Minimum 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: Minimum Largest Element (Min-Max Greedy Optimization)
* ============================================================================================
*
* ---------------------
* 1. Logical Breakdown & Mathematical Formulation
* ---------------------
* We want to perform B additions such that \(max(A_{modified})\) is minimized.
*
* Step 1: The Greedy Insight
* - In each operation, we look at what the value of an element would be if we
* incremented it.
* - To keep the maximum low, we should pick the element whose "next value"
* (current + original) is the smallest among all possible next values.
*
* Step 2: Min-Heap Implementation
* 1. Maintain a Min-Heap of pairs: {nextValue, index}.
* - nextValue is the value the element will become after one more addition.
* - index helps us retrieve the originalValue from array A.
* 2. Initially, the heap contains {A[i] * 2, i} for all i.
* 3. Perform B operations:
* a. Extract the pair with the smallest nextValue.
* b. This is the element we choose to increment.
* c. Update its state: nextValue = nextValue + A[index].
* d. Push the updated pair back into the heap.
*
* Step 3: Final Calculation
* - After B operations, the answer is the maximum value currently present in the array.
* - Note: The current value of an element at index is nextValue - A[index].
*
* ---------------------
* 2. Complexity Analysis
* ---------------------
* Time Complexity:
* O((N + B) log N)
* - Initial heap build: O(N log N).
* - B operations of extract-min and insert: O(B log N).
*
* Space Complexity:
* O(N) to store the state of N elements in the Priority Queue.
*
* ---------------------
* 3. Base Case Verification
* ---------------------
* - If B = 0: The answer is simply the current max of array A.
* - If all A[i] are equal: Operations will be distributed evenly across all indices.
* ============================================================================================
*/
public class MinimumLargestElement {
public int solve(int[] a3, int A) {
// TODO: Implement solution
return 0;
}
}