There are n cars traveling at different speeds in the same direction along a one-lane road. You are given an array cars of length n, where cars[i] = [positioni, speedi] represents:

• positioni is the distance between the ith car and the beginning of the road in meters. It is guaranteed that positioni < positioni+1.
• speedi is the initial speed of the ith car in meters per second.

For simplicity, cars can be considered as points moving along the number line. Two cars collide when they occupy the same position. Once a car collides with another car, they unite and form a single car fleet. The cars in the formed fleet will have the same position and the same speed, which is the initial speed of the slowest car in the fleet.

Return an array answer, where answer[i] is the time, in seconds, at which the ith car collides with the next car, or -1 if the car does not collide with the next car. Answers within 10-5 of the actual answers are accepted.

Example 1:

Input: cars = [[1,2],[2,1],[4,3],[7,2]]
Output: [1.00000,-1.00000,3.00000,-1.00000]
Explanation: After exactly one second, the first car will collide with the second car, and form a car fleet with speed 1 m/s. After exactly 3 seconds, the third car will collide with the fourth car, and form a car fleet with speed 2 m/s.


Example 2:

Input: cars = [[3,4],[5,4],[6,3],[9,1]]
Output: [2.00000,1.00000,1.50000,-1.00000]


Constraints:

• 1 <= cars.length <= 105
• 1 <= positioni, speedi <= 106
• positioni < positioni+1

## Solution: Monotonic Stack

Key observation: If my speed is slower than the speed of the previous car, not only mine but also all cars behind me will NEVER be able to catch/collide with the previous car. Such that we can throw it away.

Maintain a stack that stores the indices of cars with increasing speed.

Process car from right to left, for each car, pop the stack (throw the fastest car away) if any of the following conditions hold.
1) speed <= stack.top().speed
2) There are more than one car before me and it takes more than to collide the fastest car than time the fastest took to collide.

Time complexity: O(n)
Space complexity: O(n)

## C++

You are given two arrays of integers nums1 and nums2, possibly of different lengths. The values in the arrays are between 1 and 6, inclusive.

In one operation, you can change any integer’s value in any of the arrays to any value between 1 and 6, inclusive.

Return the minimum number of operations required to make the sum of values in nums1 equal to the sum of values in nums2. Return -1​​​​​ if it is not possible to make the sum of the two arrays equal.

Example 1:

Input: nums1 = [1,2,3,4,5,6], nums2 = [1,1,2,2,2,2]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed.
- Change nums2[0] to 6. nums1 = [1,2,3,4,5,6], nums2 = [6,1,2,2,2,2].
- Change nums1[5] to 1. nums1 = [1,2,3,4,5,1], nums2 = [6,1,2,2,2,2].
- Change nums1[2] to 2. nums1 = [1,2,2,4,5,1], nums2 = [6,1,2,2,2,2].


Example 2:

Input: nums1 = [1,1,1,1,1,1,1], nums2 = [6]
Output: -1
Explanation: There is no way to decrease the sum of nums1 or to increase the sum of nums2 to make them equal.


Example 3:

Input: nums1 = [6,6], nums2 = [1]
Output: 3
Explanation: You can make the sums of nums1 and nums2 equal with 3 operations. All indices are 0-indexed.
- Change nums1[0] to 2. nums1 = [2,6], nums2 = [1].
- Change nums1[1] to 2. nums1 = [2,2], nums2 = [1].
- Change nums2[0] to 4. nums1 = [2,2], nums2 = [4].


Constraints:

• 1 <= nums1.length, nums2.length <= 105
• 1 <= nums1[i], nums2[i] <= 6

## Solution: Greedy

Assuming sum(nums1) < sum(nums2),
sort both arrays
* scan nums1 from left to right, we need to increase the value form the smallest one.
* scan nums2 from right to left, we need to decrease the value from the largest one.
Each time, select the one with the largest delta to change.

e.g. nums1[i] = 2, nums[j] = 4, delta1 = 6 – 2 = 4, delta2 = 4 – 1 = 3, Increase 2 to 6 instead of decreasing 4 to 1.

Time complexity: O(mlogm + nlogn)
Space complexity: O(1)

## C++

You would like to make dessert and are preparing to buy the ingredients. You have n ice cream base flavors and m types of toppings to choose from. You must follow these rules when making your dessert:

• There must be exactly one ice cream base.
• You can add one or more types of topping or have no toppings at all.
• There are at most two of each type of topping.

You are given three inputs:

• baseCosts, an integer array of length n, where each baseCosts[i] represents the price of the ith ice cream base flavor.
• toppingCosts, an integer array of length m, where each toppingCosts[i] is the price of one of the ith topping.
• target, an integer representing your target price for dessert.

You want to make a dessert with a total cost as close to target as possible.

Return the closest possible cost of the dessert to target. If there are multiple, return the lower one.

Example 1:

Input: baseCosts = [1,7], toppingCosts = [3,4], target = 10
Output: 10
Explanation: Consider the following combination (all 0-indexed):
- Choose base 1: cost 7
- Take 1 of topping 0: cost 1 x 3 = 3
- Take 0 of topping 1: cost 0 x 4 = 0
Total: 7 + 3 + 0 = 10.


Example 2:

Input: baseCosts = [2,3], toppingCosts = [4,5,100], target = 18
Output: 17
Explanation: Consider the following combination (all 0-indexed):
- Choose base 1: cost 3
- Take 1 of topping 0: cost 1 x 4 = 4
- Take 2 of topping 1: cost 2 x 5 = 10
- Take 0 of topping 2: cost 0 x 100 = 0
Total: 3 + 4 + 10 + 0 = 17. You cannot make a dessert with a total cost of 18.


Example 3:

Input: baseCosts = [3,10], toppingCosts = [2,5], target = 9
Output: 8
Explanation: It is possible to make desserts with cost 8 and 10. Return 8 as it is the lower cost.


Example 4:

Input: baseCosts = [10], toppingCosts = [1], target = 1
Output: 10
Explanation: Notice that you don't have to have any toppings, but you must have exactly one base.

Constraints:

• n == baseCosts.length
• m == toppingCosts.length
• 1 <= n, m <= 10
• 1 <= baseCosts[i], toppingCosts[i] <= 104
• 1 <= target <= 104

## Solution: DP / Knapsack

Pre-compute the costs of all possible combinations of toppings.

Time complexity: O(sum(toppings) * 2 * (m + n)) ~ O(10^6)
Space complexity: O(sum(toppings)) ~ O(10^5)

## Solution 2: DFS

Combination

Time complexity: O(3^m * n)
Space complexity: O(m)

## C++

You are given an array items, where each items[i] = [typei, colori, namei] describes the type, color, and name of the ith item. You are also given a rule represented by two strings, ruleKey and ruleValue.

The ith item is said to match the rule if one of the following is true:

• ruleKey == "type" and ruleValue == typei.
• ruleKey == "color" and ruleValue == colori.
• ruleKey == "name" and ruleValue == namei.

Return the number of items that match the given rule.

Example 1:

Input: items = [["phone","blue","pixel"],["computer","silver","lenovo"],["phone","gold","iphone"]], ruleKey = "color", ruleValue = "silver"
Output: 1
Explanation: There is only one item matching the given rule, which is ["computer","silver","lenovo"].


Example 2:

Input: items = [["phone","blue","pixel"],["computer","silver","phone"],["phone","gold","iphone"]], ruleKey = "type", ruleValue = "phone"
Output: 2
Explanation: There are only two items matching the given rule, which are ["phone","blue","pixel"] and ["phone","gold","iphone"]. Note that the item ["computer","silver","phone"] does not match.

Constraints:

• 1 <= items.length <= 104
• 1 <= typei.length, colori.length, namei.length, ruleValue.length <= 10
• ruleKey is equal to either "type""color", or "name".
• All strings consist only of lowercase letters.

## Solution: Brute Force

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

## C++

You are given two strings, word1 and word2. You want to construct a string in the following manner:

• Choose some non-empty subsequence subsequence1 from word1.
• Choose some non-empty subsequence subsequence2 from word2.
• Concatenate the subsequences: subsequence1 + subsequence2, to make the string.

Return the length of the longest palindrome that can be constructed in the described manner. If no palindromes can be constructed, return 0.

subsequence of a string s is a string that can be made by deleting some (possibly none) characters from s without changing the order of the remaining characters.

palindrome is a string that reads the same forward as well as backward.

Example 1:

Input: word1 = "cacb", word2 = "cbba"
Output: 5
Explanation: Choose "ab" from word1 and "cba" from word2 to make "abcba", which is a palindrome.

Example 2:

Input: word1 = "ab", word2 = "ab"
Output: 3
Explanation: Choose "ab" from word1 and "a" from word2 to make "aba", which is a palindrome.

Example 3:

Input: word1 = "aa", word2 = "bb"
Output: 0
Explanation: You cannot construct a palindrome from the described method, so return 0.

Constraints:

• 1 <= word1.length, word2.length <= 1000
• word1 and word2 consist of lowercase English letters.

## Solution: DP

Let s = word1 + word2, build dp table on s. We just need to make sure there’s at least one char from each string.

Time complexity: O((m+n)^2)
Space complexity: O((m+n)^2)

## C++

O(m+n) Space complexity