Largest Substring Between Two Equal Characters | 3 Ways | C++ | JAVA | Leetcode 1624
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This is the 17th Video of our Playlist "Leetcode Easy".
In this video we will try to solve an easy but good problem - Largest Substring Between Two Equal Characters (Leetcode 1624).
We will solve it in 3 ways.
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 : Largest Substring Between Two Equal Characters
Company Tags : Amazon, Google, Meta
My solutions on Github(C++ & JAVA) : https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/strings/Easy%20Tagged/Largest%20Substring%20Between%20Two%20Equal%20Characters.cpp
Leetcode Link : https://leetcode.com/problems/largest-substring-between-two-equal-characters/
My DP Concepts Playlist : https://youtu.be/7eLMOE1jnls
My Graph Concepts Playlist : https://youtu.be/5JGiZnr6B5w
My GitHub Repo for interview preparation : https://github.com/MAZHARMIK/Interview_DS_Algo
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Approach-1 Summary : It uses a nested loop to iterate through pairs of characters in the string, and whenever it finds a pair that is equal, it calculates the distance between them (the difference of their indices) and updates the result with the maximum distance encountered so far. The final result represents the maximum distance between any two equal characters in the string. If no equal characters are found, the function returns -1.
Approach-2 Summary : It uses an unordered map (unordered_map) to store the last index where each character was encountered during the iteration. For each character in the string, it checks whether the character is already in the map. If it is not, it adds the character to the map with its current index. If the character is already present in the map, it calculates the distance between the current index and the last index where the character was found and updates the result with the maximum distance encountered so far. The final result represents the maximum distance between any two equal characters in the string. If no equal characters are found, the function returns -1. This approach has a linear time complexity of O(n), where n is the length of the input string.
Approach-3 Summary : It utilizes a vector (count) to keep track of the last index where each character, represented by its ASCII value ('a' to 'z'), was encountered during the iteration. For each character in the string, it checks whether the character is already in the vector. If it is not, it adds the character to the vector with its current index. If the character is already present in the vector, it calculates the distance between the current index and the last index where the character was found and updates the result with the maximum distance encountered so far. The final result represents the maximum distance between any two equal characters in the string. If no equal characters are found, the function returns -1. This approach has a linear time complexity of O(n), where n is the length of the input string.
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✨ Timelines✨
00:00 - Introduction
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