Count the Number of Powerful Integers | Detailed Thought Process | Leetcode 2999 | codestorywithMIK
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Hi Everyone, this is the 77th video of our Playlist "Leetcode Easy".
Now we will be solving a good and easy problem - Count the Number of Powerful Integers | Detailed Thought Process | Leetcode 2999 | codestorywithMIK
The video contains my detailed thought process and a simple approach to this Hard Problem.
Problem Name : Count the Number of Powerful Integers | Detailed Thought Process | Leetcode 2999 | codestorywithMIK
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My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/strings/Count the Number of Powerful Integers.cpp
Leetcode Link : https://leetcode.com/problems/count-the-number-of-powerful-integers
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Video Summary :
To solve the problem, the idea is to count how many numbers between start and finish end with a given suffix and have all other digits (the prefix) using only digits from 0 to a given limit.
The approach starts by converting the range endpoints to strings and focusing on counting how many valid numbers are less than or equal to a given bound. For each digit in the prefix (before the suffix), we count all combinations where that digit is strictly less than the corresponding digit in the bound, multiplying by the number of valid combinations that can follow. If the digit is greater than the limit, all further digits would have exceeded the constraint, so we break early.
After processing the prefix, we check if the suffix of the number is greater than or equal to the target suffix and add one more to the count if so.
Finally, we compute the result by subtracting the count up to start - 1 from the count up to finish, effectively getting the number of valid integers in the range [start, finish].
✨ Timelines✨
00:00 - Introduction
0:23 - Motivation
0:41 - Problem Explanation
3:40 - How to find possible count
5:34 - Thought Process
12:00 - Example Dry Run
27:49 - Another Important Example
31:13 - One Edge Case
38:00 - Coding it up
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