• Operation 1: Swap any two existing characters.
  • For example, abcde -> aecdb
• Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character.
  • For example, aacabb -> bbcbaa (all a‘s turn into b‘s, and all b‘s turn into a‘s)

You can use the operations on either string as many times as necessary.

Given two strings, word1 and word2, return true if word1 and word2 are close, and false otherwise.

Example 1:

Input: word1 = "abc", word2 = "bca"
Output: true
Explanation: You can attain word2 from word1 in 2 operations.
Apply Operation 1: "abc" -> "acb"
Apply Operation 1: "acb" -> "bca"

Example 2:

Input: word1 = "a", word2 = "aa"
Output: false
Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.

Example 3:

Input: word1 = "cabbba", word2 = "abbccc"
Output: true
Explanation: You can attain word2 from word1 in 3 operations.
Apply Operation 1: "cabbba" -> "caabbb"
Apply Operation 2: "caabbb" -> "baaccc"
Apply Operation 2: "baaccc" -> "abbccc"

Example 4:

Input: word1 = "cabbba", word2 = "aabbss"
Output: false
Explanation: It is impossible to attain word2 from word1, or vice versa, in any amount of operations.

Constraints:

• 1 <= word1.length, word2.length <= 105
• word1 and word2 contain only lowercase English letters.

Solution: Hashtable

Two strings are close:
1. Have the same length, ccabbb => 6 == aabccc => 6
2. Have the same char set, ccabbb => (a, b, c) == aabccc => (a, b, c)
3. Have the same sorted char counts ccabbb => (1, 2, 3) == aabccc => (1, 2, 3)

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

// Author: Huahua
class Solution {
public:
  bool closeStrings(string word1, string word2) {
    const int l1 = word1.length();
    const int l2 = word2.length();
    if (l1 != l2) return false;
    vector<int> f1(128), f2(128);
    vector<int> s1(128), s2(128);
    for (char c: word1) ++f1[c], s1[c] = 1;
    for (char c: word2) ++f2[c], s2[c] = 1;
    sort(begin(f1), end(f1));
    sort(begin(f2), end(f2));
    return f1 == f2 && s1 == s2;
  }
};

# Author: Huahua
class Solution:
  def closeStrings(self, word1: str, word2: str) -> bool:   
    c1, c2 = Counter(word1), Counter(word2)
    return all([len(word1) == len(word2), 
                c1.keys() == c2.keys(),
                sorted(c1.values()) == sorted(c2.values())])

The post 花花酱 LeetCode 1657. Determine if Two Strings Are Close appeared first on Huahua's Tech Road.

In one step, you can select any non-empty sub-array of arr and reverse it. You are allowed to make any number of steps.

Return True if you can make arr equal to target, or False otherwise.

Example 1:

Input: target = [1,2,3,4], arr = [2,4,1,3]
Output: true
Explanation: You can follow the next steps to convert arr to target:
1- Reverse sub-array [2,4,1], arr becomes [1,4,2,3]
2- Reverse sub-array [4,2], arr becomes [1,2,4,3]
3- Reverse sub-array [4,3], arr becomes [1,2,3,4]
There are multiple ways to convert arr to target, this is not the only way to do so.

Example 2:

Input: target = [7], arr = [7]
Output: true
Explanation: arr is equal to target without any reverses.

Example 3:

Input: target = [1,12], arr = [12,1]
Output: true

Example 4:

Input: target = [3,7,9], arr = [3,7,11]
Output: false
Explanation: arr doesn't have value 9 and it can never be converted to target.

Example 5:

Input: target = [1,1,1,1,1], arr = [1,1,1,1,1]
Output: true

Constraints:

• target.length == arr.length
• 1 <= target.length <= 1000
• 1 <= target[i] <= 1000
• 1 <= arr[i] <= 1000

Solution: Counting

target and arr must have same elements.

Time complexity: O(n)
Space complexity: O(1001)

// Author: Huahua
class Solution {
public:
  bool canBeEqual(vector<int>& target, vector<int>& arr) {
    vector<int> t(1001);
    vector<int> a(1001);
    for (int i = 0; i < target.size(); ++i) {
      ++t[target[i]];
      ++a[arr[i]];
    }
    return t == a;
  }
};

# Author: Huahua
class Solution:
  def canBeEqual(self, target: List[int], arr: List[int]) -> bool:
    return Counter(target) == Counter(arr)

The post 花花酱 LeetCode 1460. Make Two Arrays Equal by Reversing Sub-arrays appeared first on Huahua's Tech Road.

In a string S of lowercase letters, these letters form consecutive groups of the same character.

For example, a string like S = "abbxxxxzyy" has the groups "a", "bb", "xxxx", "z" and "yy".

Call a group large if it has 3 or more characters.  We would like the starting and ending positions of every large group.

The final answer should be in lexicographic order.

Example 1:

Input: "abbxxxxzzy"
Output: [[3,6]]
Explanation: "xxxx" is the single large group with starting 3 and ending positions 6.

Example 2:

Input: "abc"
Output: []
Explanation: We have "a","b" and "c" but no large group.

Example 3:

Input: "abcdddeeeeaabbbcd"
Output: [[3,5],[6,9],[12,14]]

Note:  1 <= S.length <= 1000

Solution: Brute Force

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua
// Running time: 18 ms
class Solution {
public:
  vector<vector<int>> largeGroupPositions(string S) {
    constexpr int kLargeGroupSize = 3;
    const int n = S.size();
    vector<vector<int>> ans;
    int c = 0;
    for (int i = 0; i <= n; ++i, ++c) {
      if (i == n || S[i] != S[i - 1]) {
        if (c >= kLargeGroupSize) ans.push_back({i - c, i - 1});
        c = 0;
      }
    }
    return ans;
  }
};

The post 花花酱 LeetCode 830. Positions of Large Groups appeared first on Huahua's Tech Road.

There is a table courses with columns: student and class

Please list out all classes which have more than or equal to 5 students.

For example, the table:

+---------+------------+
| student | class      |
+---------+------------+
| A       | Math       |
| B       | English    |
| C       | Math       |
| D       | Biology    |
| E       | Math       |
| F       | Computer   |
| G       | Math       |
| H       | Math       |
| I       | Math       |
+---------+------------+

Should output:

+---------+
| class   |
+---------+
| Math    |
+---------+

Note:
The students should not be counted duplicate in each course.

Solution

SQL

# Author: Huahua
# Running time: 1756 ms
select class from courses group by class having count(distinct student) >= 5

The post 花花酱 LeetCode 596. Classes More Than 5 Students appeared first on Huahua's Tech Road.