09/13/18

# HackerRank ‘CamelCase’ Solution

##### 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.

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


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09/12/18

# HackerRank ‘Super Reduced String’ Solution

##### 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

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)


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