Posts tagged as “math”

花花酱 LeetCode 939. Minimum Area Rectangle

By zxi on November 11, 2018

Problem

Given a set of points in the xy-plane, determine the minimum area of a rectangle formed from these points, with sides parallel to the x and y axes.

If there isn’t any rectangle, return 0.

Example 1:

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

Example 2:

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

Note:

1 <= points.length <= 500
0 <= points[i][0] <= 40000
0 <= points[i][1] <= 40000
All points are distinct.

Solution 1: HashTable + Brute Force

Try all pairs of points to form a diagonal and see whether pointers of another diagonal exist or not.

Assume two points are (x0, y0), (x1, y1) x0 != x1 and y0 != y1. The other two points will be (x0, y1), (x1, y0)

Time complexity: O(n^2)

Space complexity: O(n)

C++

// Author: Huahua, running time: 96 ms

class Solution {

public:

int minAreaRect(vector<vector<int>>& points) {

unordered_map<int, unordered_set<int>> s;

for (const auto& point : points)

s[point[0]].insert(point[1]);

const int n = points.size();

int min_area = INT_MAX;

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

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

int x0 = points[i][0];

int y0 = points[i][1];

int x1 = points[j][0];

int y1 = points[j][1];

if (x0 == x1 || y0 == y1) continue;

int area = abs(x0 - x1) * abs(y0 - y1);

if (area > min_area) continue;

if (!s[x1].count(y0) || !s[x0].count(y1)) continue;

min_area = area;

}

return min_area == INT_MAX ? 0 : min_area;

}

};

C++ / 1D hashtable

// Author: Huahua, 68 ms

class Solution {

public:

int minAreaRect(vector<vector<int>>& points) {

unordered_set<int> s;

for (const auto& point : points)

s.insert((point[0] << 16) | point[1]);

const int n = points.size();

int min_area = INT_MAX;

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

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

int x0 = points[i][0];

int y0 = points[i][1];

int x1 = points[j][0];

int y1 = points[j][1];

if (x0 == x1 || y0 == y1) continue;

int area = abs(x0 - x1) * abs(y0 - y1);

if (area > min_area) continue;

int p0 = (x1 << 16) | y0;

int p1 = (x0 << 16) | y1;

if (!s.count(p0) || !s.count(p1)) continue;

min_area = area;

}

return min_area == INT_MAX ? 0 : min_area;

}

};

花花酱 LeetCode 470. Implement Rand10() Using Rand7()

By zxi on October 15, 2018

Problem

Given a function rand7 which generates a uniform random integer in the range 1 to 7, write a function rand10 which generates a uniform random integer in the range 1 to 10.

Do NOT use system’s Math.random().

Example 1:

Input: 1
Output: [7]

Example 2:

Input: 2
Output: [8,4]

Example 3:

Input: 3
Output: [8,1,10]

Note:

rand7 is predefined.
Each testcase has one argument: n, the number of times that rand10 is called.

Follow up:

What is the expected value for the number of calls to rand7() function?
Could you minimize the number of calls to rand7()?

Solution:

Time complexity: O(1)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int rand10() {

int index = INT_MAX;

while (index >= 40)

index = 7 * (rand7() - 1) + (rand7() - 1);

return index % 10 + 1;

}

};

花花酱 LeetCode 923. 3Sum With Multiplicity

By zxi on October 13, 2018

Problem

Given an integer array A, and an integer target, return the number of tuples i, j, k such that i < j < k and A[i] + A[j] + A[k] == target.

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

Example 1:

Input: A = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation: 
Enumerating by the values (A[i], A[j], A[k]):
(1, 2, 5) occurs 8 times;
(1, 3, 4) occurs 8 times;
(2, 2, 4) occurs 2 times;
(2, 3, 3) occurs 2 times.

Example 2:

Input: A = [1,1,2,2,2,2], target = 5
Output: 12
Explanation: 
A[i] = 1, A[j] = A[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.

Note:

3 <= A.length <= 3000
0 <= A[i] <= 100
0 <= target <= 300

Solution: Math / Combination

Time complexity: O(n + |target|^2)

Space complexity: O(|target|)

C++

// Author: Huahua, runtime: 4 ms

class Solution {

public:

int threeSumMulti(vector<int>& A, int target) {

constexpr int kMaxN = 100;

constexpr int kMod = 1e9 + 7;

vector<long> c(kMaxN + 1, 0);

for (int a : A) ++c[a];

long ans = 0;

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

for (int j = i; j <= target; ++j) {

const int k = target - i - j;

if (k < 0 || k >= c.size() || k < j) continue;

if (!c[i] || !c[j] || !c[k]) continue;

if (i == j && j == k)

ans += (c[i] - 2) * (c[i] - 1) * c[i] / 6;

else if (i == j && j != k)

ans += c[i] * (c[i] - 1) / 2 * c[k];

else if (i != j && j == k)

ans += c[i] * (c[j] - 1) * c[j] / 2;

else

ans += c[i] * c[j] * c[k];

}

return ans % kMod;

}

};

花花酱 LeetCode 908. Smallest Range I

By zxi on September 23, 2018

Problem

Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and add xto A[i].

After this process, we have some array B.

Return the smallest possible difference between the maximum value of B and the minimum value of B.

Example 1:

Input: A = [1], K = 0
Output: 0
Explanation: B = [1]

Example 2:

Input: A = [0,10], K = 2
Output: 6
Explanation: B = [2,8]

Example 3:

Input: A = [1,3,6], K = 3
Output: 0
Explanation: B = [3,3,3] or B = [4,4,4]

Note:

1 <= A.length <= 10000
0 <= A[i] <= 10000
0 <= K <= 10000

Solution 0: Brute Force (TLE)

Try all pairs

Time complexity: O(n^2)

Space complexity: O(1)

Solution 1: Math

Time complexity: O(n)

Space complexity: O(1)

Find the min/max element of the array.

min + k v.s. max – k

ans = max(0, (max – min) – 2 * k))

C++

// Author: Huahua

class Solution {

public:

int smallestRangeI(vector<int>& A, int K) {

int a_min = *min_element(begin(A), end(A));

int a_max = *max_element(begin(A), end(A));

return max(0, (a_max - a_min) - 2 * K);

}

};

Python3

# Author: Huahua, 72 ms

class Solution:

def smallestRangeI(self, A, K):

return max(0, max(A) - min(A) - 2 * K);

花花酱 LeetCode 223. Rectangle Area

By zxi on September 19, 2018

Problem

Find the total area covered by two rectilinear rectangles in a 2D plane.

Each rectangle is defined by its bottom left corner and top right corner as shown in the figure.

Rectangle Area

Example:

Input: A = -3, B = 0, C = 3, D = 4, E = 0, F = -1, G = 9, H = 2
Output: 45

Note:

Assume that the total area is never beyond the maximum possible value of int.

Solution:

area1 + area2 – overlapped area

Time complexity: O(1)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {

int area1 = (C - A) * (D - B);

int area2 = (G - E) * (H - F);

int x1 = max(A, E);

int x2 = max(x1, min(C, G));

int y1 = max(B, F);

int y2 = max(y1, min(D, H));

return area1 + area2 - (x2 - x1) * (y2 - y1);

}

};

Java

// Author: Huahua

class Solution {

public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {

int area1 = (C - A) * (D - B);

int area2 = (G - E) * (H - F);

int x1 = Math.max(A, E);

int x2 = Math.max(x1, Math.min(C, G));

int y1 = Math.max(B, F);

int y2 = Math.max(y1, Math.min(D, H));

return area1 + area2 - (x2 - x1) * (y2 - y1);

}

Python3

# Author: Huahua

class Solution:

def computeArea(self, A, B, C, D, E, F, G, H):

area1 = (C - A) * (D - B);

area2 = (G - E) * (H - F);

x1 = max(A, E);

x2 = max(x1, min(C, G));

y1 = max(B, F);

y2 = max(y1, min(D, H));

return area1 + area2 - (x2 - x1) * (y2 - y1)