## 1046. Last Stone Weight

Solution: Simulation (priority_queue)

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

## 1047. Remove All Adjacent Duplicates In String

Solution: Stack / Deque

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

## 1048. Longest String Chain

Solution: DP

dp[i] := max length of chain of (A[0] ~ A[i-1])

dp[i] = max{dp[j] + 1} if A[j] is prederrsor of A[i], 1 <= j <i

Time complexity: O(n^2*l)
Space complexity: O(n)

## 1049. Last Stone Weight II

Solution: DP / target sum

Time complexity: O(n * S) = O(n * 100)

Space complexity: O(S) = O(100)

## C++

iven two integer arrays A and B, return the maximum length of an subarray that appears in both arrays.

Example 1:

Input:
A: [1,2,3,2,1]
B: [3,2,1,4,7]
Output: 3
Explanation:
The repeated subarray with maximum length is [3, 2, 1].


Note:

1. 1 <= len(A), len(B) <= 1000
2. 0 <= A[i], B[i] < 100

## Solution: DP

dp[i][j] := max length of (A[0:i], B[0:j])

dp[i][j] = dp[i – 1][j – 1] + 1 if A[i-1] == B[j-1] else 0

Time complexity: O(m*n)
Space complexity: O(m*n) -> O(n)

## C++ S:O(min(m,n))

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Solution: DFS

The key to this problem is how to remove/avoid duplicates efficiently.

For the same depth, among the same numbers, only the first number can be used.

Time complexity: O(2^n * n)
Space complexity: O(n)

## C++

Given an integer array A, you partition the array into (contiguous) subarrays of length at most K.  After partitioning, each subarray has their values changed to become the maximum value of that subarray.

Return the largest sum of the given array after partitioning.

Example 1:

Input: A = [1,15,7,9,2,5,10], K = 3
Output: 84
Explanation: A becomes [15,15,15,9,10,10,10]

Note:

1. 1 <= K <= A.length <= 500
2. 0 <= A[i] <= 10^6

## Solution: DP

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

dp[i] := max sum of A[1] ~ A[i]
init: dp[0] = 0
transition: dp[i] = max{dp[i – k] + max(A[i-k:i]) * k}, 1 <= k <= min(i, K)
ans: dp[n]

A = | 2 | 1 | 4 | 3 |
K = 3
dp[0] = 0
dp[1] = max(dp[0] + 2 * 1) = 2
dp[2] = max(dp[0] + 2 * 2, dp[1] + 1 * 1) = max(4, 3) = 4
dp[3] = max(dp[0] + 4 * 3, dp[1] + 4 * 2, dp[2] + 4 * 1) = max(12, 10, 8) = 12
dp[4] = max(dp[1] + 4 * 3, dp[2] + 4 * 2, dp[3] + 3 * 1) = max(14, 12, 15) = 15
best = | 4 | 4 | 4 | 3 |


## C++

On an infinite plane, a robot initially stands at (0, 0) and faces north.  The robot can receive one of three instructions:

• "G": go straight 1 unit;
• "L": turn 90 degrees to the left;
• "R": turn 90 degress to the right.

The robot performs the instructions given in order, and repeats them forever.

Return true if and only if there exists a circle in the plane such that the robot never leaves the circle.

Example 1:

Input: "GGLLGG"
Output: true
Explanation:
The robot moves from (0,0) to (0,2), turns 180 degrees, and then returns to (0,0).
When repeating these instructions, the robot remains in the circle of radius 2 centered at the origin.


Example 2:

Input: "GG"
Output: false
Explanation:
The robot moves north indefinitely.


Example 3:

Input: "GL"
Output: true
Explanation:
The robot moves from (0, 0) -> (0, 1) -> (-1, 1) -> (-1, 0) -> (0, 0) -> ...


Note:

1. 1 <= instructions.length <= 100
2. instructions[i] is in {'G', 'L', 'R'}

Solution: Simulation

When instructions end, if the robot is back to (0,0) or is not facing north (which guarantees it will come back to 0, 0 for another 1 or 3 rounds)

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

## C++

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