Connect Ropes
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: Connect Ropes (Minimum Cost to Connect)
* ============================================================================================
*
* ---------------------
* 1. Problem Description
* ---------------------
* Given an array A representing lengths of ropes. Connect all ropes into one.
* The cost of connecting two ropes of length L1 and L2 is (L1 + L2).
* Task: Find the minimum total cost to connect all ropes.
*
* ---------------------
* 2. Logical Breakdown & Mathematical Formulation
* ---------------------
* This problem is equivalent to building a Huffman Tree. To minimize the total cost,
* we must always pick the two shortest ropes available to connect.
*
* Algorithm:
* 1. Initialization:
* - Insert all rope lengths from array A into a Min-Priority Queue (Min-Heap).
* 2. Iterative Connection:
* - While there is more than one rope in the heap:
* a. Extract the two smallest elements: min1 and min2.
* b. Calculate the connection cost: currentCost = min1 + min2.
* c. Add currentCost to the totalCost accumulator.
* d. Insert currentCost back into the Min-Heap (it's now a single rope).
* 3. Result:
* - Return totalCost.
*
* ---------------------
* 3. Constraints & Complexity Analysis
* ---------------------
* Constraints:
* 1 <= A.length <= 10^5
* 1 <= A[i] <= 10^3
*
* Time Complexity:
* O(N log N)
* - Initial heap construction: O(N log N) or O(N).
* - (N-1) extractions and insertions: each takes O(log N).
*
* Space Complexity:
* O(N) to store the ropes in the Priority Queue.
* ============================================================================================
/
public class ConnectRopes {
public int solve(int[] A) {
// TODO: Implement solution
return 0;
}
}