Input: root = [4,2,6,1,3,null,null]
Output: 1
Explanation:
Note that root is a TreeNode object, not an array.

The given tree [4,2,6,1,3,null,null] is represented by the following diagram:

          4
        /   \
      2      6
     / \    
    1   3  

while the minimum difference in this tree is 1, it occurs between node 1 and node 2, also between node 3 and node 2.

Solution 1: In order traversal 

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua
// Running time: 783
class Solution {
public:
  int minDiffInBST(TreeNode* root) {
    min_diff_ = INT_MAX;
    prev_ = nullptr;
    minDiff(root);
    return min_diff_;
  }
private:
  int min_diff_;
  int* prev_;
  void minDiff(TreeNode* root) {
    if (root == nullptr) return;    
    minDiff(root->left);
    if (prev_ != nullptr)
      min_diff_ = min(min_diff_, abs(root->val - *prev_));
    prev_ = &root->val;
    minDiff(root->right);
  }
};

Related Problems:

• 花花酱 LeetCode 530. Minimum Absolute Difference in BST

The post 花花酱 LeetCode 783. Minimum Distance Between BST Nodes appeared first on Huahua's Tech Road.

Now given an M x N matrix, return True if and only if the matrix is Toeplitz.
Example 1:

Input: matrix = [[1,2,3,4],[5,1,2,3],[9,5,1,2]]
Output: True
Explanation:
1234
5123
9512

In the above grid, the diagonals are "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]", and in each diagonal all elements are the same, so the answer is True.

Example 2:

Input: matrix = [[1,2],[2,2]]
Output: False
Explanation:
The diagonal "[1, 2]" has different elements.

Note:

1. matrix will be a 2D array of integers.
2. matrix will have a number of rows and columns in range [1, 20].
3. matrix[i][j] will be integers in range [0, 99].

Idea:

Check m[i][j] with m[i-1][j-1]

Solution: Brute Force

Time complexity: O(n*m)

Space complexity: O(1)

C++

// Author: Huahua
// Running time: 32 ms
class Solution {
public:
  bool isToeplitzMatrix(vector<vector<int>>& matrix) {
    for (int i = 1; i < matrix.size(); ++i)
      for (int j = 1; j < matrix[0].size(); ++j)
        if (matrix[i][j] != matrix[i - 1][j - 1]) return false;
    return true;
  }
};

Java

// Author: Huahua
// Running time: 29 ms
class Solution {
  public boolean isToeplitzMatrix(int[][] matrix) {
    for (int i = 1; i < matrix.length; ++i)
      for (int j = 1; j < matrix[0].length; ++j)
        if (matrix[i][j] != matrix[i - 1][j - 1]) return false;
    return true;
  }
}

Python3

"""
Author: Huahua
Running time: 99 ms
"""
class Solution:
  def isToeplitzMatrix(self, matrix):
    m = len(matrix)
    n = len(matrix[0])
    for i in range(1, m):
      for j in range(1, n):
        if matrix[i][j] != matrix[i - 1][j - 1]: return False
    return True

The post 花花酱 LeetCode 766. Toeplitz Matrix appeared first on Huahua's Tech Road.

题目大意：求给定范围内，数的二进制形式中1的个数为素数个的数字的个数。

Given two integers L and R, find the count of numbers in the range [L, R] (inclusive) having a prime number of set bits in their binary representation.

(Recall that the number of set bits an integer has is the number of 1s present when written in binary. For example, 21 written in binary is 10101 which has 3 set bits. Also, 1 is not a prime.)

Example 1:

Input: L = 6, R = 10
Output: 4
Explanation:
6 -> 110 (2 set bits, 2 is prime)
7 -> 111 (3 set bits, 3 is prime)
9 -> 1001 (2 set bits , 2 is prime)
10->1010 (2 set bits , 2 is prime)

Example 2:

Input: L = 10, R = 15
Output: 5
Explanation:
10 -> 1010 (2 set bits, 2 is prime)
11 -> 1011 (3 set bits, 3 is prime)
12 -> 1100 (2 set bits, 2 is prime)
13 -> 1101 (3 set bits, 3 is prime)
14 -> 1110 (3 set bits, 3 is prime)
15 -> 1111 (4 set bits, 4 is not prime)

Note:

1. L, R will be integers L <= R in the range [1, 10^6].
2. R - L will be at most 10000.

Solution 1: Brute Force

C++

// Author: Huahua
// Running time: 22 ms
class Solution {
public:
  int countPrimeSetBits(int L, int R) {
    int ans = 0;
    for (int n = L; n <= R; ++n)
      if (isPrime(bits(n))) ++ans;
    return ans;
  }
private:
  bool isPrime(int n) {
    if (n <= 1) return false;
    if (n == 2) return true;      
    for (int i = 2; i <= sqrt(n); ++i)
      if (n % i == 0) return false;
    return true;
  }
  
  int bits(int n) {
    int s = 0;
    while (n) {
      s += n & 1;
      n >>= 1;
    }        
    return s;
  }
};

// Author: Huahua
// Running time: 16 ms
class Solution {
public:
  int countPrimeSetBits(int L, int R) {
    int ans = 0;
    set<int> primes{2, 3, 5, 7, 11, 13, 17, 19};
    for (int n = L; n <= R; ++n)
      if (primes.count(__builtin_popcountll(n))) ++ans;
    return ans;
  }
};

// Author: Huahua
// Running time: 30 ms
class Solution {
public:
  int countPrimeSetBits(int L, int R) {
    int ans = 0;
    unordered_set<int> primes{2, 3, 5, 7, 11, 13, 17, 19};
    for (int n = L; n <= R; ++n)
      if (primes.count(__builtin_popcountll(n))) ++ans;
    return ans;
  }
};

// Author: Huahua
// Running time: 9 ms
class Solution {
public:
  int countPrimeSetBits(int L, int R) {
    int ans = 0;
    vector<int> p(20, 0);
    p[2] = p[3] = p[5] = p[7] = p[11] = p[13] = p[17] = p[19] = 1;
    for (int n = L; n <= R; ++n)
      if (p[__builtin_popcountll(n)]) ++ans;
    return ans;
  }
};

// Author: Huahua
// Running time: 8 ms
class Solution {
public:
  int countPrimeSetBits(int L, int R) {
    constexpr int magic = 665772;
    int ans = 0;
    for (int n = L; n <= R; ++n)
      if (magic & (1 << __builtin_popcountll(n))) ++ans;
    return ans;
  }
};

Java

// Author: Huahua
// Running time: 39 ms
class Solution {
  public int countPrimeSetBits(int L, int R) {
    int ans = 0;
    for (int n = L; n <= R; ++n)
      if (isPrime(bits(n))) ++ans;
    return ans;
  }
  
  private boolean isPrime(int n) {
    if (n <= 1) return false;
    if (n == 2) return true;      
    for (int i = 2; i <= (int)Math.sqrt(n); ++i)
      if (n % i == 0) return false;
    return true;
  }
  
  private int bits(int n) {
    int s = 0;
    while (n != 0) {
      s += n & 1;
      n >>= 1;
    }        
    return s;
  }
}

Python2

"""
Author: Huahua
Running time: 1618 ms
"""
class Solution(object):
  def countPrimeSetBits(self, L, R):
    def isPrime(n):
      if n <= 1: return False
      if n == 2: return True
      for i in range(2, int(math.sqrt(n)) + 1):
        if n % i == 0: return False
      return True

    def bits(n):
      s = 0
      while n != 0:
        s += n & 1
        n >>= 1
      return s

    ans = 0
    for n in range(L, R + 1):
      if isPrime(bits(n)): ans += 1
    return ans

Python2

"""
Author: Huahua
Running time: 1163 ms
"""
class Solution(object):
  def countPrimeSetBits(self, L, R):
    def bits(n):
      s = 0
      while n != 0:
        s += n & 1
        n >>= 1
      return s
    
    primes = set([2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31])

    ans = 0
    for n in range(L, R + 1):
      if bits(n) in primes: ans += 1
    return ans

"""
Author: Huahua
Running time: 667 ms
"""
class Solution(object):   
  def countPrimeSetBits(self, L, R):
    def bits(n):
      s = 0
      while n != 0:
        n &= n - 1;
        s += 1
      return s
        
    ans = 0
    while L <= R:
      if (665772 >> bits(L)) & 1 > 0: ans += 1
      L += 1

    return ans

The post 花花酱 LeetCode 762. Prime Number of Set Bits in Binary Representation appeared first on Huahua's Tech Road.

题目大意：让你实现开根号函数，只需要返回整数部分。

Problem:

Implement int sqrt(int x).

Compute and return the square root of x.

x is guaranteed to be a non-negative integer.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842...

Solution 1: Brute force

Time complexity: sqrt(x)

// Author: Huahua
// Running time: 40 ms
class Solution {
public:
  int mySqrt(int x) {
    if (x <= 1) return x;
    for (long long s = 1; s <= x; ++s)
      if (s * s > x) return s - 1;
    return -1;
  }
};

// Author: Huahua
// Running time: 162 ms
class Solution {
public:
  int mySqrt(int x) {
    if (x <= 1) return x;
    for (int s = 1; s <= x; ++s)
      if (s > x / s) return s - 1;
    return -1;
  }
};

// Author: Huahua
// Running time: 71 ms
class Solution {
  public int mySqrt(int x) {
    if (x <= 1) return x;
    for (long s = 1; s <= x; ++s)
      if (s * s > x) return (int)s - 1;
    return -1;
  }
}

"""
Author: Huahua
TLE: 602 / 1017 test cases passed.
"""
class Solution:
  def mySqrt(self, x):
    if x <= 1: return x
    s = 1
    while True:
      if s*s > x: return s - 1
      s += 1
    return -1

Solution 2: Binary search

Time complexity: O(logn)

// Author: Huahua
// Running time: 12 ms
class Solution {
public:
  int mySqrt(int x) {      
    long l = 1;
    long r = static_cast<long>(x) + 1;
    while (l < r) {
      long m = l + (r - l) / 2;
      if (m * m > x) { 
        r = m;
      } else {
        l = m + 1;
      }
    }
    // l: smallest number such that l * l > x
    return l - 1;
  }
};

// Author: Huahua
// Running time: 47 ms
class Solution {
  public int mySqrt(int x) {      
    int l = 1;
    int r = x;
    while (l <= r) {
      int m = l + (r - l) / 2;
      if (m > x / m) {
        r = m - 1;
      } else {
        l = m + 1;
      }
    }
    return r;
  }
}

"""
Author: Huahua
Running time: 119 ms
"""
class Solution:
  def mySqrt(self, a):
    l = 1
    r = a
    while l <= r:
      m = l + (r - l) // 2
      if m * m > a:
        r = m - 1
      else:
        l = m + 1
    return r;

Solution 3: Newton’s method

// Author: Huahua
// Running time: 28 ms
class Solution {
public:
    int mySqrt(int a) {
      constexpr double epsilon = 1e-2;
      double x = a;
      while (x * x - a > epsilon) {
        x = (x + a / x) / 2.0;
      }
      return x;
    }
};

// Author: Huahua
// Running time: 27 ms
class Solution {
public:
  int mySqrt(int a) {
    // f(x) = x^2 - a, find root of f(x)
    // Newton's method
    // f'(x) = 2x
    // x' = x - f(x) / f'(x) = x - (1/2*x - a/(2*x))
    //    = (x + a / x) / 2
    int x = a;
    while (x > a / x)
      x = (x + a / x) / 2;
    return x;
  }
};

// Author: Huahua
// Running time: 44 ms
class Solution {
  public int mySqrt(int a) {    
    long x = a;
    while (x * x > a)
      x = (x + a / x) / 2;    
    return (int)x;
  }
}

"""
Author: Huahua
Running time: 100 ms
"""
class Solution:
  def mySqrt(self, a):
    x = a
    while x * x > a:
      x = (x + a // x) // 2
    return x

The post 花花酱 LeetCode. 69 Sqrt(x) appeared first on Huahua's Tech Road.

题目大意：给你一个字符串和一些要加粗的单词，返回加粗后的HTML代码。

Problem:

Given a set of keywords words and a string S, make all appearances of all keywords in S bold. Any letters between <b> and </b> tags become bold.

The returned string should use the least number of tags possible, and of course the tags should form a valid combination.

For example, given that words = ["ab", "bc"] and S = "aabcd", we should return "a<b>abc</b>d". Note that returning "a<b>a<b>b</b>c</b>d" would use more tags, so it is incorrect.

Note:

1. words has length in range [0, 50].
2. words[i] has length in range [1, 10].
3. S has length in range [0, 500].
4. All characters in words[i] and S are lowercase letters.

Solution:

C++

Time complexity: O(nL^2)

Space complexity: O(n + d)

d: size of dictionary

L: max length of the word which is 10.

// Author: Huahua
// Running time: 13 ms
class Solution {
public:
    string boldWords(vector<string>& words, string S) {
      const int kMaxWordLen = 10;      
      unordered_set<string> dict(words.begin(), words.end());
      
      int n = S.length();
      vector<int> bolded(n, 0);
      for (int i = 0; i < n; ++i)
        for (int l = 1; l <= min(n - i, kMaxWordLen); ++l)
          if (dict.count(S.substr(i, l)))
            std::fill(bolded.begin() + i, bolded.begin() + i + l, 1);
      
      string ans;
      for (int i = 0; i < n; ++i) {
        if (bolded[i] && (i == 0 || !bolded[i - 1])) ans += "<b>";
        ans += S[i];
        if (bolded[i] && (i == n - 1 || !bolded[i + 1])) ans += "</b>";
      }
      
      return ans;
    }
};

The post 花花酱 LeetCode 758. Bold Words in String appeared first on Huahua's Tech Road.

Given two lists Aand B, and B is an anagram of A. B is an anagram of A means B is made by randomizing the order of the elements in A.

We want to find an index mapping P, from A to B. A mapping P[i] = j means the ith element in A appears in B at index j.

These lists A and B may contain duplicates. If there are multiple answers, output any of them.

For example, given

A = [12, 28, 46, 32, 50]
B = [50, 12, 32, 46, 28]

We should return

[1, 4, 3, 2, 0]

as P[0] = 1 because the 0th element of A appears at B[1], and P[1] = 4 because the 1st element of Aappears at B[4], and so on.

Solution:

C++

Time complexity: O(nlogn)

Space complexity: O(n)

// Author: Huahua
// Running time: 6 ms
class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        map<int, int> indices;
        for (int i = 0; i < B.size(); ++i)
            indices[B[i]] = i;
        vector<int> ans;
        for (const int a : A)
            ans.push_back(indices[a]);
        return ans;
    }
};

C++ / No duplication

// Author: Huahua
// Running time: 6 ms
class Solution {
public:
    vector<int> anagramMappings(vector<int>& A, vector<int>& B) {
        map<int, vector<int>> indices;
        for (int i = 0; i < B.size(); ++i)
          indices[B[i]].push_back(i);
        vector<int> ans;
        for (const int a : A) {
          ans.push_back(indices[a].back());
          indices[a].pop_back();
        }
        return ans;
    }
};

The post 花花酱 LeetCode 760. Find Anagram Mappings appeared first on Huahua's Tech Road.

题目大意：给你一个正整数，输出和它互补的数（翻转所有的bits）。

Problem:

Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.

Note:

1. The given integer is guaranteed to fit within the range of a 32-bit signed integer.
2. You could assume no leading zero bit in the integer’s binary representation.

Example 1:

Input: 5
Output: 2
Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.

Example 2:

Input: 1
Output: 0
Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.

Idea:

Bit

Solution:

C++

// Author: Huahua
// Running time: 3 ms
class Solution {
public:
    int findComplement(int num) {        
        int mask = ~0;        
        while (num & mask) mask <<= 1;
        return ~num ^ mask;
    }
};

The post 花花酱 LeetCode 476. Number Complement appeared first on Huahua's Tech Road.

Given a positive integer num, write a function which returns True if num is a perfect square else False.

Note: Do not use any built-in library function such as sqrt.

Example 1:

Input: 16
Returns: True

Example 2:

Input: 14
Returns: False

Idea:

Binary search

Solution:

C++

Time complexity: O(log(num))

Space complexity: O(1)

// Author: Huahua
// Runtime: 0 ms
class Solution {
public:
    bool isPerfectSquare(int num) {        
        long l = 1;
        long r = num + 1;
        while (l < r) {
            long m = l + (r - l) / 2;
            long t = m * m;
            if (t == num)
                return true;
            else if (t > num)
                r = m;
            else
                l = m + 1;
        }
        
        return false;
    }
};

The post 花花酱 LeetCode 367. Valid Perfect Square appeared first on Huahua's Tech Road.

Problem:

Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target.

Example 1:

Input: 
    5
   / \
  3   6
 / \   \
2   4   7

Target = 9

Output: True

Example 2:

Input: 
    5
   / \
  3   6
 / \   \
2   4   7

Target = 28

Output: False

Solution:

C++

// Author: Huahua
// Running time: 36 ms
class Solution {
public:
    bool findTarget(TreeNode* root, int k) {
        vector<int> nums;
        inorder(root, nums);
        int l = 0;
        int r = nums.size() - 1;
        while (l < r) {
            int sum = nums[l] + nums[r];
            if (sum == k) return true;
            else if (sum < k)
                ++l;
            else
                --r;
        }
        
        return false;
    }
private:
    void inorder(TreeNode* root, vector<int>& nums) {
        if (root == nullptr) return;
        inorder(root->left, nums);
        nums.push_back(root->val);
        inorder(root->right, nums);
    }
};

The post 花花酱 LeetCode 653. Two Sum IV – Input is a BST appeared first on Huahua's Tech Road.

Problem:

In a given integer array nums, there is always exactly one largest element.

Find whether the largest element in the array is at least twice as much as every other number in the array.

If it is, return the index of the largest element, otherwise return -1.

Example 1:

Input: nums = [3, 6, 1, 0]
Output: 1
Explanation: 6 is the largest integer, and for every other number in the array x,
6 is more than twice as big as x.  The index of value 6 is 1, so we return 1.

Example 2:

Input: nums = [1, 2, 3, 4]
Output: -1
Explanation: 4 isn't at least as big as twice the value of 3, so we return -1.

Note:

1. nums will have a length in the range [1, 50].
2. Every nums[i] will be an integer in the range [0, 99].

Solution:

C++

// Author: Huahua
// Running time: 6 ms
class Solution {
public:
    int dominantIndex(vector<int>& nums) {
        auto it1 = max_element(nums.begin(), nums.end());
        const int max1 = *it1;
        *it1 = -1;
        const int max2 = *max_element(nums.begin(), nums.end());
        return max1 >= max2 * 2 ? distance(nums.begin(), it1) : -1;
    }
};

The post 花花酱 LeetCode 748. Largest Number At Least Twice of Others appeared first on Huahua's Tech Road.

题目大意: 给你一只股票每天的价格，如果只能做一次交易（一次买进一次卖出）问你最多能赚多少钱。

Problem:

Say you have an array for which the i^th element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:

Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)

Example 2:

Input: [7, 6, 4, 3, 1]
Output: 0

In this case, no transaction is done, i.e. max profit = 0.

Idea:

DP

Solution 1:

C++

// Author: Huahua
// Runtime: 6 ms
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        const int n = prices.size();
        if (n < 1) return 0;
        vector<int> min_prices(n);
        vector<int> max_profit(n);
        min_prices[0] = prices[0];
        max_profit[0] = 0;
        for (int i = 1; i < n; ++i) {
            min_prices[i] = min(min_prices[i - 1], 
                                prices[i]);
            
            max_profit[i] = max(max_profit[i - 1], 
                                prices[i] - min_prices[i - 1]);
        }
        return max_profit[n - 1];
    }
};

C++ / reduce to maximum subarray

// Author: Huahua
// Runtime: 6 ms
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        if (n < 2) return 0;
        vector<int> gains(n - 1, 0);
        for (int i = 1; i < n; ++i)
            gains[i - 1] = prices[i] - prices[i - 1];
        return max(0, maxSubArray(gains));
    }
private:
    // From LC 53. Maximum Subarray
    int maxSubArray(vector<int>& nums) {
        vector<int> f(nums.size());
        f[0] = nums[0];
        
        for (int i = 1; i < nums.size(); ++i)
            f[i] = max(f[i - 1] + nums[i], nums[i]);
        
        return *std::max_element(f.begin(), f.end());
    }
};

Related Problems:

• 花花酱 LeetCode 309. Best Time to Buy and Sell Stock with Cooldown

The post 花花酱 LeetCode 121. Best Time to Buy and Sell Stock appeared first on Huahua's Tech Road.

题目大意: 给你一个由字母和数字组成车牌。另外给你一些单词，让你找一个最短的单词能够覆盖住车牌中的字母（不考虑大小写）。如果有多个解，输出第一个解。

Problem:

Find the minimum length word from a given dictionary words, which has all the letters from the string licensePlate. Such a word is said to complete the given string licensePlate

Here, for letters we ignore case. For example, "P" on the licensePlate still matches "p" on the word.

It is guaranteed an answer exists. If there are multiple answers, return the one that occurs first in the array.

The license plate might have the same letter occurring multiple times. For example, given a licensePlate of "PP", the word "pair" does not complete the licensePlate, but the word "supper" does.

Example 1:

Input: licensePlate = "1s3 PSt", words = ["step", "steps", "stripe", "stepple"]
Output: "steps"
Explanation: The smallest length word that contains the letters "S", "P", "S", and "T".
Note that the answer is not "step", because the letter "s" must occur in the word twice.
Also note that we ignored case for the purposes of comparing whether a letter exists in the word.

Example 2:

Input: licensePlate = "1s3 456", words = ["looks", "pest", "stew", "show"]
Output: "pest"
Explanation: There are 3 smallest length words that contains the letters "s".
We return the one that occurred first.

Note:

1. licensePlate will be a string with length in range [1, 7].
2. licensePlate will contain digits, spaces, or letters (uppercase or lowercase).
3. words will have a length in the range [10, 1000].
4. Every words[i] will consist of lowercase letters, and have length in range [1, 15].

Idea:

Hashtable

Solution:

C++

Time complexity: 时间复杂度 O(N*26), N is number of words.

Space complexity: 空间复杂度 O(26) = O(1)

// Author: Huahua
// Runtime: 15 ms
class Solution {
public:
    string shortestCompletingWord(string licensePlate, vector<string>& words) {
        vector<int> l(26, 0);
        for (const char ch : licensePlate)
            if (isalpha(ch)) ++l[tolower(ch) - 'a'];
        string ans;
        int min_l = INT_MAX;
        for (const string& word : words) {
            if (word.length() >= min_l) continue;
            if (!matches(l, word)) continue;
            min_l = word.length();
            ans = word;
        }
        return ans;
    }
private:
    bool matches(const vector<int>& l, const string& word) {
        vector<int> c(26, 0);
        for (const char ch : word)
            ++c[tolower(ch) - 'a'];        
        for (int i = 0; i < 26; ++i)
            if (c[i] < l[i]) return false;
        return true;
    }
};

The post 花花酱 LeetCode 748. Shortest Completing Word appeared first on Huahua's Tech Road.

Problem:

Given a binary search tree with non-negative values, find the minimum absolute difference between values of any two nodes.

Example:

Input:

   1
    \
     3
    /
   2

Output:
1

Explanation:
The minimum absolute difference is 1, which is the difference between 2 and 1 (or between 2 and 3).

Note: There are at least two nodes in this BST.

Idea:

Sorting via inorder traversal gives us sorted values, compare current one with previous one to reduce space complexity from O(n) to O(h).

Solution:

C++ O(n) space

// Author: Huahua
// Runtime: 19 ms
class Solution {
public:
    int getMinimumDifference(TreeNode* root) {
        std::vector<int> sorted;
        inorder(root, sorted);
        int min_diff = sorted.back();
        for (int i = 1; i < sorted.size(); ++i)
            min_diff = min(min_diff, sorted[i] - sorted[i - 1]);
        return min_diff;
    }
private:
    void inorder(TreeNode* root, std::vector<int>& sorted) {
        if (!root) return;
        inorder(root->left, sorted);
        sorted.push_back(root->val);
        inorder(root->right, sorted);
    }
};

C++ O(h) space

// Author: Huahua
// Runtime: 16 ms
class Solution {
public:
    int getMinimumDifference(TreeNode* root) {
        min_diff_ = INT_MAX;
        prev_ = nullptr;
        inorder(root);
        return min_diff_;
    }
private:
    void inorder(TreeNode* root) {
        if (!root) return;
        inorder(root->left);
        if (prev_) min_diff_ = min(min_diff_, root->val - *prev_);
        prev_ = &root->val;
        inorder(root->right);
    }
    
    int* prev_;
    int min_diff_;
};

Java

// Author: Huahua
// Runtime: 16 ms
class Solution {
    public int getMinimumDifference(TreeNode root) {
        prev_ = null;
        min_diff_ = Integer.MAX_VALUE;
        inorder(root);
        return min_diff_;
    }
    
    private void inorder(TreeNode root) {
        if (root == null) return;
        inorder(root.left);
        if (prev_ != null) min_diff_ = Math.min(min_diff_, root.val - prev_);
        prev_ = root.val;
        inorder(root.right);
    }
    
    private Integer prev_;
    private int min_diff_;
}

Python

"""
Author: Huahua
Runtime: 92 ms
"""
class Solution(object):
    def getMinimumDifference(self, root):        
        def inorder(root):
            if not root: return
            inorder(root.left)
            if self.prev is not None: self.min_diff = min(self.min_diff, root.val - self.prev)
            self.prev = root.val
            inorder(root.right)
        
        self.prev = None
        self.min_diff = float('inf')
        inorder(root)
        return self.min_diff

Related Problems:

• [解题报告] LeetCode 98. Validate Binary Search Tree

The post 花花酱 LeetCode 530. Minimum Absolute Difference in BST appeared first on Huahua's Tech Road.

Problem:

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

Idea:

DP

Time complexity: O(n)

Space complexity: O(n) -> O(1)

Solution:

C++ / Recursion + Memorization

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua
// Runtime: 0 ms
class Solution {
public:
    int rob(vector<int>& nums) {
        const int n = nums.size();
        m_ = vector<int>(n, -1);
        return rob(nums, n - 1);
    }
private:
    int rob(const vector<int>& nums, int i) {
        if (i < 0) return 0;
        if (m_[i] >= 0) return m_[i];
        return m_[i] = max(rob(nums, i - 2) + nums[i], 
                           rob(nums, i - 1));
    }
    
    vector<int> m_;
};

C++ / DP

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua
// Runtime: 3 ms
class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.empty()) return 0;
        vector<int> dp(nums.size(), 0);
        for (int i = 0; i < nums.size() ; ++i)
            dp[i] = max((i > 1 ? dp[i - 2] : 0) + nums[i], 
                        (i > 0 ? dp[i - 1] : 0));
        return dp.back();
    }
};

C++ / O(1) Space

// Author: Huahua
// Runtime: 3 ms
class Solution {
public:
    int rob(vector<int>& nums) {
        if (nums.empty()) return 0;
        int dp2 = 0;
        int dp1 = 0;
        for (int i = 0; i < nums.size() ; ++i) {
            int dp = max(dp2 + nums[i], dp1);
            dp2 = dp1;
            dp1 = dp;
        }
        return dp1;
    }
};

The post 花花酱 LeetCode 198. House Robber appeared first on Huahua's Tech Road.

Problem:

Given two sentences words1, words2 (each represented as an array of strings), and a list of similar word pairs pairs, determine if two sentences are similar.

For example, “great acting skills” and “fine drama talent” are similar, if the similar word pairs are pairs = [["great", "fine"], ["acting","drama"], ["skills","talent"]].

Note that the similarity relation is not transitive. For example, if “great” and “fine” are similar, and “fine” and “good” are similar, “great” and “good” are not necessarily similar.

However, similarity is symmetric. For example, “great” and “fine” being similar is the same as “fine” and “great” being similar.

Also, a word is always similar with itself. For example, the sentences words1 = ["great"], words2 = ["great"], pairs = [] are similar, even though there are no specified similar word pairs.

Finally, sentences can only be similar if they have the same number of words. So a sentence like words1 = ["great"] can never be similar to words2 = ["doubleplus","good"].

Note:

• The length of words1 and words2 will not exceed 1000.
• The length of pairs will not exceed 2000.
• The length of each pairs[i] will be 2.
• The length of each words[i] and pairs[i][j] will be in the range [1, 20].

题目大意：

给你两个句子（由单词数组表示）和一些近义词对，问你这两个句子是否相似，即每组相对应的单词都要相似。

注意相似性不能传递，比如给只你”great”和”fine”相似、”fine”和”good”相似，这种情况下”great”和”good”不相似。

Idea:

Use hashtable to store mapping from word to its similar words.

Solution: HashTable

Time complexity: O(|pairs| + |words1|)

Space complexity: O(|pairs|)

// Author: Huahua
// Runtime: 9 ms
class Solution {
public:
    bool areSentencesSimilar(vector<string>& words1, vector<string>& words2, vector<pair<string, string>> pairs) {
        if (words1.size() != words2.size()) return false;
        
        unordered_map<string, unordered_set<string>> similar_words;
        for (const auto& pair : pairs) {            
            similar_words[pair.first].insert(pair.second);
            similar_words[pair.second].insert(pair.first);
        }
        
        for (int i = 0; i < words1.size(); ++i) {
            if (words1[i] == words2[i]) continue;
            if (!similar_words[words1[i]].count(words2[i])) return false;
        }
        
        return true;
    }
};

// Author: Huahua
// Runtime: 10 ms
class Solution {
    public boolean areSentencesSimilar(String[] words1, String[] words2, String[][] pairs) {
        if (words1.length != words2.length) return false;
        
        Map<String, Set<String>> similar_words = new HashMap<>();
        
        for (String[] pair : pairs) {
            if (!similar_words.containsKey(pair[0]))
                similar_words.put(pair[0], new HashSet<>());
            if (!similar_words.containsKey(pair[1]))
                similar_words.put(pair[1], new HashSet<>());
            similar_words.get(pair[0]).add(pair[1]);
            similar_words.get(pair[1]).add(pair[0]);
        }
        
        for (int i = 0; i < words1.length; ++i) {
            if (words1[i].equals(words2[i])) continue;
            if (!similar_words.containsKey(words1[i])) return false;
            if (!similar_words.get(words1[i]).contains(words2[i])) return false;
        }
        
        return true;
    }
}

"""
Author: Huahua
Runtime: 62 ms
"""
class Solution:
    def areSentencesSimilar(self, words1, words2, pairs):
        if len(words1) != len(words2): return False
        
        similar_words = {}
        
        for w1, w2 in pairs:
            if not w1 in similar_words: similar_words[w1] = set()
            if not w2 in similar_words: similar_words[w2] = set()
            similar_words[w1].add(w2)
            similar_words[w2].add(w1)
        
        for w1, w2 in zip(words1, words2):
            if w1 == w2: continue
            if w1 not in similar_words: return False
            if w2 not in similar_words[w1]: return False
        
        return True

Related Problems:

• [解题报告] LeetCode 737. Sentence Similarity II

The post 花花酱 LeetCode 734. Sentence Similarity appeared first on Huahua's Tech Road.