Misha And Candies
Logical Breakdown
- [ ] Subproblem Identification:
- [ ] Optimal Substructure:
- [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
- [ ] Optimization: (Matrix Exponentiation if 'A' is huge)
Mathematical Rigor
State Definition
Let \(dp[i]\) be the state for the \(i\)-th subproblem.
Recurrence Relation
Visualization
Complexity Analysis
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Iterative DP | \(O(N)\) | \(O(1)\) |
Code Reference
package com.dsa.heaps;
/
* ============================================================================================
* PROBLEM: Misha and Candies
* ============================================================================================
*
* ---------------------
* 1. Problem Description
* ---------------------
* Misha has N boxes of candies. In each step:
* 1. Pick the box with the minimum number of candies (must be <= B).
* 2. Eat floor(minCandies / 2) candies.
* 3. Add the remaining candies (minCandies - eaten) to the next minimum box.
* 4. Remove the original box from consideration.
* Task: Calculate total candies eaten until no box with <= B candies remains.
*
* ---------------------
* 2. Logical Breakdown & Mathematical Formulation
* ---------------------
* This problem is a simulation of a greedy process using a Min-Heap.
*
* Algorithm:
* 1. Initialization:
* - Insert all candy counts from the boxes into a Min-Priority Queue.
*
* 2. Iterative Consumption:
* While the Min-Heap is not empty:
* a. Extract the minimum box minBox.
* b. If minBox > B, stop the process (Misha won't eat from this box).
* c. Calculate eaten: \(eaten = \lfloor minBox / 2 \rfloor\).
* d. Add eaten to totalEaten.
* e. Calculate remaining: \(rem = minBox - eaten\).
* f. If the heap is not empty:
* - Extract the next minimum box nextMin.
* - Add rem to nextMin.
* - If the updated nextMin <= B, push it back into the heap.
* (If > B, it remains out of the heap as Misha won't pick it anymore).
*
* ---------------------
* 3. Constraints & Complexity Analysis
* ---------------------
* Time Complexity:
* O(N log N)
* - Initial heap construction: O(N log N).
* - Each box is extracted once: O(N log N).
*
* Space Complexity:
* O(N) to store the candy counts in the Priority Queue.
* ============================================================================================
*/
public class MishaAndCandies {
public int solve(int[] a1, int A) {
// TODO: Implement solution
return 0;
}
}