The divisibility array div of word is an integer array of length n such that:

• div[i] = 1 if the numeric value of word[0,...,i] is divisible by m, or
• div[i] = 0 otherwise.

Return the divisibility array of word.

Example 1:

Input: word = "998244353", m = 3
Output: [1,1,0,0,0,1,1,0,0]
Explanation: There are only 4 prefixes that are divisible by 3: "9", "99", "998244", and "9982443".

Example 2:

Input: word = "1010", m = 10
Output: [0,1,0,1]
Explanation: There are only 2 prefixes that are divisible by 10: "10", and "1010".

Constraints:

• 1 <= n <= 105
• word.length == n
• word consists of digits from 0 to 9
• 1 <= m <= 109

Solution: Big Integer Math

r = (r * 10 + word[i]) % m

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  vector<int> divisibilityArray(string word, int m) {
    vector<int> ans;
    long long r = 0;
    for (char c : word) {
      r = (r * 10 + c - '0') % m;
      ans.push_back(r == 0);
    }
    return ans;
  }
};

The post 花花酱 LeetCode 2575. Find the Divisibility Array of a String appeared first on Huahua's Tech Road.

• If the current number is even, you have to divide it by 2.
• If the current number is odd, you have to add 1 to it.

It’s guaranteed that you can always reach to one for all testcases.

Example 1:

Input: s = "1101"
Output: 6
Explanation: "1101" corressponds to number 13 in their decimal representation.
Step 1) 13 is odd, add 1 and obtain 14. 
Step 2) 14 is even, divide by 2 and obtain 7.
Step 3) 7 is odd, add 1 and obtain 8.
Step 4) 8 is even, divide by 2 and obtain 4.  
Step 5) 4 is even, divide by 2 and obtain 2. 
Step 6) 2 is even, divide by 2 and obtain 1.  

Example 2:

Input: s = "10"
Output: 1
Explanation: "10" corressponds to number 2 in their decimal representation.
Step 1) 2 is even, divide by 2 and obtain 1.  

Example 3:

Input: s = "1"
Output: 0

Constraints:

• 1 <= s.length <= 500
• s consists of characters ‘0’ or ‘1’
• s[0] == '1'

Solution: Simulation

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int numSteps(string s) {
    int ans = 0;
    int carry = 0;
    // The highest bit must be 1, 
    // process to the 2nd highest bit
    for (int i = s.length() - 1; i > 0; --i) {
      if (s[i] - '0' + carry == 1) {
        ans += 2; // odd: +1, even: / 2
        carry = 1; // always has a carry
      } else {
        ans += 1; // even: / 2
        // carray remains e.g. 1 + 1 = 10, or 0 + 0 = 00
      }
    }
    // If there is a carry, then it's even, one more step.
    return ans + carry;
  }
};

# Author: Huahua
class Solution:
  def numSteps(self, s: str) -> int:
    ans, carry = 0, 0
    for i in range(1, len(s)):
      if ord(s[-i]) - ord('0') + carry == 1:
        ans += 2
        carry = 1
      else:
        ans += 1
    return ans + carry

The post 花花酱 LeetCode 1404. Number of Steps to Reduce a Number in Binary Representation to One appeared first on Huahua's Tech Road.

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

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

Solution: Simulation

Simulate multiplication one digit at a time.

Time complexity: O(l1*l2)

Space complexity: O(l1 + l2)

C++

// Author: Huahua
// Running time: 4 ms
class Solution {
public:
  string multiply(string num1, string num2) {
    const int l1 = num1.length();
    const int l2 = num2.length();
    string ans(l1 + l2, '0');
    for (int i = l1 - 1; i >= 0; --i)
      for (int j = l2 - 1; j >= 0; --j) {
        int sum = (ans[i + j + 1] - '0') + (num1[i] - '0') * (num2[j] - '0');        
        ans[i + j + 1] = (sum % 10) + '0';
        ans[i + j] += sum / 10;
      }
    for (int i = 0; i < ans.length(); ++i)
      if (ans[i] != '0' || i == ans.length() - 1) return ans.substr(i);
    return "";
  }
};

The post 花花酱 LeetCode 43. Multiply Strings appeared first on Huahua's Tech Road.