Special Median
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: Special Median
* ============================================================================================
*
* ---------------------
* 1. Logical Breakdown & Mathematical Formulation
* ---------------------
* An array is special if \(A[i] = Median(A[0...i-1])\) or \(A[i] = Median(A[i+1...N-1])\).
* * Step 1: Running Median Strategy
* - Finding the median of a dynamic set is optimized using a Max-Heap (left half)
* and a Min-Heap (right half).
* - For a set of size \(k\):
* - If \(k\) is odd: Median is the root of the larger heap.
* - If \(k\) is even: Median is \((MaxHeap.peek() + MinHeap.peek()) / 2.0\).
*
* Step 2: Forward Pass (Prefix Medians)
* 1. Iterate from \(i = 0\) to \(N-1\).
* 2. Before adding \(A[i]\) to the heaps, check if \(A[i]\) equals the current median
* of the elements already in the heaps (\(A[0...i-1]\)).
* 3. Special case: At \(i=0\), there is no left median.
*
* Step 3: Backward Pass (Suffix Medians)
* 1. Clear heaps and iterate from \(i = N-1\) down to 0.
* 2. Before adding \(A[i]\), check if \(A[i]\) equals the current median of \(A[i+1...N-1]\).
*
* ---------------------
* 2. Complexity Analysis
* ---------------------
* Time Complexity:
* O(N log N)
* - Two passes (forward and backward).
* - Each element involves heap insertion/rebalancing: O(log N).
*
* Space Complexity:
* O(N) to store the elements in the heaps.
*
* ---------------------
* 3. Base Case & Constraints
* ---------------------
* - N = 1: The problem implies checking neighbors. By definition, a single element
* cannot satisfy the prefix/suffix condition as the sets are empty.
* - Precision: Use double for median calculations to prevent integer division errors.
* ============================================================================================
/
public class SpecialMedian {
public int solve(int[] A) {
// TODO: Implement solution
return 0;
}
}