Ath Largest Element
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: Ath Largest Element in Subarrays [1 : i]
* ============================================================================================
*
* ---------------------
* 1. Logical Breakdown & Mathematical Formulation
* ---------------------
* We need to find the Ath largest element dynamically as we iterate through array B.
* * Step 1: Min-Heap as a Filter
* - A Min-Heap of size A allows us to maintain the A largest elements.
* - In such a heap, the root is the smallest of these A elements.
* - This root is, by definition, the Ath largest element of the entire set
* processed so far.
*
* Step 2: Stream Processing
* 1. Initialize an empty Min-Priority Queue.
* 2. Create a result array res of size N.
* 3. For each index i from 0 to N-1:
* a. If heap size < A:
* - Add B[i] to the heap.
* - If heap size is still < A, set res[i] = -1.
* - If heap size reached A, set res[i] = heap.peek().
* b. If heap size == A:
* - If B[i] > heap.peek():
* - The new element is larger than our current Ath largest.
* - Remove the current smallest (heap.poll()) and add B[i].
* - res[i] = heap.peek().
*
* ---------------------
* 2. Complexity Analysis
* ---------------------
* Time Complexity:
* O(N log A)
* - We iterate through the array of size N.
* - Each insertion/extraction in the heap takes O(log A) time.
*
* Space Complexity:
* O(A) to store the top A elements in the Priority Queue.
*
* ---------------------
* 3. Base Case Verification
* ---------------------
* - If i < A-1: The subarray has fewer than A elements, resulting in -1.
* - If A = 1: The problem reduces to finding the running maximum of the array.
* ============================================================================================
/
public class AthLargestElement {
public int[] solve(int A, int[] b3) {
// TODO: Implement solution
return new int[0];
}
}