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Updated one old, added 2 new coding problems and added new training site
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Diff for: ‎Other/fancy_sequence.py

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'''
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Fancy Sequence
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Write an API that generates fancy sequences using the append, addAll, and multAll operations.
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Implement the Fancy class:
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- Fancy() Initializes the object with an empty sequence.
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- void append(val) Appends an integer val to the end of the sequence.
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- void addAll(inc) Increments all existing values in the sequence by an integer inc.
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- void multAll(m) Multiplies all existing values in the sequence by an integer m.
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- int getIndex(idx) Gets the current value at index idx (0-indexed) of the sequence modulo 109 + 7. If the index is greater or equal than the length of the sequence, return -1.
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Input: ["Fancy", "append", "addAll", "append", "multAll", "getIndex", "addAll", "append", "multAll", "getIndex", "getIndex", "getIndex"]
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[[], [2], [3], [7], [2], [0], [3], [10], [2], [0], [1], [2]]
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Output: [null, null, null, null, null, 10, null, null, null, 26, 34, 20]
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Output explanation:
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Fancy fancy = new Fancy();
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fancy.append(2); // fancy sequence: [2]
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fancy.addAll(3); // fancy sequence: [2+3] -> [5]
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fancy.append(7); // fancy sequence: [5, 7]
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fancy.multAll(2); // fancy sequence: [5*2, 7*2] -> [10, 14]
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fancy.getIndex(0); // return 10
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fancy.addAll(3); // fancy sequence: [10+3, 14+3] -> [13, 17]
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fancy.append(10); // fancy sequence: [13, 17, 10]
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fancy.multAll(2); // fancy sequence: [13*2, 17*2, 10*2] -> [26, 34, 20]
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fancy.getIndex(0); // return 26
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fancy.getIndex(1); // return 34
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fancy.getIndex(2); // return 20
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=========================================
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The program is meant to be called many times (more than a million), and because of that, each operation should be constant time.
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When getIndex is called all we need to do is to take the first (getIndex index) adding+multiplying values and the last adding+multiplying values
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and using these 2 values do math and determine the real adding+multlplying values.
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When there is a new adding value, just increase the adding value with that one.
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When there is a new multiplying value, multiply the both adding and multiplying values.
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Example x=append(8): (((x + 5) + 6) + 7) * 2 = (x + 17) * 2 = x * 2 + 34
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So total multiplayer is 2.
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And total adder is 18 (from 36/2).
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The result is (8+18)*2 = 52.
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Example y=append(9) after addAll(6) operation:
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(y + 7) * 2 = y * 2 + 14
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The total multiplayer is the same, 2 (the previous is 1, the last is 2 that's equal to 2/1=2).
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But the total adder is 7, the total last adder is 18 (from 36/2) minus the previous 11 (from 5+6 with no multiplyers till that moment).
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The result is (9+7)*2 = 32.
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Time Complexity (The whole program, N operations): O(N)
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Time Complexity (Only one operation): O(1)
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Space Complexity: O(N)
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'''
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############
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# Solution #
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############
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class Fancy:
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def __init__(self):
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self.appending = []
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self.adding = []
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self.multiplying = []
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def append(self, val: int) -> None:
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self.appending.append(val)
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if len(self.appending) > 1:
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self.adding.append(self.adding[-1])
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self.multiplying.append(self.multiplying[-1])
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else:
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self.adding.append(0)
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self.multiplying.append(1)
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def addAll(self, inc: int) -> None:
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if len(self.appending) == 0:
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return
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self.adding[-1] += inc
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def multAll(self, m: int) -> None:
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if len(self.appending) == 0:
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return
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self.adding[-1] *= m
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self.multiplying[-1] *= m
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def getIndex(self, idx: int) -> int:
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length = len(self.appending)
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if idx >= length:
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return -1
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prevAdding = 0
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prevMultiplying = 1
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if idx > 0:
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prevMultiplying = self.multiplying[idx-1]
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prevAdding = self.adding[idx-1]
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currMultiplying = self.multiplying[-1] // prevMultiplying
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currAdding = self.adding[-1] - (prevAdding * currMultiplying)
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return (self.appending[idx] * currMultiplying + currAdding) % 1000000007
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###########
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# Testing #
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###########
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# Test 1
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fancy = Fancy()
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fancy.append(2) # fancy sequence: [2]
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fancy.addAll(3) # fancy sequence: [2+3] -> [5]
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fancy.append(7) # fancy sequence: [5, 7]
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fancy.multAll(2) # fancy sequence: [5*2, 7*2] -> [10, 14]
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print(fancy.getIndex(0)) # return 10
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fancy.addAll(3) # fancy sequence: [10+3, 14+3] -> [13, 17]
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fancy.append(10) # fancy sequence: [13, 17, 10]
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fancy.multAll(2) # fancy sequence: [13*2, 17*2, 10*2] -> [26, 34, 20]
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print(fancy.getIndex(0)) # return 26
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print(fancy.getIndex(1)) # return 34
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print(fancy.getIndex(2)) # return 20

Diff for: ‎README.md

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@@ -196,6 +196,7 @@ If the problems from [LeetCode](https://leetcode.com/) are not enough and you ne
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- [Codewars](http://www.codewars.com/)
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- [Wolfram Challenges](https://challenges.wolfram.com/)
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- [Google's Coding Competitions](https://codingcompetitions.withgoogle.com/)
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- [Google Foobar](https://foobar.withgoogle.com/)
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- [Cyber-dojo](https://cyber-dojo.org/)
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- [CodingBat](http://codingbat.com/)
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- [CodeKata](http://codekata.com/)

Diff for: ‎Strings/strong_password_checker.py

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'''
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Strong Password Checker
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A password is considered strong if the below conditions are all met:
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- It has at least 6 characters and at most 20 characters.
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- It contains at least one lowercase letter, at least one uppercase letter, and at least one digit.
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- It does not contain three repeating characters in a row (i.e., "...aaa..." is weak, but "...aa...a..." is strong, assuming other conditions are met).
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- Given a string password, return the minimum number of steps required to make password strong. if password is already strong, return 0.
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In one step, you can:
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- Insert one character to password,
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- Delete one character from password, or
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- Replace one character of password with another character.
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Input: password = "a"
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Output: 5
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Input: password = "aA1"
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Output: 3
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Input: password = "1337C0d3"
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Output: 0
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=========================================
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Explanation in some kind of pseudo code:
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if length < 6
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(3 different types characters, if all 5 are same, than 2 new separators should be added or 1 new added and 1 changed; or if all are the same type, 2 new types should be added, or 1 new and 1 changed)
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if length == 5 && (5 repeating or 1 char type)
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return 2
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return 6-length
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if length > 20
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delete all till 20
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*now you're in legnth > 5 && length <=20
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separate groups by changing a character (dividing all groups >=3 with 3)
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if char type 1 and changeChar here are <2
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changeChar += 2 - changeChar
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return deletingChar + changeChar
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first, make the string between 6 and 20 characters (when making this, separate/lower the groups bigger than 3)
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- when less than 6 characters Array.add((group+1)/2, 1, group/2) - ADDING CHARACTERS - or just use the upper logic when length < 3;
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- when more than 6 characters Array.add(group-1) - DELETING CHARACTERS, start from the one with groups bigger than 3, after that delete any)!!! Deleting priority 1. mod with 3 = 0 -> 2. mod with 3 = 1 -> 3. mod with 3 = 2
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second, when the string is between 6 and 20 characters, change character in groups with same characters
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3 -> 1 operation (*** -> *|*) => 3/3 = 1
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4 -> 1 operation (**** -> **|*) => 4/3 = 1
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5 -> 1 operations (***** -> **|**) => 5/3 = 1
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6 -> 2 operations (****** -> **|*** -> **|**|) => 6/3 = 2
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7 -> 2 operations (******* -> **|**** -> **|**|*) => 7/3 = 2
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8 -> 2 operations (******** -> **|***** -> **|**|**) => 8/3 = 2
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9 -> 3 operations (********* -> **|****** -> **|**|**|) => 9/3 = 3
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Time Complexity: O(N)
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Space Complexity: O(N)
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'''
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############
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# Solution #
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############
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def strongPasswordChecker(password):
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groups = []
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length = len(password)
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last = 0
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lower = upper = digit = 0
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for curr in range(length):
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if password[curr].islower():
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lower = 1
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elif password[curr].isupper():
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upper = 1
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elif password[curr].isdigit():
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digit = 1
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if password[last] != password[curr]:
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groups.append(curr - last);
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last = curr
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groups.append(curr + 1 - last);
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lud = lower + upper + digit
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addSteps = 0
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# adding: under 6 chars
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if length < 6:
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if length == 5 and groups[0] == 5:
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addSteps = 2 # add + change, or add + add - it's same
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else:
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addSteps = 6 - length
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ludLeft = 3 - lud
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if addSteps < ludLeft:
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addSteps = ludLeft
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return addSteps
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# deleting: over 20 chars
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deleteSteps = 0
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groupsLength = len(groups)
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while length > 20:
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found = False
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for priority in range(3):
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for gidx in range(groupsLength):
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if groups[gidx] > 2 and groups[gidx] % 3 == priority:
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groups[gidx] -= 1
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found = True
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break
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if found:
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break
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if not found:
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lastGroupIdx = groupsLength - 1
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groups[lastGroupIdx] -= 1
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if groups[lastGroupIdx] == 0:
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groups.pop(lastGroupIdx)
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groupsLength -= 1
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length -= 1
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deleteSteps += 1
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# changing: between 6 and 20 chars
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changeSteps = 0
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for gidx in range(groupsLength):
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changeSteps += groups[gidx] // 3
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ludLeft = 3 - lud
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if changeSteps < ludLeft:
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changeSteps = ludLeft
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return deleteSteps + changeSteps
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###########
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# Testing #
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###########
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# Test 1
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# Correct result => 5
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print(strongPasswordChecker('a'))
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# Test 2
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# Correct result => 3
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print(strongPasswordChecker('aA1'))
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# Test 3
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# Correct result => 0
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print(strongPasswordChecker('1337C0d3'))

Diff for: ‎Strings/zigzag_conversion.py

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Middle rows have 2 times more elements than the first and last row.
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Time Complexity: O(N)
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Space Complexity: O(N)
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Collect all parts in separate bucket, in the end merge these buckets.
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Time Complexity: O(N)
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Space Complexity: O(N)
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'''
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############
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# Solution #
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############
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##############
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# Solution 1 #
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##############
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def convert(s, num_rows):
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if num_rows == 1:
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return res
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##############
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# Solution 2 #
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##############
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def convert_2(word, numRows):
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numLetters = len(word)
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bucket = [""] * numRows
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fullCycle = 2 * (numRows - 1)
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if numRows == 1:
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fullCycle = 1
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for pos in range(0, numLetters):
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posCycle = pos % fullCycle
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if posCycle >= numRows:
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posCycle = fullCycle - posCycle
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bucket[posCycle] += word[pos]
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result = ""
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for part in bucket:
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result += part
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return result
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###########
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# Testing #
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###########
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# Test 1
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# Correct result => 'PAHNAPLSIIGYIR'
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print(convert('PAYPALISHIRING', 3))
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print(convert_2('PAYPALISHIRING', 3))
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# Test 2
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# Correct result => 'PINALSIGYAHRPI'
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print(convert('PAYPALISHIRING', 4))
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print(convert_2('PAYPALISHIRING', 4))

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