Short Problem Definition:

Find the number of ways that a given integer, X , can be expressed as the sum of the Nth powers of unique, natural numbers.

For example, if X = 13 and N = 2, we have to find all combinations of unique squares adding up to 13. The only solution is 2^2 + 3^2.

Link

The Power Sum

Complexity:

time complexity is O(N!)

space complexity is O(1)

Execution:

This solution does not use DP or memoisation, but it does not time out. We know that it needs to explore all P! possible combinations where P is floor(sqrt(X, N)). Since maximum X is 1000, this should still compute in real time.

Keep in mind that all integers in the solution have to be unique. So for X = 3, the solution 1^2 + 1^2 + 1^2 is not acceptable.

Did I mention that I dislike both NP-Hard problems and recursion? No? So, now you know.

Solution:

def powerSum(X, N, current = 1):
    pw = pow(current, N)
    if pw > X:
        return 0
    elif pw == X:
        return 1
    else:
        return powerSum(X, N, current+1) + powerSum(X-pw, N, current+1)

Short Problem Definition:

Whenever George asks Lily to hang out, she’s busy doing homework. George wants to help her finish it faster, but he’s in over his head! Can you help George understand Lily’s homework so she can hang out with him?

Consider an array of m distinct integers, arr = [a[0], a[1], …, a[n-1]]. George can swap any two elements of the array any number of times. An array is beautiful if the sum of |a[i] – a[i-1] among 0 < i < n is minimal.

Link

Lily’s Homework

Complexity:

time complexity is O(N*log(N))

space complexity is O(N)

Execution:

Let us rephrase the problem to a sorting problem: Find the number of swaps to make the array sorted. We need to consider both ascending and descending arrays. The solution assumes no duplicates.

Solution:

def cntSwaps(arr):
    positions = sorted(list(enumerate(arr)), key=lambda e: e[1])
    swaps = 0
    
    for idx in xrange(len(arr)):
        while True:
            if (positions[idx][0] == idx):
                break
            else:
                swaps += 1
                swapped_idx = positions[idx][0]
                positions[idx], positions[swapped_idx] = positions[swapped_idx], positions[idx]
    
    return swaps

def lilysHomework(arr):
    return min(cntSwaps(arr), cntSwaps(arr[::-1]))

Short Problem Definition:

A gene is represented as a string of length N (where  is divisible by 4), composed of the letters A, C, G, and T. It is considered to be steady if each of the four letters occurs exactly 1/4 times. For example, GACT and AAGGCCTT are both steady genes.

Link

Bear and Steady Gene

Complexity:

time complexity is O(N)

space complexity is O(1)

Execution:

A bit of trivia: I’ve looked at this problem a couple times 2 years ago and I could not figure it out. Sometimes it is good to take a long break 🙂

In the first pass I establish how many occurrences there are of each character. Then I establish how many are missing to make the string steady. Remember that the number of occurrences needs to be evenly distributed.

The second pass takes advantage of the caterpillar method. I am looking for the shortest string that contains all the extra characters. They need to be replaced by the missing characters. All others can just be replaced with the same character. The caterpillar extends as long as it does not hit one of the extra characters. It shortens if it contains more extra characters than are needed.

Solution:

def steadyGene(gene):
    min_length_string = len(gene)
    
    occurences = dict()
    occurences['A'] = 0
    occurences['G'] = 0
    occurences['C'] = 0
    occurences['T'] = 0
    
    expected = len(gene) // 4
    
    for g in gene:
        occurences[g] += 1
    
    for x in occurences:
        occurences[x] = max(0, occurences[x] - expected)
    
    if occurences['A'] == 0 and occurences['G'] == 0 and occurences['C'] == 0 and occurences['T'] == 0:
        return 0
    
    found = dict()
    found['A'] = 0
    found['G'] = 0
    found['C'] = 0
    found['T'] = 0
    
    tail = 0
    head = 0
    
    while head != len(gene):
        found[gene[head]] += 1
        if found['A'] >= occurences['A'] and \
        found['C'] >= occurences['C'] and \
        found['G'] >= occurences['G'] and \
        found['T'] >= occurences['T']:
            # this is a valid candidate
            min_length_string = min(min_length_string, head-tail+1)
            
            # try to shorten it
            while found[gene[tail]] > occurences[gene[tail]]:
                found[gene[tail]] -= 1
                tail += 1
                min_length_string = min(min_length_string, head-tail+1)
            
            
        head += 1
    
    return min_length_string

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