Posts tagged as “math”

花花酱 LeetCode 902. Numbers At Most N Given Digit Set

By zxi on September 13, 2018

Problem

We have a sorted set of digits D, a non-empty subset of {'1','2','3','4','5','6','7','8','9'}. (Note that '0' is not included.)

Now, we write numbers using these digits, using each digit as many times as we want. For example, if D = {'1','3','5'}, we may write numbers such as '13', '551', '1351315'.

Return the number of positive integers that can be written (using the digits of D) that are less than or equal to N.

Example 1:

Input: D = ["1","3","5","7"], N = 100
Output: 20
Explanation: 
The 20 numbers that can be written are:
1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.

Example 2:

Input: D = ["1","4","9"], N = 1000000000
Output: 29523
Explanation: 
We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers,
81 four digit numbers, 243 five digit numbers, 729 six digit numbers,
2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers.
In total, this is 29523 integers that can be written using the digits of D.

Note:

D is a subset of digits '1'-'9' in sorted order.
1 <= N <= 10^9

Solution -1: DFS (TLE)

Time complexity: O(|D|^log10(N))

Space complexity: O(n)

// Author: Huahua

class Solution {

public:

int atMostNGivenDigitSet(vector<string>& D, int N) {

int ans = 0;

dfs(D, 0, N, ans);

return ans;

}

private:

void dfs(const vector<string>& D, int cur, int N, int& ans) {

if (cur > N) return;

if (cur > 0 && cur <= N) ++ans;

for (const string& d : D)

dfs(D, cur * 10 + d[0] - '0', N, ans);

}

};

Solution 1: Math

Time complexity: O(log10(N))

Space complexity: O(1)

Suppose N has n digits.

less than n digits

We can use all the numbers from D to construct numbers of with length 1,2,…,n-1 which are guaranteed to be less than N.

e.g. n = 52125, D = [1, 2, 5]

format X: e.g. 1, 2, 5 counts = |D| ^ 1

format XX: e.g. 11,12,15,21,22,25,51,52,55, counts = |D|^2

format XXX: counts = |D|^3

format XXXX: counts = |D|^4

exact n digits

if all numbers in D != N[0], counts = |d < N[0] | d in D| * |D|^(n-1), and we are done.

e.g. N = 34567, D = [1,2,8]

we can make:

X |3|^1
XX |3| ^ 2
XXX |3| ^ 3
XXXX |3| ^ 3
1XXXX, |3|^4
2XXXX, |3|^4
we can’t do 8XXXX

Total = (3^1 + 3^2 + 3^3 + 3^4) + 2 * |3|^ 4 = 120 + 162 = 282

N = 52525, D = [1,2,5]

However, if d = N[i], we need to check the next digit…

X |3|^1
XX |3| ^ 2
XXX |3| ^ 3
XXXX |3| ^ 3
1XXXX, |3|^4
2XXXX, |3|^4
5????
- 51XXX |3|^3
- 52???
  - 521XX |3|^2
  - 522XX |3|^2
  - 525??
    - 5251X |3|^1
    - 5252?
      - 52521 |3|^0
      - 52522 |3|^0
      - 52525 +1

total = (120) + 2 * |3|^4 + |3|^3 + 2*|3|^2 + |3|^1 + 2 * |3|^0 + 1 = 120 + 213 = 333

if every digit of N is from D, then we also have a valid solution, thus need to + 1.

C++

class Solution {

public:

int atMostNGivenDigitSet(vector<string>& D, int N) {

const string s = to_string(N);

const int n = s.length();

int ans = 0;

for (int i = 1; i < n; ++i)

ans += pow(D.size(), i);

for (int i = 0; i < n; ++i) {

bool prefix = false;

for (const string& d : D) {

if (d[0] < s[i]) {

ans += pow(D.size(), n - i - 1);

} else if (d[0] == s[i]) {

prefix = true;

break;

}

if (!prefix) return ans;

}

// plus one for solution N itself, all the digits are from D.

return ans + 1;

}

};

Java

// Author: Huahua, 4 ms

class Solution {

public int atMostNGivenDigitSet(String[] D, int N) {

char[] s = String.valueOf(N).toCharArray();

int n = s.length;

int ans = 0;

for (int i = 1; i < n; ++i)

ans += (int)Math.pow(D.length, i);

for (int i = 0; i < n; ++i) {

boolean prefix = false;

for (String d : D) {

if (d.charAt(0) < s[i]) {

ans += Math.pow(D.length, n - i - 1);

} else if (d.charAt(0) == s[i]) {

prefix = true;

break;

}

if (!prefix) return ans;

}

return ans + 1;

}

Python3

# Author: Huahua, 36 ms

class Solution:

def atMostNGivenDigitSet(self, D, N):

s = str(N)

n = len(s)

ans = sum(len(D) ** i for i in range(1, n))

for i, c in enumerate(s):

ans += len(D) ** (n - i - 1) * sum(d < c for d in D)

if c not in D: return ans

return ans + 1

花花酱 LeetCode 9. Palindrome Number

By zxi on September 12, 2018

Problem

Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward.

Example 1:

Input: 121
Output: true

Example 2:

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

Follow up:

Could you solve it without converting the integer to a string?

Solution 1: Convert to string (cheating)

Time complexity: O(log10(x))

Space complexity: O(log10(x))

C++

// Author: Huahua

class Solution {

public:

bool isPalindrome(int x) {

string s = to_string(x);

return s == string(rbegin(s), rend(s));

}

};

Solution 2: Digit by Digit

Every time we compare the first and last digits of x, if they are not the same, return false. Otherwise, remove first and last digit and continue this process.

How can we achieve that via int math?

e.g. x = 9999, t = pow((10, int)log10(x)) = 1000

first digit: x / t, last digit: x % 10

then x = (x – x / t * t) / 10 removes first and last digits.

t /= 100 since we removed two digits.

x / t = 9 = 9 = x % 10, 9999 => 99

9 = 9, 99 => “”

Time complexity: O(log10(x) / 2)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

bool isPalindrome(int x) {

if (x < 0) return false;

int d = static_cast<int>(log10(x) + 1);

int t = pow(10, d - 1);

for (int i = 0; i < d / 2; ++i) {

if (x / t != x % 10) return false;

x = (x - x / t * t) / 10;

t /= 100;

}

return true;

}

};

花花酱 LeetCode 7. Reverse Integer

By zxi on September 12, 2018

Problem

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2³¹, 2³¹− 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Solution: Simulation

Reverse digit by digit. Be careful about the overflow and negative numbers (especially in Python)

Time complexity: O(log(x)) ~ O(1)

Space complexity: O(log(x)) ~ O(1)

C++

// Author: Huahua

class Solution {

public:

int reverse(int x) {

int ans = 0;

while (x != 0) {

int r = x % 10;

if (ans > INT_MAX / 10 || ans < INT_MIN / 10) return 0;

ans = ans * 10 + r;

x /= 10;

}

return ans;

}

};

Java

class Solution {

public int reverse(int x) {

int ans = 0;

while (x != 0) {

int r = x % 10;

if (ans > Integer.MAX_VALUE / 10 || ans < Integer.MIN_VALUE / 10) return 0;

ans = ans * 10 + r;

x /= 10;

}

return ans;

}

Python3

class Solution:

def reverse(self, x):

min_val = -(2**31);

max_val = 2**31 - 1;

sign = -1 if x < 0 else 1

x = abs(x)

ans = 0

while x != 0:

ans = ans * 10 + (x % 10)

x //= 10

ans *= sign

if ans > max_val or ans < min_val: return 0

return ans

花花酱 LeetCode 633. Sum of Square Numbers

By zxi on August 22, 2018

Problem

Given a non-negative integer c, your task is to decide whether there’re two integers a and b such that a² + b² = c.

Example 1:

Input: 5
Output: True
Explanation: 1 * 1 + 2 * 2 = 5

Example 2:

Input: 3
Output: False

Solution: Math

Time complexity: O(sqrt(c))

Space complexity: O(1)

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

bool judgeSquareSum(int c) {

int m = sqrt(c);

for (int a = 0; a <= m; ++a) {

int b = round(sqrt(c - a * a));

if (a * a + b * b == c) return true;

}

return false;

}

};

花花酱 LeetCode 891. Sum of Subsequence Widths

By zxi on August 20, 2018

Problem

Given an array of integers A, consider all non-empty subsequences of A.

For any sequence S, let the width of S be the difference between the maximum and minimum element of S.

Return the sum of the widths of all subsequences of A.

As the answer may be very large, return the answer modulo 10^9 + 7.

Example 1:

Input: [2,1,3]
Output: 6
Explanation:
Subsequences are [1], [2], [3], [2,1], [2,3], [1,3], [2,1,3].
The corresponding widths are 0, 0, 0, 1, 1, 2, 2.
The sum of these widths is 6.

Note:

1 <= A.length <= 20000
1 <= A[i] <= 20000

Solution: Math

Sort the array, for A[i]:

i numbers <= A[i]. A[i] is the upper bound of 2^i subsequences.
n – i – 1 numbers >= A[i]. A[i] is the lower bound of 2^(n – i – 1) subsequences.
A[i] contributes A[i] * 2^i – A[i] * 2^(n – i – 1) to the ans.

\(ans = \sum\limits_{i=0}^{n-1}A_{i}2^{i} – A_{i}2^{n – i – 1} =\sum\limits_{i=0}^{n-1}(A_i – A_{n-i-1})2^{i}\)

Time complexity: O(nlogn)

Space complexity: O(1)

// Author: Huahua

// Running time: 56 ms

class Solution {

public:

int sumSubseqWidths(vector<int>& A) {

constexpr long kMod = 1e9 + 7;

const int n = A.size();

sort(begin(A), end(A));

long ans = 0;

long p = 1;

for (int i = 0; i < n; ++i) {

ans = (ans + (A[i] - A[n - i - 1]) * p) % kMod;

p = (p << 1) % kMod;

}

return (ans + kMod) % kMod;

}

};

Time complexity: O(n)

Space complexity: O(n)

Counting sort

// Author: Huahua

// Running time: 16 / 44 ms (beats 100%)

static auto x = [](){std::ios::sync_with_stdio(false);cin.tie(nullptr);return nullptr;}();

class Solution {

public:

int sumSubseqWidths(vector<int>& A) {

constexpr long kMod = 1e9 + 7;

const int n = A.size();

countingSort(A);

long ans = 0;

long p = 1;

for (int i = 0; i < n; ++i) {

ans = (ans + (A[i] - A[n - i - 1]) * p) % kMod;

p = (p << 1) % kMod;

}

return (ans + kMod) % kMod;

}

private:

void countingSort(vector<int>& A) {

int max_v = INT_MIN;

int min_v = INT_MAX;

for (int a : A) {

if (a > max_v) max_v = a;

if (a < min_v) min_v = a;

}

vector<int> c(max_v - min_v + 1);

for (int a : A)

++c[a - min_v];

int i = 0;

int j = 0;

while (i < c.size()) {

if (--c[i] >= 0)

A[j++] = i + min_v;

else

++i;

}

};