Given two non-negative integers low and high. Return the count of odd numbers between low and high (inclusive).
Example 1:
Input: low = 3, high = 7
Output: 3
Explanation: The odd numbers between 3 and 7 are [3,5,7].
Example 2:
Input: low = 8, high = 10
Output: 1
Explanation: The odd numbers between 8 and 10 are [9].
Constraints:
0 <= low <= high <= 10^9
Solution: Math
The count of odd numbers between [1, low – 1] is low / 2 e.g. low = 6, we have [1,3,5] in range [1, 5] and count is 6/2 = 3. The count of odd numbers between [1, high] is (high + 1) / 2 e.g. high = 7, we have [1,3,5,7] in range [1, 7] and count is (7+1) / 2 = 4
Then the count of odd numbers in range [low, high] = count(1, high) – count(1, low-1) e.g. in range [6, 7] we only have [7], count: 4 – 3 = 1
We have a wooden plank of the length nunits. Some ants are walking on the plank, each ant moves with speed 1 unit per second. Some of the ants move to the left, the other move to the right.
When two ants moving in two different directions meet at some point, they change their directions and continue moving again. Assume changing directions doesn’t take any additional time.
When an ant reaches one end of the plank at a time t, it falls out of the plank imediately.
Given an integer n and two integer arrays left and right, the positions of the ants moving to the left and the right. Return the moment when the last ant(s) fall out of the plank.
Example 1:
Input: n = 4, left = [4,3], right = [0,1]
Output: 4
Explanation: In the image above:
-The ant at index 0 is named A and going to the right.
-The ant at index 1 is named B and going to the right.
-The ant at index 3 is named C and going to the left.
-The ant at index 4 is named D and going to the left.
Note that the last moment when an ant was on the plank is t = 4 second, after that it falls imediately out of the plank. (i.e. We can say that at t = 4.0000000001, there is no ants on the plank).
Example 2:
Input: n = 7, left = [], right = [0,1,2,3,4,5,6,7]
Output: 7
Explanation: All ants are going to the right, the ant at index 0 needs 7 seconds to fall.
Example 3:
Input: n = 7, left = [0,1,2,3,4,5,6,7], right = []
Output: 7
Explanation: All ants are going to the left, the ant at index 7 needs 7 seconds to fall.
Example 4:
Input: n = 9, left = [5], right = [4]
Output: 5
Explanation: At t = 1 second, both ants will be at the same intial position but with different direction.
Example 5:
Input: n = 6, left = [6], right = [0]
Output: 6
Constraints:
1 <= n <= 10^4
0 <= left.length <= n + 1
0 <= left[i] <= n
0 <= right.length <= n + 1
0 <= right[i] <= n
1 <= left.length + right.length <= n + 1
All values of left and right are unique, and each value can appear only in one of the two arrays.
Solution: Keep Walking
When two ants A –> and <– B meet at some point, they change directions <– A B –>, we can swap the ids of the ants as <– B A–>, so it’s the same as walking individually and passed by. Then we just need to find the max/min of the left/right arrays.
Given an array of integers arr of even length n and an integer k.
We want to divide the array into exactly n / 2 pairs such that the sum of each pair is divisible by k.
Return True If you can find a way to do that or False otherwise.
Example 1:
Input: arr = [1,2,3,4,5,10,6,7,8,9], k = 5
Output: true
Explanation: Pairs are (1,9),(2,8),(3,7),(4,6) and (5,10).
Example 2:
Input: arr = [1,2,3,4,5,6], k = 7
Output: true
Explanation: Pairs are (1,6),(2,5) and(3,4).
Example 3:
Input: arr = [1,2,3,4,5,6], k = 10
Output: false
Explanation: You can try all possible pairs to see that there is no way to divide arr into 3 pairs each with sum divisible by 10.
Example 4:
Input: arr = [-10,10], k = 2
Output: true
Example 5:
Input: arr = [-1,1,-2,2,-3,3,-4,4], k = 3
Output: true
Constraints:
arr.length == n
1 <= n <= 10^5
n is even.
-10^9 <= arr[i] <= 10^9
1 <= k <= 10^5
Solution: Mod and Count
Count the frequency of (x % k + k) % k. f[0] should be even (zero is also even) f[1] = f[k -1] ((1 + k – 1) % k == 0) f[2] = f[k -2] ((2 + k – 2) % k == 0) … Time complexity: O(n) Space complexity: O(k)
Given an integer n, return a list of all simplified fractions between 0 and 1 (exclusive) such that the denominator is less-than-or-equal-to n. The fractions can be in any order.
Example 1:
Input: n = 2
Output: ["1/2"]
Explanation: "1/2" is the only unique fraction with a denominator less-than-or-equal-to 2.
Example 2:
Input: n = 3
Output: ["1/2","1/3","2/3"]
Example 3:
Input: n = 4
Output: ["1/2","1/3","1/4","2/3","3/4"]
Explanation: "2/4" is not a simplified fraction because it can be simplified to "1/2".
Example 4:
Input: n = 1
Output: []
Constraints:
1 <= n <= 100
Solution: GCD
if gcd(a, b) == 1 then a/b is a simplified frication.
std::gcd is available since c++17.
Time complexity: O(n^2logn) Space complexity: O(1)
Given an integer array nums, return the sum of divisors of the integers in that array that have exactly four divisors.
If there is no such integer in the array, return 0.
Example 1:
Input: nums = [21,4,7]
Output: 32
Explanation:
21 has 4 divisors: 1, 3, 7, 21
4 has 3 divisors: 1, 2, 4
7 has 2 divisors: 1, 7
The answer is the sum of divisors of 21 only.
Constraints:
1 <= nums.length <= 10^4
1 <= nums[i] <= 10^5
Solution: Math
If a number is a perfect square (e.g. 9 = 3 * 3), it will have odd number of divisors. (9: 1, 3, 9).
Time complexity: O(sum(sqrt(num_i)) Space complexity: O(1)
