# Problem

Given an array A, partition it into two (contiguous) subarrays left and right so that:

• Every element in left is less than or equal to every element in right.
• left and right are non-empty.
• left has the smallest possible size.

Return the length of left after such a partitioning.  It is guaranteed that such a partitioning exists.

Example 1:

Input: [5,0,3,8,6]
Output: 3
Explanation: left = [5,0,3], right = [8,6]

Example 2:

Input: [1,1,1,0,6,12]
Output: 4
Explanation: left = [1,1,1,0], right = [6,12]

Note:

1. 2 <= A.length <= 30000
2. 0 <= A[i] <= 10^6
3. It is guaranteed there is at least one way to partition A as described.

# Solution 1: BST

Time complexity: O(nlogn)

Space complexity: O(n)

# Solution 2: Greedy

Time complexity: O(n)

Space complexity: O(1)

# Problem

We are given two arrays A and B of words.  Each word is a string of lowercase letters.

Now, say that word b is a subset of word a if every letter in b occurs in aincluding multiplicity.  For example, "wrr" is a subset of "warrior", but is not a subset of "world".

Now say a word a from A is universal if for every b in Bb is a subset of a.

Return a list of all universal words in A.  You can return the words in any order.

Example 1:

Example 2:

Example 3:

Example 4:

Example 5:

Note:

1. 1 <= A.length, B.length <= 10000
2. 1 <= A[i].length, B[i].length <= 10
3. A[i] and B[i] consist only of lowercase letters.
4. All words in A[i] are unique: there isn’t i != j with A[i] == A[j].

# Solution: Hashtable

Find the max requirement for each letter from B.

Time complexity: O(|A|+|B|)

Space complexity: O(26)

# Problem

n a deck of cards, each card has an integer written on it.

Return true if and only if you can choose X >= 2 such that it is possible to split the entire deck into 1 or more groups of cards, where:

• Each group has exactly X cards.
• All the cards in each group have the same integer.

Example 1:

Input: [1,2,3,4,4,3,2,1]
Output: true
Explanation: Possible partition [1,1],[2,2],[3,3],[4,4]

Example 2:

Input: [1,1,1,2,2,2,3,3]
Output: false
Explanation: No possible partition.

Example 3:

Input: [1]
Output: false
Explanation: No possible partition.

Example 4:

Input: [1,1]
Output: true
Explanation: Possible partition [1,1]

Example 5:

Input: [1,1,2,2,2,2]
Output: true
Explanation: Possible partition [1,1],[2,2],[2,2]

Note:

1. 1 <= deck.length <= 10000
2. 0 <= deck[i] < 10000

# Solution 1: HashTable + Brute Force

Try all possible Xs. 2 ~ deck.size()

Time complexity: ~O(n)

Space complexity: O(1)

# Solution 2: HashTable + GCD

Time complexity: O(nlogn)

Space complexity: O(1)

# Problem

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:

• The number of elements initialized in nums1 and nums2 are m and n respectively.
• You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2.

Example:

Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3

Output: [1,2,2,3,5,6]

# Solution:

Fill nums1 from back to front

Time complexity: O(m + n)

Space complexity: O(1) in-place

# Problem

Given an array A of integers, for each integer A[i] we need to choose either x = -K or x = K, and add x to A[i] (only once).

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: 3
Explanation: B = [4,6,3]

Note:

1. 1 <= A.length <= 10000
2. 0 <= A[i] <= 10000
3. 0 <= K <= 10000

# Solution: Greedy

Sort the array and compare adjacent numbers.

Time complexity: O(nlogn)

Space complexity: O(1)

## Python3

Mission News Theme by Compete Themes.