Assign Mice To Holes
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;
import java.util.*;
/* * ============================================================================================ * PROBLEM: Assign Mice to Holes (Minimizing Maximum Displacement) * ============================================================================================ * * --------------------- * 1. Logical Breakdown & Mathematical Formulation * --------------------- * We have N mice at positions \(A_i\) and N holes at positions \(B_i\). * Goal: Assign each mouse to a unique hole to minimize \(max(|A_i - B_{assigned}|)\). * * Step 1: The Greedy Proof (Exchange Argument) * - Suppose we have two mice at \(m_1, m_2\) and two holes at \(h_1, h_2\) such that * \(m_1 < m_2\) and \(h_1 < h_2\). * - To minimize the max travel time, it is always better (or equal) to pair * \((m_1, h_1)\) and \((m_2, h_2)\) rather than \((m_1, h_2)\) and \((m_2, h_1)\). * - Crossing paths increases the distance at least one mouse has to travel. * * Step 2: Algorithm * 1. Sort the mouse positions array A in ascending order. * 2. Sort the hole positions array B in ascending order. * 3. Iterate through both arrays simultaneously from \(i = 0\) to \(N-1\): * - Calculate the time for the current pair: \(time_i = |A[i] - B[i]|\). * - Track the maximum time encountered: \(ans = max(ans, time_i)\). * 4. Return \(ans\). * * --------------------- * 2. Complexity Analysis * --------------------- * Time Complexity: * O(N log N) * - Sorting both arrays dominates the time complexity. * - The linear scan to find the maximum difference is O(N). * * Space Complexity: * O(1) or O(N) * - O(1) if sorting is done in-place. * - O(N) if we consider the space required by the sorting algorithm (like Dual-Pivot Quicksort). * * --------------------- * 3. Base Case Verification * --------------------- * - N=1: Max time is simply \(|A[0] - B[0]|\). * - Mice and holes at same positions: Max time is 0. * ============================================================================================ / public class AssignMiceToHoles { public int solve(int A) { // TODO: Implement solution return 0; }
public int mice(int[] a1, int[] b1) {
return 1;
}
}