Short Problem Definition:

Sherlock considers a string to be valid if all characters of the string appear the same number of times. It is also valid if he can remove just 1 character at 1 index in the string, and the remaining characters will occur the same number of times. Given a string , determine if it is valid. If so, return YES, otherwise return NO.

Link

Sherlock and The Valid String

Complexity:

time complexity is O(N)

space complexity is O(N)

Execution:

This is one of the easier medium problems. Create a character occurrence map. Then create an occurrence-occurrence map. If all the occurrences are the same (size is 1) the string is valid. If there is exactly one character that occurs exactly once, it is also valid. Otherwise invalid

Solution:

def isValid(S):
    char_map = Counter(S)
    char_occurence_map = Counter(char_map.values())

    if len(char_occurence_map) == 1:
        return True

    if len(char_occurence_map) == 2:
        for v in char_occurence_map.values():
            if v == 1:
                return True

    return False

Short Problem Definition:

Amanda has a string of lowercase letters that she wants to copy to a new string. She can perform the following operations with the given costs. She can perform them any number of times to construct a new string p:

• Append a character to the end of string p at a cost of 1 dollar.
• Choose any substring of p and append it to the end of  at no charge.

Link

String Construction

Complexity:

time complexity is O(N)

space complexity is O(N)

Execution:

The solution sounds too easy, but it is still very simple. A substring of length 1 is still a substring. Each character in the final string needs to be copied once for 1$. Each other occurrence of that string can be copied for 0$. Aka just count the number of distinct letters in the expected string.

Solution:

def stringConstruction(s):
    return len(set(s))

Short Problem Definition:

A weighted string is a string of lowercase English letters where each letter has a weight. Character weights are 1 to  26 from a to z…

Link

Weighted Uniform String

Complexity:

time complexity is O(N)

space complexity is O(N)

Execution:

Parsing the string for every query is suboptimal, so I first preprocess the string. Now we know that uniform strings contain the same characters. A string can be of length 1. Do a single pass of the string and create all uniform substrings.

Solution:

def weightedUniformStrings(s, queries):
    weights = set()
    prev = -1
    length = 0
    for c in s:
        weight = ord(c) - ord('a') + 1
        weights.add(weight)
        if prev == c:
            length += 1
            weights.add(length*weight)
        else:
            prev = c
            length = 1
    
    rval = []
    for q in queries:
        if q in weights:
            rval.append("Yes")
        else:
            rval.append("No")
    return rval

Short Problem Definition:

We say that a string contains the word hackerrank if a subsequence of its characters spell the word hackerrank. For example, if string s = haacckkerrannkk  it does contain hackerrank, but s = haacckkerannk does not. In the second case, the second r is missing. If we reorder the first string as , it no longer contains the subsequence due to ordering.

Link

HackerRank in a String!

Complexity:

time complexity is O(N)

space complexity is O(1)

Execution:

Keep two pointers. One to the expected string (needle) and one to the input string. If you find the needle in the haystack before you run out of characters, you are good.

Solution:

def hackerrankInString(s):
    needle = 'hackerrank'
    idx_in_needle = 0
    for c in s:
        if c == needle[idx_in_needle]:
            idx_in_needle += 1
        if idx_in_needle == len(needle):
            break
            
    if idx_in_needle == len(needle):
        return "YES"
    else: 
        return "NO"

Short Problem Definition:

Sami’s spaceship crashed on Mars! She sends a series of SOS messages to Earth for help.

Link

Mars Exploration

Complexity:

time complexity is O(N)

space complexity is O(1)

Execution:

We know that the message is basically a lot of concatenated SOS strings. There is no magic to this one.

Solution:

S = raw_input().strip()

errors = 0

for i in xrange(len(S)):
    if i % 3 == 0 and S[i] != 'S':
        errors += 1
    if i % 3 == 1 and S[i] != 'O':
        errors += 1
    if i % 3 == 2 and S[i] != 'S':
        errors += 1
        
        
print errors

Short Problem Definition:

Alice wrote a sequence of words in CamelCase as a string of letters, , having the following properties:

• It is a concatenation of one or more words consisting of English letters.
• All letters in the first word are lowercase.
• For each of the subsequent words, the first letter is uppercase and rest of the letters are lowercase.

Given s , print the number of words in s on a new line.

For example, s = OneTwoThree . There are 3 words in the string.

Link

CamelCase

Complexity:

time complexity is O(N)

space complexity is O(1)

Execution:

Since the input always has at least 1 character, we can assume that there will always be at least one word. Each upper case later identifies the next word. So the result is number of capital letters + 1.

Solution:

s = raw_input().strip()

cnt = 1

for c in s:
    if c.isupper():
        cnt += 1
        
print cnt

Short Problem Definition:

Steve has a string of lowercase characters in range ascii[‘a’..’z’]. He wants to reduce the string to its shortest length by doing a series of operations. In each operation he selects a pair of adjacent lowercase letters that match, and he deletes them. For instance, the string aab could be shortened to b in one operation.

Steve’s task is to delete as many characters as possible using this method and print the resulting string. If the final string is empty, print Empty String

Link

Super Reduced String

Complexity:

time complexity is O(N)

space complexity is O(N)

Execution:

The solution creates a stack of values. If the top value on the stack is equivalent to the next value, simply remove both. This solution assumes that there are only ever 2 values next to each other. If any adjacent values were to be removed, I would require more loops.

Solution:

i = raw_input()

s = []

for c in i:
    if not s:
        s.append(c)
    else:
        if s[-1] == c:
            s.pop()
        else:
            s.append(c)
            
if not s:
    print "Empty String"
else:
    print ''.join(s)