Sixlets
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: Sixlets * Group: 07. Recursion */ / * ============================================================================================ * PROBLEM: SIXLETS * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given an array of integers A of size N and an integer B. * Return the number of non-empty subsequences of A that satisfy two conditions: * 1. The subsequence has a length of exactly B. * 2. The sum of the elements in the subsequence is <= 1000. * * --------------------- * 2. Logical Breakdown & Mathematical Formulation * --------------------- * Let F(index, currentSum, currentSize) be the function that returns the count of valid * subsequences formed using elements from A[index...N-1]. * * Decision Tree (Recursion): * At each index 'i', we have two choices: * * 1. PICK A[i]: * - Valid only if (currentSum + A[i] <= 1000). * - New State: F(i + 1, currentSum + A[i], currentSize + 1) * * 2. LEAVE A[i]: * - Always valid. * - New State: F(i + 1, currentSum, currentSize) * * Recurrence Relation: * Count = F(i + 1, sum + A[i], size + 1) + F(i + 1, sum, size) * * Base Cases (Stopping Conditions): * 1. If currentSize == B: * Return 1 (We found a valid subsequence of length B with sum <= 1000). * 2. If index == N: * Return 0 (We ran out of elements without forming a subsequence of size B). * * --------------------- * 3. Constraints & Complexity Analysis * --------------------- * Constraints: * 1 <= N <= 20 * 1 <= A[i] <= 1000 * 1 <= B <= N * * Time Complexity: * O(2^N) - In the worst case, we explore all subsets. * Since N <= 20, 2^20 approx 10^6 operations, which is well within the 1-second time limit. * * Space Complexity: * O(N) - Recursion stack depth. * * Optimization Note: * While dynamic programming (DP) could be used (State: Index, Sum, Count), the constraints * on N are small enough that simple recursion is faster to implement and sufficiently efficient. * * --------------------- * 4. Input / Output Format * --------------------- * Input: * - Integer Array A. * - Integer B (Required size). * * Output: * - Return integer count of valid subsequences. * * --------------------- * 5. Examples * --------------------- * Input 1: A = [1, 2, 8], B = 2 * Pairs of size 2: * - {1, 2} -> Sum 3 (<= 1000) -> Valid * - {1, 8} -> Sum 9 (<= 1000) -> Valid * - {2, 8} -> Sum 10 (<= 1000) -> Valid * Output 1: 3 * * Input 2: A = [5, 17, 1000, 11], B = 4 * Only one subsequence of size 4: {5, 17, 1000, 11}. * Sum = 1033 (> 1000). * Output 2: 0 * * ============================================================================================ */ public class Sixlets { public int solve(int[] A, int a) { // TODO: Implement solution return 0; } }