✅ Free leetcode questions which is 99% similar with the first question :

class Solution:
    def countSubstrings(self, s: str, c: str) -> int:
        count = s.count(c)
        return count * (count + 1) // 2

https://leetcode.com/problems/count-substrings-starting-and-ending-with-given-character/description/

/*
You are given a string s consisting only of lowercase English letters.
We call a substring special if it contains no character which has occurred at least twice (in other words, it does not contain a repeating character).
Your task is to count the number of special substrings.
For example, in the string "pop", the substring "po" is a special substring, however, "pop" is not special (since 'p' has occurred twice).
Return the number of special substrings.
A substring is a contiguous sequence of characters within a string. For example, "abc" is a substring of "abcd", but "acd" is not.

Example 1:
Input: s = "abcd"


Output: 10
Explanation: Since each character occurs once, every substring is a special substring.
We have 4 substrings of length one, 3 of length two, 2 of length three, and 1 substring of length four. So overall there are 4 + 3 + 2 + 1 = 10 special substrings.

Example 2:
Input: s = "ooo"
Output: 3
Explanation: Any substring with a length of at least two contains a repeating character. So we have to count the number of substrings of length one, which is 3.

Example 3:
Input: s = "abab"
Output: 7
Explanation: Special substrings are as follows (sorted by their start positions):
Special substrings of length 1: "a", "b", "a", "b"
Special substrings of length 2: "ab", "ba", "ab"
And it can be shown that there are no special substrings with a length of at least three. So the answer would be 4 + 3 = 7.
l r
a b c d a

Constraints:
1 <= s.length <= 10^5
s consists of lowercase English letters

This one took me much longer time not gonna lie.😮‍💨

✅ Solution : this is as efficient as it can get

python 
def countSpaceialSubString(s):
    seen = set()
    l = 0
    count = 0
    
    for r in range(len(s)):
        if s[r] in seen:
            while s[l] != s[r]:
                seen.remove(s[l])
                l += 1
            l += 1
        
        seen.add(s[r])
        count += r - l + 1
    
    return count 
        
        
print(countSpaceialSubString("abcd"))
print(countSpaceialSubString("ooo"))
print(countSpaceialSubString("abab"))
print(countSpaceialSubString("abcabc"))

Question: Can You Make This String a Palindrome?

A palindrome is a string that reads the same forwards and backwards. Given a string, determine if it's possible to rearrange the characters to form a palindrome.

Examples:

1. Input: "civic"

   • Output: True

   • Explanation: The string is already a palindrome.

2. Input: "ivicc"

   • Output: True

   • Explanation: Rearranging the characters can form the palindrome "civic".

3. Input: "hello"

   • Output: False

   • Explanation: No rearrangement can form a palindrome.

4. Input: "aabbcc"

   • Output: True

   • Explanation: Rearranging the characters can form the palindrome "abcba".

5. Input: "racecar"

   • Output: True

   • Explanation: The string is already a palindrome.

Challenge:
Write a function that takes a string as input and returns True if the string can be rearranged to form a palindrome, and False otherwise.

Question Description
Given a string s and an integer k, return the number of substrings in s of length k with no repeated characters.
Example 1:
Input: s = "unonleetcode", k = 5
Output: 2
Explanation: There are 6 substrings they are: 'havef','avefu','vefun','efuno','etcod','tcode'.

Example 2:
Input: s = "home", k = 5
Output: 0
Explanation: Notice k can be larger than the length of s. In this case, it is not possible to find any substring.

Example 3:
Input: s = "havefunonleetcode", k = 5
Output: 6
Explanation: There are 6 substrings they are: 'havef','avefu','vefun','efuno','etcod','tcode'.

✅ Solution: as I have told you this pattern repeats a lot so you got this

def subStringK(s, k):
    seen = set()
    count = 0
    l = 0
    
    for r in range(len(s)):
        if s[r] in seen:
            while s[l] != s[r]:
                seen.remove(s[l])
                l += 1
            l += 1
        
        if r - l + 1 == k:
            count += 1
            seen.remove(s[l])
            l += 1
        
        seen.add(s[r])
    return count


print(subStringK("unonleetcode", 5))
print(subStringK("havefunonleetcode", 5))
print(subStringK("aaabbaaa", 2))
print(subStringK("aaabbaaa", 100))

✅ Hello everyone, today was my interview date, and I was asked the following question: At first, I thought I could use the two pointers technique to solve it, but then I realized that that would make the algorithm inefficient. Then I noticed that the number of 1s will be the length of the subarray with grouped 1s. This changed my approach to a fixed sliding window, and then the rest was easy. My interviewer was very nice and guided me the whole way.

'''
Given a binary array data, return the minimum number of swaps required to group all 1’s
present in the array together in any place in the array.


Example 1:

Input: data = [1,0,1,0,1]
Output: 1
Explanation: There are 3 ways to group all 1's together:
[1,1,1,0,0] using 1 swap.
[0,1,1,1,0] using 2 swaps.
[0,0,1,1,1] using 1 swap.
The minimum is 1.

Example 2:
Input: data = [0,0,0,1,0]
Output: 0
Explanation: Since there is only one 1 in the array, no swaps are needed.

Example 3:
Input: data = [1,0,1,0,1,0,0,1,1,0,1] count_ones = 6 count_zeros = 3 curr_zeros = 3 min of count_zeros and curr_zeros
l
r
time comp = O(n)
space comp = O(1)

Output: 3
Explanation: One possible solution that uses 3 swaps is [0,0,0,0,0,1,1,1,1,1,1].


Constraints:

1 <= data.length <= 10**5
data[i] is either 0 or 1.
'''

"""
1. count 1's store one count_ones
2. assign count_zeros = inf curr_zeros = 0
3. l, r = 0
4. check for a valid window
5. update curr_zeros
6. take the min of the count_zeros and curr_zeros
7. check if the values at the indexes are zeros if so decrement curr_zeros
8. update pointers
9. return count_zeros
"""
# my code

def minNumberOfSwaps(arr):
    count_ones = arr.count(1)
    count_zeros, curr_zeros = float('inf'), 0
    l = 0
    
    for r in range(len(arr)):
        if arr[r] == 0:
            curr_zeros += 1
        # check for a valid window
        if r - l + 1 == count_ones:
            count_zeros = min(count_zeros, curr_zeros)
            if arr[l] == 0:
                curr_zeros -= 1
            l += 1

    return count_zeros if count_zeros != float('inf') else 0
       

"""
1= 6
curr_zeros = 3
count_zeros = 3
1,0,1,0,1,0,0,1,1,0,1
l
r
"""


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"""
You are given a string s consisting only of the letters 'a' and 'b', and an integer k.
What is the minimum number of characters you need to change to obtain a substring of length ≥ k where all characters are the same?

Example 1:
s = “aabaabaa”, k = 3
Output: 1
Explanation: s can be transformed to “aaaaabaa”

Example 2:
s = “bbabbaba”, k = 8
Output: 3
Explanation: s can be transformed to “bbbbbbbb”

Constraints:
1 <= s.length <= 10^5
1 <= k <= s.length

aaab, k = 4
aaaa
abs(3 - 1) = 2
"""

Given an array of integer arrays arrays where each arrays[i] is sorted in strictly increasing order,
return an integer array representing the longest common subsequence among all the arrays.
A subsequence is a sequence that can be derived from another sequence by deleting some elements (possibly none)
without changing the order of the remaining elements.

Example 1:
Input: arrays = [[1,3,4],
                 [1,4,7,9]]
Output: [1,4]
Explanation: The longest common subsequence in the two arrays is [1,4].
Example 2:
Input: arrays = [[2,3,6,8],
                 [1,2,3,5,6,7,10],
                 [2,3,4,6,9]]
Output: [2,3,6]
Explanation: The longest common subsequence in all three arrays is [2,3,6].

Example 3:
Input: arrays = [[1,2,3,4,5],
                 [6,7,8]]
Output: []
Explanation: There is no common subsequence between the two arrays.


Constraints:
2 <= arrays.length <= 100
1 <= arrays[i].length <= 100
1 <= arrays[i][j] <= 100
arrays[i] is sorted in strictly increasing order.

'''

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Register Here


Requirement:

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(Any department)

Interest for puzzles, games, problem solving


Join us @aastucpc

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