234/10$(corresponding to 1,11,21,...1221) + $\min (4,1)$(corresponding to 1231) =$124$

No of $\text{’1’}$ in $\text{tens}$ place = $(1234/100)\ast 10$(corresponding to 10,11,12,...,110,111,...1919) +$\min (21,10)$(corresponding to 1210,1211,...1219)=$130$

No of $\text{’1’}$ in $\text{hundreds}$ place = $(1234/1000)\ast 100$(corresponding to 100,101,12,...,199) +$\min (135,100)$(corresponding to 1100,1101...1199)=$200$

No of $\text{’1’}$ in $\text{thousands}$ place = $(1234/10000)\ast 10000$ +$\min (235,1000)$(corresponding to 1000,1001,...1234)=$235$

Therefore, Total = $124+130+200+235=689$.

Herein, one formula has been devised, but many other formulae can be devised for faster implementations, but the essence and complexity remains the same. The users are encouraged to try to divise their own version of solution using the mathematical concepts.

Algorithm

• Iterate over $i$ from $1$ to $n$ incrementing by $10$ each time:
• Add $(n/(i\ast 10))\ast i$ to $\text{countr}$ representing the repetition of groups of $$i$ sizes after each $(i\ast 10)$ interval.
• Add $\min (\max ((\text{n mod (i*10)})-i+1,0),i)$ to $\text{countr}$ representing the additional digits dependant on the digit in $i$th place as described in intuition.

C++

int countDigitOne(int n){    int countr = 0;    for (long long i = 1; i <= n; i *= 10) {        long long divider = i * 10;        countr += (n / divider) * i + min(max(n % divider - i + 1, 0LL), i);    }    return countr;}

Complexity analysis

• Time complexity: O(log_{10}(n))O(log​10​​(n)).

• No of iterations equal to the number of digits in n which is log_{10}(n)log​10​​(n)

• Space complexity: $O(1)$ space required.