Posts published in “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 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;

}

};

花花酱 LeetCode 628. Maximum Product of Three Numbers

By zxi on August 17, 2018

Problem

Given an integer array, find three numbers whose product is maximum and output the maximum product.

Example 1:

Input: [1,2,3]
Output: 6

Example 2:

Input: [1,2,3,4]
Output: 24

Note:

The length of the given array will be in range [3,10⁴] and all elements are in the range [-1000, 1000].
Multiplication of any three numbers in the input won’t exceed the range of 32-bit signed integer.

Idea:

Find the top 3 numbers t1, t2, t3, and bottom 2 numbers, b1, b2.

If all numbers are positives, answer must be t1 * t2 * t3.

Since the number can go negative, the answer must be either t1*t2*t3 or b1 * b2 * t1, if b1 and b2 are both negatives.

ex. nums: [5, 1, -6, 3, -1]

t1, t2, t3: 5, 3, 1

b1, b2: -6, -1

t1 * t2 * t3 = 15

t1 * b1 * b2 = 30

Solution 1: Manual Tracking

Time complexity: O(n)

Space complexity: O(1)

// Author: Huahua

// Running time: 28 ms

class Solution {

public:

int maximumProduct(vector<int>& nums) {

int max1 = INT_MIN;

int max2 = INT_MIN;

int max3 = INT_MIN;

int min1 = INT_MAX;

int min2 = INT_MAX;

for (const int num: nums) {

if (num > max1) {

max3 = max2;

max2 = max1;

max1 = num;

} else if (num > max2) {

max3 = max2;

max2 = num;

} else if (num > max3) {

max3=num;

}

if (num < min1) {

min2 = min1;

min1 = num;

} else if (num < min2) {

min2=num;

}

return max(max1*max2*max3, max1*min1*min2);

}

};

Solution 2: Sorting

Time complexity: O(nlogn)

Space complexity: O(1)

// Author: Huahua

// Running time: 48 ms

class Solution {

public:

int maximumProduct(vector<int>& nums) {

int n = nums.size();

sort(nums.rbegin(), nums.rend());

return max(nums[0] * nums[1] * nums[2], nums[0] * nums[n - 1] * nums[n - 2]);

}

};

Solution 3: Two Heaps (Priority Queues)

Time complexity: O(nlog3)

Space complexity: O(2 + 3)

// Author: Huahua

// Running time: 40 ms

class Solution {

public:

int maximumProduct(vector<int>& nums) {

priority_queue<int, vector<int>, less<int>> min_q; // max_heap

priority_queue<int, vector<int>, greater<int>> max_q; // min_heap

for (int num : nums) {

max_q.push(num);

if (max_q.size() > 3) max_q.pop();

min_q.push(num);

if (min_q.size() > 2) min_q.pop();

}

int max3 = max_q.top(); max_q.pop();

int max2 = max_q.top(); max_q.pop();

int max1 = max_q.top(); max_q.pop();

int min2 = min_q.top(); min_q.pop();

int min1 = min_q.top(); min_q.pop();

return max(max1 * max2 * max3, max1 * min1 * min2);

}

};

花花酱 LeetCode 463. Island Perimeter

By zxi on August 17, 2018

Problem

You are given a map in form of a two-dimensional integer grid where 1 represents land and 0 represents water. Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells). The island doesn’t have “lakes” (water inside that isn’t connected to the water around the island). One cell is a square with side length 1. The grid is rectangular, width and height don’t exceed 100. Determine the perimeter of the island.

Example:

[[0,1,0,0],
 [1,1,1,0],
 [0,1,0,0],
 [1,1,0,0]]

Answer: 16
Explanation: The perimeter is the 16 yellow stripes in the image below:

Solution: Counting

If a land has 0 neighbour, it contributes 4 to the perimeter

If a land has 1 neighbour, it contributes 3 to the perimeter

If a land has 2 neighbours, it contributes 2 to the perimeter

If a land has 3 neighbours, it contributes 1 to the perimeter

If a land has 4 neighbours, it contributes 0 to the perimeter

perimeter = area * 4 – total_neighbours

Time complexity: O(mn)

Space complexity: O(1)

// Author: Huahua

// Running time: 148 ms

class Solution {

public:

int islandPerimeter(vector<vector<int>>& grid) {

if (grid.empty()) return 0;

int m = grid.size();

int n = grid[0].size();

int area = 0;

int conn = 0;

for (int y = 0; y < m; ++y)

for (int x = 0; x < n; ++x)

if (grid[y][x] == 1) {

++area;

if (y > 0 && grid[y - 1][x] == 1) ++conn;

if (x > 0 && grid[y][x - 1] == 1) ++conn;

}

return area*4 - conn*2;

}

};