Count the Number of Consistent Strings | Using Bit Manipulation | Leetcode 1684 | codestorywithMIK
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This is the 54th Video of our Playlist "Leetcode Easy : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve an easy problem : Count the Number of Consistent Strings | Using Bit Manipulation | Leetcode 1684 | codestorywithMIK
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Count the Number of Consistent Strings | Using Bit Manipulation | Leetcode 1684 | codestorywithMIK
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Summary :
Approach-1 (Using HashSet)
Time Complexity: O(m + n*k) where m is the length of allowed, n is the number of words, and k is the average length of each word.
Space Complexity: O(m) for storing the allowed characters in an unordered set.
Description: This approach uses a HashSet to store allowed characters. For each word, it checks if all its characters are present in the set. If a character is missing, the word is considered inconsistent. This method is efficient in character lookup due to the O(1) average time complexity of set operations.
Approach-2 (Using Boolean Array)
Time Complexity: O(m + n*k)
Space Complexity: O(m) (actually O(26), which is constant, but generally expressed as O(m) for larger alphabet sizes).
Description: A boolean array of size 26 is used to mark the presence of allowed characters. For each word, we check if all characters are present by referencing the boolean array. If a character is absent, the word is inconsistent. This approach avoids hashing and offers a more memory-efficient solution for a fixed alphabet size.
Approach-3 (Using Bit Manipulation)
Time Complexity: O(m + n*k)
Space Complexity: O(1)
Description: This method uses bit manipulation to create a bitmask representing the allowed characters. Each bit in the mask corresponds to a character from 'a' to 'z'. For each word, the method checks if all characters are allowed by comparing their corresponding bits in the mask. It’s highly space-efficient since the bitmask uses only 32 bits (constant space).
Key Differences:
Space Efficiency: The bit manipulation approach (Approach-3) is the most space-efficient with O(1) space, while the other two require O(m) space.
✨ Timelines✨
00:00 - Introduction
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