花花酱 LeetCode 2575. Find the Divisibility Array of a String

By zxi on February 25, 2023

You are given a 0-indexed string word of length n consisting of digits, and a positive integer m.

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)

C++

// 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;

}

};

花花酱 LeetCode 1404. Number of Steps to Reduce a Number in Binary Representation to One

By zxi on April 5, 2020

Given a number s in their binary representation. Return the number of steps to reduce it to 1 under the following rules:

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)

C++

// 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;

}

};

Python3

# 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

花花酱 LeetCode 43. Multiply Strings

By zxi on July 26, 2018

Problem

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

Example 1:

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

Example 2:

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

Note:

The length of both num1 and num2 is < 110.
Both num1 and num2 contain only digits 0-9.
Both num1 and num2 do not contain any leading zero, except the number 0 itself.
You must not use any built-in BigInteger library or convert the inputs to integer directly.

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 "";

}

};

Posts tagged as “big integer”

花花酱 LeetCode 2575. Find the Divisibility Array of a String

Solution: Big Integer Math

C++

花花酱 LeetCode 1404. Number of Steps to Reduce a Number in Binary Representation to One

Solution: Simulation

C++

Python3

花花酱 LeetCode 43. Multiply Strings

Problem

Solution: Simulation