[Dynamic Programming] Russian Doll Envelopes - Medium

Russian Doll Envelopes Problem

Russian Doll Envelopes is a dynamic programming problem. It is about finding the most envelopes we can nest inside each other. Each envelope has a width and a height. One envelope can fit into another if both its width and height are smaller. To solve this, we need to sort the envelopes. Then we use a longest increasing subsequence (LIS) strategy to find the biggest group of envelopes that can nest.

In this article, we will look closely at the Russian Doll Envelopes problem. We will explain the input and output needs. We will also check different dynamic programming ways to solve this problem in Java, Python, and C++. We will talk about binary search improvements for these languages. We will give a code walkthrough for the solutions. Lastly, we will answer common questions to help you understand the Russian Doll Envelopes problem better.

• [Dynamic Programming] Russian Doll Envelopes Problem Explanation
• Understanding the Input and Output Requirements
• Dynamic Programming Approach for Russian Doll Envelopes in Java
• Dynamic Programming Approach for Russian Doll Envelopes in Python
• Dynamic Programming Approach for Russian Doll Envelopes in C++
• Binary Search Optimization for Russian Doll Envelopes in Java
• Binary Search Optimization for Russian Doll Envelopes in Python
• Binary Search Optimization for Russian Doll Envelopes in C++
• Code Walkthrough for Russian Doll Envelopes Solutions
• Frequently Asked Questions

Understanding the Input and Output Requirements

In the Russian Doll Envelopes problem, we want to find out the most envelopes we can nest inside each other. An envelope can hold another envelope only if both the width and height of the inside envelope are less than those of the outside envelope.

Input Requirements

• Input: A 2D array of integers called envelopes. Each element shows an envelope. It has two numbers:

  • envelopes[i][0]: the width of the i-th envelope.
  • envelopes[i][1]: the height of the i-th envelope.

  Example:

  envelopes = [[5,4],[6,4],[6,7],[2,3]]

Output Requirements

• Output: An integer that shows the maximum number of envelopes that can be nested inside each other.

Constraints

• The number of envelopes is from 1 to 1000.
• The width and height of each envelope are from 1 to 10^4.

Example

For the input:

envelopes = [[5,4],[6,4],[6,7],[2,3]]

The output should be:

3

This means a maximum of 3 envelopes can be nested. The specific nesting is: [[2,3] -> [5,4] -> [6,7]].

We can solve this problem well by using dynamic programming. We can also use sorting or binary search to help find the longest increasing subsequence by looking at the sizes of the envelopes.

Dynamic Programming Approach for Russian Doll Envelopes in Java

To solve the Russian Doll Envelopes problem using dynamic programming in Java, we take a clear approach. The main goal is to find the longest increasing subsequence of envelopes based on their width and height.

Steps to Implement the Dynamic Programming Approach:

1. Sort Envelopes: First, we sort the envelopes. We sort mainly by width in increasing order. If two envelopes have the same width, we sort by height in decreasing order. This way, we do not count envelopes with the same width.

2. Dynamic Programming Array: We create a DP array. Here dp[i] shows the maximum number of envelopes that can be nested ending with envelope i.

3. Fill the DP Array: For each envelope, we look at all previous envelopes to see if the current envelope can fit inside the previous ones.

Java Implementation:

import java.util.Arrays;

public class RussianDollEnvelopes {

    public int maxEnvelopes(int[][] envelopes) {
        // Sort the envelopes
        Arrays.sort(envelopes, (a, b) -> {
            if (a[0] == b[0]) return b[1] - a[1]; // Descending order for height
            return a[0] - b[0]; // Ascending order for width
        });

        // DP array to store the maximum envelopes ending at each index
        int n = envelopes.length;
        int[] dp = new int[n];
        int maxCount = 0;

        // Fill the DP array
        for (int i = 0; i < n; i++) {
            dp[i] = 1; // Each envelope can at least hold itself
            for (int j = 0; j < i; j++) {
                if (envelopes[i][0] > envelopes[j][0] && envelopes[i][1] > envelopes[j][1]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
            maxCount = Math.max(maxCount, dp[i]); // Update the maximum count
        }

        return maxCount;
    }

    public static void main(String[] args) {
        RussianDollEnvelopes solution = new RussianDollEnvelopes();
        int[][] envelopes = {{5, 4}, {6, 7}, {2, 3}, {1, 2}, {7, 8}};
        System.out.println("Maximum number of envelopes that can be nested: " + solution.maxEnvelopes(envelopes));
    }
}

Explanation of the Code:

• Sorting: We sort the envelopes using a custom rule. We sort by width in increasing order. If widths are equal, we sort by height in decreasing order.
• DP Array Initialization: We set up a DP array. Each element starts as 1 because each envelope can at least hold itself.
• Nested Loops: The outer loop goes through each envelope. The inner loop checks all earlier envelopes to find if they can fit.
• Result: The maximum number in the DP array gives the answer. This shows the maximum number of envelopes that can fit inside each other.

This way, we use dynamic programming to solve the Russian Doll Envelopes problem in Java. For more details on dynamic programming, you can read Dynamic Programming - Fibonacci with Memoization.

Dynamic Programming Approach for Russian Doll Envelopes in Python

The Russian Doll Envelopes problem is a well-known challenge in dynamic programming. We need to find the most envelopes that can fit inside each other. One envelope fits into another if its width and height are both bigger than the other envelope’s width and height.

Problem Definition

We have a list of envelopes shown as pairs of numbers. Each pair has a width and a height. Our goal is to find the maximum number of envelopes that can nest.

Steps for the Dynamic Programming Approach

1. Sort Envelopes: First, we sort the envelopes by width. If two envelopes have the same width, we sort them by height but in reverse order. This keeps us from counting envelopes with the same width.
2. Apply Longest Increasing Subsequence (LIS): After we sort, we need to find the longest increasing subsequence based on the heights of the envelopes.

Python Implementation

Here is how we can implement the solution in Python:

def maxEnvelopes(envelopes):
    # Sort the envelopes. First by width, then by height in descending order
    envelopes.sort(key=lambda x: (x[0], -x[1]))
    
    # Extract the heights for LIS calculation
    heights = [h for _, h in envelopes]
    
    # Function to find the length of LIS
    def lis(sequence):
        dp = []
        for h in sequence:
            pos = bisect.bisect_left(dp, h)
            if pos == len(dp):
                dp.append(h)
            else:
                dp[pos] = h
        return len(dp)
    
    return lis(heights)

# Example usage:
envelopes = [[5,4],[6,4],[6,7],[2,3]]
print(maxEnvelopes(envelopes))  # Output: 3

Explanation of the Code

• We sort the input envelopes first by width and then by height in reverse order.
• The lis function finds the length of the longest increasing subsequence. It uses binary search to keep the process fast with O(n log n) time.
• In the end, the function gives us the maximum number of envelopes we can nest.

This dynamic programming method works well for the Russian Doll Envelopes problem in Python. It uses sorting and the longest increasing subsequence method. If you want to learn more about dynamic programming, you can check out Dynamic Programming: Longest Increasing Subsequence.

Dynamic Programming Approach for Russian Doll Envelopes in C++

The Russian Doll Envelopes problem is about finding the most envelopes we can put inside each other. Each envelope has two numbers, width and height. An envelope can fit inside another if both its width and height are smaller than the other.

Dynamic Programming Solution

1. Sort the Envelopes: First, we need to sort the envelopes by width. If two envelopes have the same width, we sort them by height from high to low. This way, we avoid putting envelopes with the same width inside each other.

2. Apply Longest Increasing Subsequence (LIS): After sorting, we will use the Longest Increasing Subsequence (LIS) method on the heights of the envelopes.

C++ Implementation

#include <vector>
#include <algorithm>

using namespace std;

int maxEnvelopes(vector<vector<int>>& envelopes) {
    // Sort envelopes: first by width, then by height in descending order
    sort(envelopes.begin(), envelopes.end(), [](const vector<int>& a, const vector<int>& b) {
        return a[0] < b[0] || (a[0] == b[0] && a[1] > b[1]);
    });

    // Vector to store the heights for LIS
    vector<int> dp;
    
    for (const auto& envelope : envelopes) {
        int height = envelope[1];
        // Use binary search to find the position of height in dp
        auto it = lower_bound(dp.begin(), dp.end(), height);
        if (it == dp.end()) {
            dp.push_back(height); // If height is larger than all, add it
        } else {
            *it = height; // Change it with the current height
        }
    }
    
    return dp.size(); // The size of dp is the most envelopes we can nest
}

Explanation of the Code

• Sorting: We sort the envelopes so we can easily find the longest increasing subsequence in heights.
• Binary Search: We use the lower_bound function to keep the dp vector updated fast. This helps keep the time at O(n log n).
• Result: The length of the dp vector tells us the maximum number of envelopes that can be nested.

This dynamic programming way helps us solve the Russian Doll Envelopes problem well in C++. For more similar problems, you can look at Longest Increasing Subsequence.

Binary Search Optimization for Russian Doll Envelopes in Java

We can solve the Russian Doll Envelopes problem in a smart way. We will use dynamic programming and binary search together. This method helps us save time and works well with large datasets.

Problem Overview

The Russian Doll Envelopes problem wants us to find the biggest number of envelopes that can fit inside each other. Each envelope is shown by two numbers. The first number is the width and the second is the height. Envelope A can fit inside envelope B only if both the width and height of A are less than those of B.

Steps to Implement Binary Search Optimization

1. Sort the Envelopes: First, we need to sort the envelopes. We sort by width in increasing order. If two envelopes have the same width, we sort them by height in decreasing order. This way, we avoid nesting envelopes that have the same width.

2. Use Dynamic Programming with Binary Search: We will create a dynamic programming array to keep track of the increasing heights. For each envelope, we will use binary search to find its position in the array and update it.

Java Implementation

Here is how we can implement the binary search optimization in Java:

import java.util.Arrays;
import java.util.Comparator;

public class RussianDollEnvelopes {
    
    public int maxEnvelopes(int[][] envelopes) {
        // Sort envelopes first by width and then by height in descending order
        Arrays.sort(envelopes, new Comparator<int[]>() {
            public int compare(int[] a, int[] b) {
                if (a[0] == b[0]) {
                    return b[1] - a[1]; // Descending order for heights
                }
                return a[0] - b[0]; // Ascending order for widths
            }
        });
        
        // DP array to hold the heights
        int[] dp = new int[envelopes.length];
        int length = 0;

        for (int[] envelope : envelopes) {
            int height = envelope[1];
            int index = binarySearch(dp, length, height);
            dp[index] = height;
            if (index == length) {
                length++;
            }
        }

        return length;
    }

    private int binarySearch(int[] dp, int length, int height) {
        int left = 0, right = length;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (dp[mid] < height) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }
    
    public static void main(String[] args) {
        RussianDollEnvelopes solution = new RussianDollEnvelopes();
        int[][] envelopes = {{5,4},{6,7},{2,3},{1,1},{7,8}};
        System.out.println("Maximum number of envelopes: " + solution.maxEnvelopes(envelopes));
    }
}

Explanation of the Code

• Sorting: The envelopes get sorted using a special method to make sure they are in the right order for nesting.
• Dynamic Programming Array (dp): This array keeps track of the heights of the envelopes that can be stacked.
• Binary Search Method: The binarySearch method finds the place to change or add in the dp array quickly.
• Time Complexity: The total time complexity is O(n log n). This is because of the sorting and binary search steps.

This optimization really helps us make the solution better for the Russian Doll Envelopes problem in Java. If you want to learn more about similar dynamic programming problems, you can check out the Longest Increasing Subsequence.

Binary Search Optimization for Russian Doll Envelopes in Python

To solve the Russian Doll Envelopes problem well, we can use a binary search optimization after sorting the envelopes. Our main goal is to find the longest increasing subsequence (LIS) of envelopes based on their sizes. This way, we can reduce the time needed to O(n log n). This is much better than the O(n^2) dynamic programming method.

Steps to Implement Binary Search Optimization

1. Sort the Envelopes: First, we sort the envelopes by their width in order from smallest to largest. If two envelopes have the same width, we sort them by height from tallest to shortest. This helps us avoid counting envelopes that cannot fit inside each other.

2. Use a List to Track the LIS: We keep a list that will hold the heights of the envelopes in the current increasing subsequence.

3. Binary Search for Insertion: For each envelope’s height, we use binary search to find the spot where it can replace an element in the LIS list. If it is bigger than all elements, we add it to the end of the list.

Python Code Implementation

Here is the Python code that shows how to use binary search optimization for the Russian Doll Envelopes problem:

from bisect import bisect_left

def maxEnvelopes(envelopes):
    # Step 1: Sort envelopes
    envelopes.sort(key=lambda x: (x[0], -x[1]))
    
    # Step 2: Initialize a list to store the LIS
    lis = []
    
    for _, height in envelopes:
        # Step 3: Perform binary search
        pos = bisect_left(lis, height)
        
        # If pos is equal to the length of lis, it means height is greater than all elements in lis
        if pos == len(lis):
            lis.append(height)
        else:
            lis[pos] = height
            
    return len(lis)

# Example usage:
envelopes = [[5,4],[6,7],[2,3],[7,8]]
result = maxEnvelopes(envelopes)
print("Maximum number of envelopes that can be Russian dolled:", result)

Explanation of the Code

• Sorting: The sort method with a special key sorts the envelopes by width and height.
• LIS Calculation: The bisect_left function finds where to put the current envelope’s height in the lis list.
• Building the LIS: If the position equals the length of lis, it means the current height is bigger than all heights in lis. We then add it to the list. If not, we replace an existing height to keep the smallest possible values for future checks.

This method quickly calculates the maximum number of envelopes that can fit inside each other, following the rules of the Russian Doll Envelopes problem.

Binary Search Optimization for Russian Doll Envelopes in C++

To make a better solution for the Russian Doll Envelopes problem in C++, we can use a binary search method after we sort the envelopes. This method makes the time it takes to solve the problem faster, reducing it to O(n log n).

Problem Recap

We have a list of envelopes that have widths and heights. Our goal is to find the most envelopes we can fit inside each other. An envelope A can go into envelope B if both the width and height of A are less than those of B.

Approach

1. Sorting: First, we sort the envelopes. We sort by width in rising order. If two envelopes have the same width, we sort by height in falling order. This stops us from counting envelopes that are the same width.

2. Dynamic Programming with Binary Search: After we sort, we can use a dynamic programming technique plus binary search to find the longest increasing sequence based on heights.

C++ Code

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int maxEnvelopes(vector<vector<int>>& envelopes) {
    if (envelopes.empty()) return 0;

    // Sort envelopes. First by width in ascending order, then by height in descending order
    sort(envelopes.begin(), envelopes.end(), [](const vector<int>& a, const vector<int>& b) {
        return a[0] == b[0] ? a[1] > b[1] : a[0] < b[0];
    });

    // Apply a dynamic programming approach to find the longest increasing subsequence in heights
    vector<int> dp;
    for (const auto& env : envelopes) {
        int height = env[1];
        auto it = lower_bound(dp.begin(), dp.end(), height);
        if (it == dp.end()) {
            dp.push_back(height); // If height is greater than all in dp, add it
        } else {
            *it = height; // Replace the found position with the new height
        }
    }

    return dp.size(); // The size of dp will be the maximum number of envelopes
}

int main() {
    vector<vector<int>> envelopes = {{5,4}, {6,4}, {6,7}, {2,3}};
    cout << "Maximum number of envelopes: " << maxEnvelopes(envelopes) << endl;
    return 0;
}

Explanation of the Code

• Sorting: The sort function puts the envelopes in order like we said.
• Dynamic Programming Array: The dp vector keeps the heights of the longest increasing sequence we found.
• Binary Search: The lower_bound function helps us quickly find the place to replace or add in the dp vector.

Time Complexity

• The sorting step takes O(n log n). The binary search inside the loop also runs in O(n log n). So the total time complexity is O(n log n). This makes it good for large inputs.

This binary search optimization for the Russian Doll Envelopes problem in C++ shows how we can mix sorting with dynamic programming for a smart solution to tough problems. For more learning on dynamic programming ideas, you can check the article on Longest Increasing Subsequence which is very useful.

Code Walkthrough for Russian Doll Envelopes Solutions

The Russian Doll Envelopes problem is about finding how many envelopes can fit inside each other. Each envelope has two sizes: width and height. An envelope can go inside another one if both its width and height are smaller.

Dynamic Programming Approach

To solve this with dynamic programming, we need to sort the envelopes. Then, we will use a version of the Longest Increasing Subsequence (LIS) algorithm.

1. Sort the envelopes by width from smallest to biggest. If two envelopes have the same width, sort them by height from biggest to smallest. This way, we do not count envelopes that have the same width.

2. Use a DP array to keep track of the longest increasing subsequence based on height.

Java Code Example

import java.util.Arrays;
import java.util.Comparator;

public class RussianDollEnvelopes {
    public int maxEnvelopes(int[][] envelopes) {
        Arrays.sort(envelopes, new Comparator<int[]>() {
            public int compare(int[] a, int[] b) {
                return a[0] == b[0] ? b[1] - a[1] : a[0] - b[0];
            }
        });

        int n = envelopes.length;
        int[] dp = new int[n];
        int maxLen = 0;

        for (int i = 0; i < n; i++) {
            dp[i] = 1; // Each envelope can at least fit itself
            for (int j = 0; j < i; j++) {
                if (envelopes[i][1] > envelopes[j][1]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
            maxLen = Math.max(maxLen, dp[i]);
        }
        return maxLen;
    }
}

Python Code Example

class Solution:
    def maxEnvelopes(self, envelopes: List[List[int]]) -> int:
        envelopes.sort(key=lambda x: (x[0], -x[1]))
        dp = []
        for _, h in envelopes:
            idx = bisect.bisect_left(dp, h)
            if idx == len(dp):
                dp.append(h)
            else:
                dp[idx] = h
        return len(dp)

C++ Code Example

#include <vector>
#include <algorithm>

using namespace std;

class Solution {
public:
    int maxEnvelopes(vector<vector<int>>& envelopes) {
        sort(envelopes.begin(), envelopes.end(), [](const vector<int>& a, const vector<int>& b) {
            return a[0] == b[0] ? a[1] > b[1] : a[0] < b[0];
        });

        vector<int> dp;
        for (const auto& envelope : envelopes) {
            int h = envelope[1];
            auto it = lower_bound(dp.begin(), dp.end(), h);
            if (it == dp.end()) {
                dp.push_back(h);
            } else {
                *it = h;
            }
        }
        return dp.size();
    }
};

Binary Search Optimization

We can make the dynamic programming solution faster by using binary search to manage the DP array better.

1. Sort envelopes just like before.

2. Use binary search to find where the current height fits in the DP array.

This change makes finding the position faster. The time needed is now O(log n).

For more on dynamic programming and its uses, we can check topics like the Longest Increasing Subsequence.

Frequently Asked Questions

What is the Russian Doll Envelopes problem in dynamic programming?

The Russian Doll Envelopes problem is a well-known challenge in dynamic programming. The goal is to find the most envelopes that can fit into each other. Each envelope has a width and height. One envelope can fit into another only if the outer envelope is bigger in both width and height. This problem is similar to the Longest Increasing Subsequence problem.

How can I implement the Russian Doll Envelopes solution in Java?

To solve the Russian Doll Envelopes problem in Java, we can use a method that combines dynamic programming and sorting. First, we should sort the envelopes by their width and height. Next, we apply a dynamic programming algorithm to find the longest increasing subsequence based on the heights of the sorted envelopes. This helps us find the maximum number of envelopes that can be nested. For detailed code, check our section on the Dynamic Programming Approach for Russian Doll Envelopes in Java.

What is the time complexity of the Russian Doll Envelopes problem?

The time complexity to solve the Russian Doll Envelopes problem using dynamic programming is O(n^2). Here, n is the number of envelopes. This happens because we use a nested loop to compare each envelope with every other envelope. But if we use a binary search after sorting the envelopes, we can lower the time complexity to O(n log n).

Can I solve the Russian Doll Envelopes problem using Python?

Yes, we can solve the Russian Doll Envelopes problem with Python. The method is to sort the envelopes by their size and then use dynamic programming to find the longest increasing subsequence. This method is easy and works well in Python. For a detailed guide, look at our section on the Dynamic Programming Approach for Russian Doll Envelopes in Python.

How is binary search utilized in optimizing the Russian Doll Envelopes solution?

We use binary search in the Russian Doll Envelopes problem to quickly find where to put an envelope’s height in a list. This list keeps track of the maximum heights of envelopes in the longest increasing subsequence. After we sort the envelopes, binary search helps us place each envelope’s height fast. This makes the solution more efficient and brings the time complexity down to O(n log n). For more information, see our section on Binary Search Optimization for Russian Doll Envelopes in Java, Python, or C++.

With these dynamic programming and optimization techniques, we can solve the Russian Doll Envelopes problem effectively. This will also help us improve our skills in competitive programming.