You are given an integer array A.  From some starting index, you can make a series of jumps.  The (1st, 3rd, 5th, …) jumps in the series are called odd numbered jumps, and the (2nd, 4th, 6th, …) jumps in the series are called even numbered jumps.

You may from index i jump forward to index j (with i < j) in the following way:

• During odd numbered jumps (ie. jumps 1, 3, 5, …), you jump to the index j such that A[i] <= A[j] and A[j] is the smallest possible value.  If there are multiple such indexes j, you can only jump to the smallest such index j.
• During even numbered jumps (ie. jumps 2, 4, 6, …), you jump to the index j such that A[i] >= A[j] and A[j] is the largest possible value.  If there are multiple such indexes j, you can only jump to the smallest such index j.
• (It may be the case that for some index i, there are no legal jumps.)

A starting index is good if, starting from that index, you can reach the end of the array (index A.length - 1) by jumping some number of times (possibly 0 or more than once.)

Return the number of good starting indexes.

Example 1:

Input: [10,13,12,14,15] 
Output: 2 

Example 2:

Input: [2,3,1,1,4] 
Output: 3 

Example 3:

Input: [5,1,3,4,2] 
Output: 3

Note:

1. 1 <= A.length <= 20000
2. 0 <= A[i] < 100000

Solution: Binary Search + DP

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

class Solution {
public:
  int oddEvenJumps(vector<int>& A) {
    const int n = A.size();
    map<int, int> m;
    vector<vector<int>> dp(n + 1, vector<int>(2));
    dp[n - 1][0] = dp[n - 1][1] = 1;
    m[A[n - 1]] = n - 1;
    int ans = 1;
    for (int i = n - 2; i >= 0; --i) {
      auto o = m.lower_bound(A[i]);
      if (o != m.end()) dp[i][0] = dp[o->second][1];
      auto e = m.upper_bound(A[i]);
      if (e != m.begin()) dp[i][1] = dp[prev(e)->second][0];
      if (dp[i][0]) ++ans;
      m[A[i]] = i;
    }
    return ans;
  }
};

The post 花花酱 LeetCode 975. Odd Even Jump appeared first on Huahua's Tech Road.

Problem

Write a class StockSpanner which collects daily price quotes for some stock, and returns the span of that stock’s price for the current day.

The span of the stock’s price today is defined as the maximum number of consecutive days (starting from today and going backwards) for which the price of the stock was less than or equal to today’s price.

For example, if the price of a stock over the next 7 days were [100, 80, 60, 70, 60, 75, 85], then the stock spans would be [1, 1, 1, 2, 1, 4, 6].

Example 1:

Input: ["StockSpanner","next","next","next","next","next","next","next"], [[],[100],[80],[60],[70],[60],[75],[85]]
Output: [null,1,1,1,2,1,4,6]
Explanation: 
First, S = StockSpanner() is initialized.  Then:
S.next(100) is called and returns 1,
S.next(80) is called and returns 1,
S.next(60) is called and returns 1,
S.next(70) is called and returns 2,
S.next(60) is called and returns 1,
S.next(75) is called and returns 4,
S.next(85) is called and returns 6.

Note that (for example) S.next(75) returned 4, because the last 4 prices
(including today's price of 75) were less than or equal to today's price.

Note:

1. Calls to StockSpanner.next(int price) will have 1 <= price <= 10^5.
2. There will be at most 10000 calls to StockSpanner.next per test case.
3. There will be at most 150000 calls to StockSpanner.next across all test cases.
4. The total time limit for this problem has been reduced by 75% for C++, and 50% for all other languages.

Solution 1: Brute Force (TLE)

Time complexity: O(n) per next call

Space complexity: O(n)

Solution 2: DP

dp[i] := span of prices[i]

j = i – 1
while j >= 0 and prices[i] >= prices[j]: j -= dp[j]
dp[i] = i – j

// Author: Huahua, 144 ms
class StockSpanner {
public:
  StockSpanner() {}

  int next(int price) {
    if (prices_.empty() || price < prices_.back()) {
      dp_.push_back(1);      
    } else {
      int j = prices_.size() - 1;
      while (j >= 0 && price >= prices_[j]) {
        j -= dp_[j];
      }
      dp_.push_back(prices_.size() - j);
    }    
    prices_.push_back(price);    
    return dp_.back();
  }
private:
  vector<int> dp_;
  vector<int> prices_;  
};

Solution 3: Monotonic Stack

Maintain a monotonic stack whose element are pairs of <price, span>, sorted by price from high to low.

When a new price comes in

1. If it’s less than top price, add a new pair (price, 1) to the stack, return 1
2. If it’s greater than top element, collapse the stack and accumulate the span until the top price is higher than the new price. return the total span

e.g. prices: 10, 6, 5, 4, 3, 7

after 3, the stack looks [(10,1), (6,1), (5,1), (4,1), (3, 1)],

when 7 arrives, [(10,1), (6,1), (5,1), (4,1), (3, 1), (7, 4 + 1)] = [(10, 1), (7, 5)]

Time complexity: O(1) amortized, each element will be pushed on to stack once, and pop at most once.

Space complexity: O(n), in the worst case, the prices is in descending order.

// Author: Huahua
class StockSpanner {
public:
  StockSpanner() {}

  int next(int price) {
    int span = 1;
    while (!s_.empty() && price >= s_.top().first) {
      span += s_.top().second;  
      s_.pop();
    }
    s_.emplace(price, span);
    return span;
  }
private:
  stack<pair<int, int>> s_; // {price, span}, ordered by price DESC.
};

// Author: Huahua
class StockSpanner {
  private Stack<Integer> prices;
  private Stack<Integer> spans;
  public StockSpanner() {
    prices = new Stack<>();
    spans = new Stack<>();
  }

  public int next(int price) {
    int span = 1;
    while (!prices.empty() && price >= prices.peek()) {
      span += spans.pop();
      prices.pop();
    }
    prices.push(price);
    spans.push(span);
    return span;
  }
}

# Author: Huahua
class StockSpanner:
  def __init__(self):
    self.s = list()

  def next(self, price):
    span = 1
    while self.s and price >= self.s[-1][0]:
      span += self.s[-1][1]
      self.s.pop()
    self.s.append((price, span))
    return span

Related Problems

• 花花酱 LeetCode 735. Asteroid Collision
• 花花酱 LeetCode 239. Sliding Window Maximum

The post 花花酱 LeetCode 901. Online Stock Span appeared first on Huahua's Tech Road.

An array A is monotone increasing if for all i <= j, A[i] <= A[j].  An array A is monotone decreasing if for all i <= j, A[i] >= A[j].

Return true if and only if the given array A is monotonic.

Solution: 

// Author: Huahua
class Solution {
public:
  bool isMonotonic(vector<int>& A) {
    bool inc = true;
    bool dec = true;
    
    for (int i = 1; i < A.size(); ++i) {
      inc &= A[i] >= A[i - 1];
      dec &= A[i] <= A[i - 1];
    }
    
    return inc || dec;
  }
};

class Solution {
  public boolean isMonotonic(int[] A) {
    boolean inc = true;
    boolean dec = true;
    
    for (int i = 1; i < A.length; ++i) {
      inc &= A[i] >= A[i - 1];
      dec &= A[i] <= A[i - 1];
    }
    
    return inc || dec;
  }
}

# Author: Huahua
class Solution:
  def isMonotonic(self, A):    
    inc = True;
    dec = True;
    
    for i in range(1, len(A)):
      inc = inc and A[i] >= A[i - 1]
      dec = dec and A[i] <= A[i - 1]    
    
    return inc or dec;

The post 花花酱 LeetCode 896. Monotonic Array appeared first on Huahua's Tech Road.