# Posts published in “Digital Operations”

Given an integer `num`, repeatedly add all its digits until the result has only one digit, and return it.

Example 1:

```Input: num = 38
Output: 2
Explanation: The process is
38 --> 3 + 8 --> 11
11 --> 1 + 1 --> 2
Since 2 has only one digit, return it.
```

Example 2:

```Input: num = 0
Output: 0
```

Constraints:

• `0 <= num <= 231 - 1`

Idea 1:

recursive

Solution 1:

``````/**
* @param {number} num
* @return {number}
*/
let nums = num.toString().split("");
let sumOfNums = sum(nums);

// base case:
if (sumOfNums  < 10) return sumOfNums;
};

function sum(arr) {
return arr.reduce((a, b) => parseInt(a) + parseInt(b), 0);
}``````

Idea 2:

mathematics algorithms in O(1) runtime

Solution 2:

``````/**
* @param {number} num
* @return {number}
*/
return num % 9 ? num % 9 : Math.min(num, 9);
};``````

Given an integer `n`, return the number of trailing zeroes in `n!`.

Follow up: Could you write a solution that works in logarithmic time complexity?

Example 1:

```Input: n = 3
Output: 0
Explanation: 3! = 6, no trailing zero.
```

Example 2:

```Input: n = 5
Output: 1
Explanation: 5! = 120, one trailing zero.
```

Example 3:

```Input: n = 0
Output: 0
```

Constraints:

• `0 <= n <= 104`

Idea:

All trailing zeros are come from even_num x 5, we have more even_num than 5, so only count factor 5.

4! = 1x2x3x4 = 24, we haven’t encountered any 5 yet, so we don’t have any trailing zero.

5! = 1x2x3x4x5 = 120, we have one trailing zero. either 2×5, or 4×5 can contribute to that zero.

9! = 362880, we only encountered 5 once, so 1 trailing zero as expected.

10! = 3628800, 2 trailing zeros, since we have two numbers that have factor 5, one is 5 and the other is 10 (2×5)

100/5 = 20, we have 20 numbers have factor 5: 5, 10, 15, 20, 25, …, 95, 100.

Is the number of trailing zero 20? No, it’s 24, why?

Within that 20 numbers, we have 4 of them: 25 (5×5), 50 (2x5x5), 75 (3x5x5), 100 (4x5x5) that have an extra factor of 5.

So, for a given number n, we are looking how many numbers <=n have factor 5, 5×5, 5x5x5, …

Summing those numbers up we got the answer.

e.g. 1000! has 249 trailing zeros:

1000/5 = 200

1000/25 = 40

1000/125 = 8

1000/625 = 1

200 + 40 + 8 + 1 = 249

Solution:

``````/**
* @param {number} n
* @return {number}
*/
var trailingZeroes = function(n) {
return n < 5 ? 0 : parseInt(n / 5) + trailingZeroes(parseInt(n / 5));
};``````

Given two non-negative integers `num1` and `num2` represented as strings, return the product of `num1` and `num2`, also represented as a string.

Note: You must not use any built-in BigInteger library or convert the inputs to integer directly.

Example 1:

```Input: num1 = "2", num2 = "3"
Output: "6"
```

Example 2:

```Input: num1 = "123", num2 = "456"
Output: "56088"
```

Constraints:

• `1 <= num1.length, num2.length <= 200`
• `num1` and `num2` consist of digits only.
• Both `num1` and `num2` do not contain any leading zero, except the number `0` itself.

Solution:

``````/**
* @param {string} num1
* @param {string} num2
* @return {string}
*/
var multiply = function(num1, num2) {
let len1 = num1.length;
let len2 = num2.length;
let carry = 0;
let val = 0;
let index = 0;
let res = Array(len1 + len2).fill(0);

for (let i = len1 - 1; i >= 0; i--) {
carry = 0;
for (let j = len2 - 1; j >= 0; j--) {
// calucate from the right position and store in result array from index 0
// reverse it later
index = len1 + len2 - 2 - i - j;
// basic math calculation
val = (num1[i] * num2[j]) + res[index] + carry;
carry = Math.floor(val / 10);
res[index] = val % 10;
}
// check if has carry 1
if (carry) res[index + 1] = carry;
}

// before reverse, remove the begining 0
while (res.length > 1 && res[res.length - 1] === 0) res.pop();
return res.reverse().join("");
};``````

Given an integer `x`, return `true` if `x` is palindrome integer.

An integer is a palindrome when it reads the same backward as forward. For example, `121` is palindrome while `123` is not.

Example 1:

```Input: x = 121
Output: true
```

Example 2:

```Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
```

Example 3:

```Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
```

Example 4:

```Input: x = -101
Output: false
```

Constraints:

• `-231 <= x <= 231 - 1`

Solution:

``````/**
* @param {number} x
* @return {boolean}
*/
var isPalindrome = function(x) {
// negative number is not a Palindrome Number
if (x < 0) return false;
let y = x;
let reverse = 0;
while (y > 0) {
let lastDigit = y % 10;
reverse = reverse * 10 + lastDigit;
y = parseInt(y / 10);
}
return reverse === x ;
};``````

Implement the `myAtoi(string s)` function, which converts a string to a 32-bit signed integer (similar to C/C++’s `atoi` function).

The algorithm for `myAtoi(string s)` is as follows:

2. Check if the next character (if not already at the end of the string) is `'-'` or `'+'`. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
3. Read in next the characters until the next non-digit charcter or the end of the input is reached. The rest of the string is ignored.
4. Convert these digits into an integer (i.e. `"123" -> 123``"0032" -> 32`). If no digits were read, then the integer is `0`. Change the sign as necessary (from step 2).
5. If the integer is out of the 32-bit signed integer range `[-231, 231 - 1]`, then clamp the integer so that it remains in the range. Specifically, integers less than `-231` should be clamped to `-231`, and integers greater than `231 - 1` should be clamped to `231 - 1`.
6. Return the integer as the final result.

Note:

• Only the space character `' '` is considered a whitespace character.
• Do not ignore any characters other than the leading whitespace or the rest of the string after the digits.

Example 1:

```Input: s = "42"
Output: 42
Explanation: The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42" (no characters read because there is no leading whitespace)
^
Step 2: "42" (no characters read because there is neither a '-' nor '+')
^
Step 3: "42" ("42" is read in)
^
The parsed integer is 42.
Since 42 is in the range [-231, 231 - 1], the final result is 42.
```

Example 2:

```Input: s = "   -42"
Output: -42
Explanation:
^
Step 2: "   -42" ('-' is read, so the result should be negative)
^
Step 3: "   -42" ("42" is read in)
^
The parsed integer is -42.
Since -42 is in the range [-231, 231 - 1], the final result is -42.
```

Example 3:

```Input: s = "4193 with words"
Output: 4193
Explanation:
Step 1: "4193 with words" (no characters read because there is no leading whitespace)
^
Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
^
Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
^
The parsed integer is 4193.
Since 4193 is in the range [-231, 231 - 1], the final result is 4193.
```

Example 4:

```Input: s = "words and 987"
Output: 0
Explanation:
Step 1: "words and 987" (no characters read because there is no leading whitespace)
^
Step 2: "words and 987" (no characters read because there is neither a '-' nor '+')
^
Step 3: "words and 987" (reading stops immediately because there is a non-digit 'w')
^
The parsed integer is 0 because no digits were read.
Since 0 is in the range [-231, 231 - 1], the final result is 0.
```

Example 5:

```Input: s = "-91283472332"
Output: -2147483648
Explanation:
Step 1: "-91283472332" (no characters read because there is no leading whitespace)
^
Step 2: "-91283472332" ('-' is read, so the result should be negative)
^
Step 3: "-91283472332" ("91283472332" is read in)
^
The parsed integer is -91283472332.
Since -91283472332 is less than the lower bound of the range [-231, 231 - 1], the final result is clamped to -231 = -2147483648.
```

Constraints:

• `0 <= s.length <= 200`
• `s` consists of English letters (lower-case and upper-case), digits (`0-9`), `' '``'+'``'-'`, and `'.'`.

Solution 1:

``````/**
* @param {string} s
* @return {number}
*/
var myAtoi = function(s) {
const len = s.length;
let max = 2147483647;
let min = -2147483648;
let numberMatch = /^[\d ()+-]+\$/;

for (let i = 0, j = i; i < len; i++) {
let current = s[i];

if (current != " " && !current.match(numberMatch)) {
return 0;
} else if (current != " " && current.match(numberMatch)) {
let result = "";

while (j < len && current.match(numberMatch)) {
result += s[j];
j++;
}
let output = parseInt(result);
if (isNaN(output)) return 0;
else if (output > max) return max;
else if (output < min) return min;
else return output;
}
}
return 0;
};``````

Solution 2:

``````/**
* @param {string} s
* @return {number}
*/
var myAtoi = function(s) {
let i = 0;
let sign = 1;
let res = 0;
let len = s.length;
let INT_MAX = 2147483647;
let INT_MIN = - INT_MAX - 1;

// skip all space
while (s[i] === ' ') i++;

// check sign
if (s[i] === '+' || s[i] === '-') {
sign = s[i] === '+' ? 1 : -1;
i++;
}

while (s[i] >= '0' && s[i] <= '9') {
res = (res * 10) + (s[i] - 0);
if (sign === 1 && res > INT_MAX) return INT_MAX;
if (sign === -1 && res > INT_MAX + 1) return INT_MIN;
i++;
}

return res * sign;
};``````

Given a signed 32-bit integer `x`, return `x` with its digits reversed. If reversing `x` causes the value to go outside the signed 32-bit integer range `[-231, 231 - 1]`, then return `0`.

Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

Example 1:

```Input: x = 123
Output: 321
```

Example 2:

```Input: x = -123
Output: -321
```

Example 3:

```Input: x = 120
Output: 21
```

Example 4:

```Input: x = 0
Output: 0
```

Constraints:

• `-231 <= x <= 231 - 1`

Solution:

``````/**
* @param {number} x
* @return {number}
*/

var reverse = function(x) {
const str      = "" + Math.abs(x);   // convert absolute value to string
const reversed = str.split("")       // get array of digit characters
.reverse()            // reverse it
.join("");            // join into string again;
const num      = parseInt(reversed); // convert to integer
const revNum = Math.sign(x) * num;   // multiple the sign
// if out of range return 0
if (revNum < Math.pow(-2, 31) || revNum > Math.pow(2, 31) - 1) return 0;
return revNum;
}``````