Squareful Arrays
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.recursion;
import java.util.*;
/ * Problem: Squareful Arrays * Group: 07. Recursion */ / * ============================================================================================ * PROBLEM: Number of Squareful Arrays * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given an array of integers A, the array is "squareful" if for every pair of adjacent elements, * their sum is a perfect square. * * Task: Find and return the number of distinct permutations of A that are squareful. * * Definition of Distinct Permutations: * Two permutations P1 and P2 are considered different if there exists an index i * such that P1[i] != P2[i]. * * --------------------- * 2. Logical Breakdown & Mathematical Formulation * --------------------- * This can be modeled as a Graph Problem or a Constraint Satisfaction Problem. * - Nodes: Indices of the array A. * - Edge (u, v): Exists if (A[u] + A[v]) is a perfect square. * - Goal: Find the number of distinct Hamiltonian Paths (paths that visit every node exactly once). * * Strategy (Backtracking with Pruning): * 1. Sort the Array: * Sorting brings duplicate elements together (e.g., [2, 2, 2]). This is essential for * handling the "distinct permutation" constraint efficiently. * * 2. Backtracking Function: * Let solve(current_index, last_value) be the recursive function. * - Try to place every unused element A[i] at the current_index. * - CONSTRAINT CHECK: Only place A[i] if (last_value + A[i]) is a perfect square. * - DUPLICATE CHECK: If A[i] == A[i-1] and A[i-1] was not used in the current depth context, * skip A[i] to avoid counting the same arrangement twice. * * 3. Base Case: * If count of placed elements == N, we found one valid permutation. Return 1. * * Mathematical Check (Perfect Square): * boolean isSquare(long sum) { * long root = (long) Math.sqrt(sum); * return root * root == sum; * } * * --------------------- * 3. Constraints & Complexity Analysis * --------------------- * Constraints: * 1 <= Length <= 12 (Very small N -> Indicates exponential complexity is acceptable) * 1 <= A[i] <= 10^9 (Sum can reach 210^9, fits in signed 32-bit int but use long for safety). * * Time Complexity: * O(N!) - Worst case (e.g., A=[0,0,0...]). * However, the "Squareful" constraint acts as a powerful pruning heuristic. * In practice, the branching factor is much smaller than N. * * Space Complexity: * O(N) - Recursion stack depth and boolean 'used' array. * * --------------------- * 4. Input / Output Format * --------------------- * Input: Integer array A. * Output: Integer count of squareful permutations. * * --------------------- * 5. Examples * --------------------- * Input 1: A = [2, 2, 2] * - 2+2=4 (Square). * - Only 1 unique permutation: [2, 2, 2]. * Output 1: 1 * * Input 2: A = [1, 17, 8] * - 1+8=9 (Square), 8+17=25 (Square). * - Valid Paths: 1 -> 8 -> 17 and 17 -> 8 -> 1. * Output 2: 2 * * ============================================================================================ / public class SquarefulArrays { public int solve(int[] A) { // TODO: Implement solution return 0; } }