664. Strange Printer | Dynamic Programming | Leetcode POTD Explained

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In this video, we dive deep into an intriguing coding challenge: Strange Printer ! 🎯

⏱ TIMESTAMPS
0:00 – Intro
1:38 - Constraints
2:12 – Understand the Test Case
6:38 - Recursion tree
15:40 – Code Explanation

🎯 Objective:
In this video, we'll tackle the intriguing "Strange Printer" problem. We'll learn how to efficiently compute the minimum number of turns needed for a printer to print a given string with some unique constraints.

💡 Concepts Covered:
- Dynamic Programming: Understand how to use DP to solve the problem optimally.
- Recursive Function: Learn how to implement and utilize recursive solutions for substring problems.
- State Compression: Explore techniques to reduce problem complexity using memoization.

📈 What You'll Learn:
- How to set up and use a DP table to store and reuse results of subproblems.
- Techniques to minimize the number of operations by considering character overlaps.
- Step-by-step implementation of a solution using C++.

🚀 Watch Now and Level Up Your Algorithm Skills!

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[codeby_naruto]

Code :-

class Solution {
public:
vector<vector<int>> dp; // DP table to store results of subproblems

// Recursive function to solve the problem for substring s[i...j]
int solve(string &s, int i, int j) {
// Base case: If the substring is of length 1, only 1 turn is needed
if (i == j) return 1;

// Return the stored result if it has been computed before
if (dp[i][j] != -1) return dp[i][j];

int minTurn = INT_MAX; // Initialize minimum turns to a large value

// Try every possible partition point k in the substring
for (int k = i; k < j; k++) {
// Calculate the minimum turns for the left and right parts of the partition
int LeftPart = solve(s, i, k);
int RightPart = solve(s, k + 1, j);

// Update the minimum turns needed considering this partition
minTurn = min(minTurn, LeftPart + RightPart);
}

// If characters at the start and end of the substring are the same, reduce the count by 1
return dp[i][j] = (s[i] == s[j]) ? minTurn - 1 : minTurn;
}

// Main function to calculate the minimum number of turns needed for the entire string
int strangePrinter(string s) {
int n = s.size();
// Initialize the DP table with -1, indicating that no subproblem has been solved yet
dp.resize(n + 1, vector<int>(n + 1, -1));

// Compute the result for the entire string from index 0 to n-1
return solve(s, 0, n - 1);
}
};

[ardtierguy]

Bhaiya ye pahle aapne 10^8 karke kaise complexity bataya next problem jab karana to please acche se samjhana bhaiya