🙏

জয়  শ্রী  রাম

🕉

---

The DP problems belonging to this category, in its simplest form, looks like below or some kind of variations of it:

• Given a target find minimum (maximum) cost / path / sum to reach the target.

The solution for this kind of problems, in a very generalized form, would often look like below:

• Choose optimal (minimal or maximal, as the case may be) path among all possible paths that lead to the current state, and then add value for the current state.
  
• routes[curr] = min(routes[curr - 1], routes[curr - 2], ... , routes[curr - k]) + cost[i]
  where current target can be reached only from (curr - 1), (curr - 2), ... (curr - k).
  
• Overall the solution would look like this :

  
  for (int curr = 1; curr <= target; curr++) {
     for (int k = 0; k < waysToReachCurrentTarget.size(); k++) {
      dp[i] = min(dp[curr], dp[waysToReachCurrentTarget[k]] + cost / path ) ;
     }
  }
  
  return dp[target];
  
  

The below problem and its solution beautifully demonstrate this approach:

Problem Statement:

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
Note: You can only move either down or right at any point in time.
Example:
Input:

[
  [1,3,1],
  [1,5,1],
  [4,2,1]
]

Output: 7
Explanation: Because the path 1→3→1→1→1 minimizes the sum.

Naive Solution:

Login to Access Content

Optimized Solution:

Login to Access Content

Problem Solving:

• Minimum Cost Tickets

Instructor: