Count Number of Balanced Permutations | Super Detailed Explanation | Leetcode 3343 |codestorywithMIK
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Hi Everyone, this is the 110th video of our Playlist "Dynamic Programming : Popular Interview Problems".
Now we will be solving a Hard and very good DP Problem - Count Number of Balanced Permutations | Super Detailed Explanation | Leetcode 3343 | codestorywithMIK
I will explain it in full detail so that it becomes easy to understand. Each line will be explained and you will know the WHY behind everything.
We will deep dive so that the thought process will be cleared.
Problem Name : Count Number of Balanced Permutations | Super Detailed Explanation | Leetcode 3343 | codestorywithMIK
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Code Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/DP/DP on Strings/Count Number of Balanced Permutations.cpp
Leetcode Link : https://leetcode.com/problems/count-number-of-balanced-permutations
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Video Summary :
The core idea behind the solution is to count permutations where the sum of digits at even indices equals the sum at odd indices. Brute-force checks all permutations, but it's inefficient for large inputs. The optimized approach uses digit frequency counting and combinatorics to avoid generating all permutations explicitly. It recursively assigns digits to even and odd positions, tracking the running sums and ensuring balance. To handle duplicate digits and large numbers, it uses modular arithmetic, factorials, and Fermat's Little Theorem for inverse factorials. Memoization avoids redundant calculations for repeated states.
✨ Timelines✨
00:00 - Introduction
0:23 - Motivation
0:36 - Problem Explanation
2:26 - Brute Force
7:10 - Thought Process and Detailed Analysis
26:52 - Writing Important Function - recursion
46:45 - Pre Compute Factorial
48:51 - Pre Compute Inverse Factorial (Using Fermat’s Little Theorem)
56:00 - Coding it up
1:10:22 - Time and Space Complexity Without Memoization
1:15:56 - Time and Space Complexity With Memoization
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