Seats
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: Seats (Minimize Total Jumps to Cluster People)
* ============================================================================================
*
* ---------------------
* 1. Logical Breakdown & Mathematical Formulation
* ---------------------
* We have indices of people \(P = \{p_1, p_2, ..., p_k\}\). We want to move them to
* a contiguous range of seats \([start, start + k - 1]\).
*
* Step 1: Transformation to Median Problem
* - If we move each person \(p_i\) to a position \(start + i\), the cost is \(|p_i - (start + i)|\).
* - This can be rewritten as \(|(p_i - i) - start|\).
* - Let \(y_i = p_i - i\). We now want to minimize \(\sum |y_i - start|\).
* - Mathematically, the value of start that minimizes this sum is the Median* of
* the array \(Y\).
*
* Step 2: Algorithm
* 1. Traverse the string and store the indices of all 'x' in a list pos.
* 2. If no 'x' or one 'x' exists, return 0.
* 3. Find the median index: mid = pos.size() / 2.
* 4. The person at pos.get(mid) will stay relatively in their place (the anchor).
* 5. For every other person at pos.get(i), the number of jumps to reach the
* cluster around the median is:
* $\(Cost = |pos.get(i) - pos.get(mid)| - |i - mid|\)$
* 6. Sum all costs modulo \(10^7 + 3\).
*
* ---------------------
* 2. Complexity Analysis
* ---------------------
* Time Complexity:
* O(N)
* - Single pass to find indices of 'x'.
* - Single pass to calculate the total jumps from the median.
*
* Space Complexity:
* O(N)
* - To store the indices of the 'x' characters in the worst case (all 'x').
*
* ---------------------
* 3. Base Case Verification
* ---------------------
* - String with no 'x': Returns 0.
* - String with all 'x' already together: Returns 0.
* ============================================================================================
/
public class Seats {
public int solve(String A) {
// TODO: Implement solution
return 0;
}
}