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K Places Apart

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

\[ dp[i] = ... \]

Visualization

graph TD A["Current State"] --> B["Previous State 1"] A --> C["Previous State 2"]

Complexity Analysis

Approach Time Complexity Space Complexity
Iterative DP \(O(N)\) \(O(1)\)

Code Reference

package com.dsa.heaps;

/* * ============================================================================================ * PROBLEM: K Places Apart (Sorting a Nearly Sorted Array) * ============================================================================================ * * --------------------- * 1. Logical Breakdown & Mathematical Formulation * --------------------- * An element at index i in the sorted array must have originated from the range * [i - B, i + B] in the original array. * * Step 1: Sliding Window with Min-Heap * - To fill the 0-th position of our sorted array, we only need to look at the first * B+1 elements of the input. The smallest among them MUST be the element * that belongs at index 0. * * Step 2: Incremental Sorting * 1. Initialize a Min-Heap and insert the first B + 1 elements: A[0...B]. * 2. For each index i from 0 to N-1: * a. The current minimum in the heap is the correct element for index i. * b. Extract the minimum and place it in the result array. * c. If there are remaining elements in the input array (i + B + 1 < N), * insert the next element into the heap. * * --------------------- * 2. Complexity Analysis * --------------------- * Time Complexity: * O(N log B) * - We iterate N times. * - Each iteration involves one heap extraction and one insertion, both taking O(log B). * * Space Complexity: * O(B) to maintain the heap of size B + 1. * * --------------------- * 3. Recurrence & Base Case * --------------------- * - Base Case: If B = 0, the array is already sorted. * - If B = N-1, the problem reduces to a standard Heap Sort, O(N log N). * ============================================================================================ / public class KPlacesApart { public int[] solve(int[] a1, int A) { // TODO: Implement solution return new int[0]; } }