# Posts tagged as “hashtable”

There are n people whose IDs go from 0 to n - 1 and each person belongs exactly to one group. Given the array groupSizes of length n telling the group size each person belongs to, return the groups there are and the people’s IDs each group includes.

You can return any solution in any order and the same applies for IDs. Also, it is guaranteed that there exists at least one solution.

Example 1:

Input: groupSizes = [3,3,3,3,3,1,3]
Output: [,[0,1,2],[3,4,6]]
Explanation:
Other possible solutions are [[2,1,6],,[0,4,3]] and [,[0,6,2],[4,3,1]].


Example 2:

Input: groupSizes = [2,1,3,3,3,2]
Output: [,[0,5],[2,3,4]]


Constraints:

• groupSizes.length == n
• 1 <= n <= 500
• 1 <= groupSizes[i] <= n

## Solution: HashMap + Greedy

hashmap: group_size -> {ids}
greedy: whenever a group of size s has s people, assign those s people to the same group.

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

## C++

On an 8×8 chessboard, there can be multiple Black Queens and one White King.

Given an array of integer coordinates queens that represents the positions of the Black Queens, and a pair of coordinates king that represent the position of the White King, return the coordinates of all the queens (in any order) that can attack the King.

Example 1:

Input: queens = [[0,1],[1,0],[4,0],[0,4],[3,3],[2,4]], king = [0,0]
Output: [[0,1],[1,0],[3,3]]
Explanation:
The queen at [0,1] can attack the king cause they're in the same row.
The queen at [1,0] can attack the king cause they're in the same column.
The queen at [3,3] can attack the king cause they're in the same diagnal.
The queen at [0,4] can't attack the king cause it's blocked by the queen at [0,1].
The queen at [4,0] can't attack the king cause it's blocked by the queen at [1,0].
The queen at [2,4] can't attack the king cause it's not in the same row/column/diagnal as the king.


Example 2:

Input: queens = [[0,0],[1,1],[2,2],[3,4],[3,5],[4,4],[4,5]], king = [3,3]
Output: [[2,2],[3,4],[4,4]]


Example 3:

Input: queens = [[5,6],[7,7],[2,1],[0,7],[1,6],[5,1],[3,7],[0,3],[4,0],[1,2],[6,3],[5,0],[0,4],[2,2],[1,1],[6,4],[5,4],[0,0],[2,6],[4,5],[5,2],[1,4],[7,5],[2,3],[0,5],[4,2],[1,0],[2,7],[0,1],[4,6],[6,1],[0,6],[4,3],[1,7]], king = [3,4]
Output: [[2,3],[1,4],[1,6],[3,7],[4,3],[5,4],[4,5]]


Constraints:

• 1 <= queens.length <= 63
• queens.length == 2
• 0 <= queens[i][j] < 8
• king.length == 2
• 0 <= king, king < 8
• At most one piece is allowed in a cell.

## Solution2: Simulation

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

## Solution 2: HashTable + Binary Search

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

Support arbitrarily large boards, e.g. 1e9 x 1e9 with 1e6 # of queens.

## C++

Given an integer array arr and an integer difference, return the length of the longest subsequence in arr which is an arithmetic sequence such that the difference between adjacent elements in the subsequence equals difference.

Example 1:

Input: arr = [1,2,3,4], difference = 1
Output: 4
Explanation: The longest arithmetic subsequence is [1,2,3,4].

Example 2:

Input: arr = [1,3,5,7], difference = 1
Output: 1
Explanation: The longest arithmetic subsequence is any single element.


Example 3:

Input: arr = [1,5,7,8,5,3,4,2,1], difference = -2
Output: 4
Explanation: The longest arithmetic subsequence is [7,5,3,1].


Constraints:

• 1 <= arr.length <= 10^5
• -10^4 <= arr[i], difference <= 10^4

## Solution: DP

dp[i] := max length of sequence ends with x
dp[x] = max(0, dp[x – diff]) + 1

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

## C++

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

    3
/ \
9  20
/  \
15   7

## Solution: Recursion

Similar to LC 105

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

## Related Problems

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
/ \
9  20
/  \
15   7

## Solution: Recursion

Preprocessing: use a hashtable to store the index of element in preorder array.

For an element in inorder array, find the pos of it in preorder array in O(1), anything to the left will be the leftchild and anything to the right will be the right child.

e.g.
buildTree([9, 3, 15, 20, 7], [3, 9, 20, 15, 7]):
root = TreeNode(9) # inorder = 9
root.left = buildTree(, )
root.right = buildTree([15, 20, 7], [20, 15, 7])
return root

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

## Related Problems

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