Russian Doll Envelopes

Logical Breakdown

• [ ] Subproblem Identification:
• [ ] Optimal Substructure:
• [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
• [ ] Optimization: (Matrix Exponentiation if 'A' is huge)

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Mathematical Rigor

State Definition

Let \(dp[i]\) be the state for the \(i\)-th subproblem.

Recurrence Relation

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

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Visualization

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

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Complexity Analysis

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

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Code Reference

package com.dsa.dynamic_programming;

import java.util.*;

/ * Problem: Russian Doll Envelopes * Group: 11. Dynamic Programming */ / * ============================================================================================ * PROBLEM: Russian Doll Envelopes * ============================================================================================ * * --------------------- * 1. Problem Description * --------------------- * Given a 2D array of integers 'envelopes' where envelopes[i] = [width, height]. * Find the maximum number of envelopes you can Russian doll (put one inside another). * Constraint: Both width and height must be strictly greater to fit. * * --------------------- * 2. Logical Breakdown * --------------------- * This is a variation of the Longest Increasing Subsequence (LIS) problem. * However, we have two dimensions (width, height). * * Strategy: Sorting + LIS * 1. Sort the envelopes based on WIDTH in Ascending order. * - If widths are equal, sort by HEIGHT in Descending order. * * Why Descending Height for same Width? * - If we have [3, 3] and [3, 4], we cannot put [3, 3] inside [3, 4] (width not strictly greater). * - By sorting heights descending ([3, 4], [3, 3]), the LIS algorithm (which only looks for * increasing heights) will naturally skip [3, 3] after picking [3, 4], preventing the invalid case. * * 2. Find LIS of Heights: * - After sorting, the problem reduces to finding the Longest Increasing Subsequence of the heights array. * - Since widths are already sorted (and handled via the secondary sort), an increasing height sequence guarantees valid nesting. * * --------------------- * 3. Complexity * --------------------- * Time Complexity: O(N log N) * - Sorting: O(N log N) * - LIS (using Binary Search): O(N log N) * Space Complexity: O(N) * * ============================================================================================ */ public class RussianDollEnvelopes { public int solve(int A) { // TODO: Implement solution return 0; }

public int maxEnvelopes(int[][] env1) {
    return 1;
}

}