O(logn) Archives - Huahua's Tech Road

花花酱 LeetCode 1688. Count of Matches in Tournament

By zxi on December 13, 2020

You are given an integer n, the number of teams in a tournament that has strange rules:

If the current number of teams is even, each team gets paired with another team. A total of n / 2 matches are played, and n / 2 teams advance to the next round.
If the current number of teams is odd, one team randomly advances in the tournament, and the rest gets paired. A total of (n - 1) / 2 matches are played, and (n - 1) / 2 + 1 teams advance to the next round.

Return the number of matches played in the tournament until a winner is decided.

Example 1:

Input: n = 7
Output: 6
Explanation: Details of the tournament: 
- 1st Round: Teams = 7, Matches = 3, and 4 teams advance.
- 2nd Round: Teams = 4, Matches = 2, and 2 teams advance.
- 3rd Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
Total number of matches = 3 + 2 + 1 = 6.

Example 2:

Input: n = 14
Output: 13
Explanation: Details of the tournament:
- 1st Round: Teams = 14, Matches = 7, and 7 teams advance.
- 2nd Round: Teams = 7, Matches = 3, and 4 teams advance.
- 3rd Round: Teams = 4, Matches = 2, and 2 teams advance.
- 4th Round: Teams = 2, Matches = 1, and 1 team is declared the winner.
Total number of matches = 7 + 3 + 2 + 1 = 13.

Constraints:

1 <= n <= 200

Solution: Simulation / Recursion

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

C++

// Ver 1

class Solution {

public:

int numberOfMatches(int n) {

int ans = 0;

while (n > 1) {

ans += n / 2 + (n & 1);

n /= 2;

}

return ans;

}

};

// Ver 2

class Solution {

public:

int numberOfMatches(int n) {

return n > 1 ? n / 2 + (n & 1) + numberOfMatches(n / 2) : 0;

}

};

花花酱 LeetCode 365. Water and Jug Problem

By zxi on March 9, 2020

You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs.

If z liters of water is measurable, you must have z liters of water contained within one or both buckets by the end.

Operations allowed:

Fill any of the jugs completely with water.
Empty any of the jugs.
Pour water from one jug into another till the other jug is completely full or the first jug itself is empty.

Example 1: (From the famous “Die Hard” example)

Input: x = 3, y = 5, z = 4
Output: True

Example 2:

Input: x = 2, y = 6, z = 5
Output: False

Solution: Math

special case 1: x == z or y == z or x + y == z: return True
special case 2: x + y < z: return False
normal case: z must be a factor of gcd(x, y)

Time complexity: O(log(min(x, y))
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

bool canMeasureWater(int x, int y, int z) {

if (x == z || y == z || x + y == z) return true;

if (x + y < z) return false;

return z % gcd(x, y) == 0;

}

};

花花酱 LeetCode 162. Find Peak Element

By zxi on January 31, 2020

A peak element is an element that is greater than its neighbors.

Given an input array nums, where nums[i] ≠ nums[i+1], find a peak element and return its index.

The array may contain multiple peaks, in that case return the index to any one of the peaks is fine.

You may imagine that nums[-1] = nums[n] = -∞.

Example 1:

Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.

Example 2:

Input: nums = [1,2,1,3,5,6,4]
Output: 1 or 5 
Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.

Solution: Binary Search

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

C++

// Author: Huahua

class Solution {

public:

int findPeakElement(const vector<int> &num) {

int l = 0, r = num.size() - 1; // preventing OOB

while (l < r) {

int m = l + (r - l) / 2;

// Find the first m s.t. num[m] > num[m + 1]

if (num[m] > num[m + 1])

r = m;

else

l = m + 1;

}

return l;

}

};

Binary search template 二分搜索模板

By zxi on December 29, 2019

Python

# Author: Huahua

# Returns the smallest m in [l, r),

# s.t. cond(m) == True

# If not found returns r.

def binarySearch(l, r)

while l < r:

m = l + (r - l) // 2

if cond(m):

r = m

else

l = m + 1

return l

C++

// Author: Huahua

// Returns the smallest m in [l, r),

// s.t. cond(m) == True

// If not found returns r.

int binarySearch(int l, int r) {

while (l < r) {

int m = l + (r - l) / 2;

if (cond(m)) r = m;

else l = m + 1;

}

return l;

}

花花酱 LeetCode 1201. Ugly Number III

By zxi on September 22, 2019

Write a program to find the n-th ugly number.

Ugly numbers are positive integers which are divisible by a or b or c.

Example 1:

Input: n = 3, a = 2, b = 3, c = 5
Output: 4
Explanation: The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3rd is 4.

Example 2:

Input: n = 4, a = 2, b = 3, c = 4
Output: 6
Explanation: The ugly numbers are 2, 3, 4, 6, 8, 9, 12... The 4th is 6.

Example 3:

Input: n = 5, a = 2, b = 11, c = 13
Output: 10
Explanation: The ugly numbers are 2, 4, 6, 8, 10, 11, 12, 13... The 5th is 10.

Example 4:

Input: n = 1000000000, a = 2, b = 217983653, c = 336916467
Output: 1999999984

Constraints:

1 <= n, a, b, c <= 10^9
1 <= a * b * c <= 10^18
It’s guaranteed that the result will be in range [1, 2 * 10^9]

Solution: Binary Search

Number of ugly numbers that are <= m are:

m / a + m / b + m / c – (m / LCM(a,b) + m / LCM(a, c) + m / LCM(b, c) + m / LCM(a, LCM(b, c))

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

C++

// Author: Huahua

class Solution {

public:

int nthUglyNumber(int n, long a, long b, long c) {

long l = 1;

long r = INT_MAX;

long ab = lcm(a, b);

long ac = lcm(a, c);

long bc = lcm(b, c);

long abc = lcm(a, bc);

while (l < r) {

long m = l + (r - l) / 2;

long k = m / a + m / b + m / c - m / ab - m / ac - m / bc + m / abc;

if (k >= n) r = m;

else l = m + 1;

}

return l;

}

};

Posts tagged as “O(logn)”

花花酱 LeetCode 1688. Count of Matches in Tournament

Solution: Simulation / Recursion

C++

花花酱 LeetCode 365. Water and Jug Problem

Solution: Math

C++

花花酱 LeetCode 162. Find Peak Element

Solution: Binary Search

C++

Binary search template 二分搜索模板

Python

C++

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花花酱 LeetCode 1201. Ugly Number III

Solution: Binary Search

C++