Maximum number of envelopes that can be put inside other bigger envelopes

Given N number of envelopes, as {W, H} pair, where W as the width and H as the height. One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope. Find the maximum number of envelopes that can be put inside another envelope and so on. Rotation of envelope is not allowed.

Examples:

Input: envelope[] = {{4, 3}, {5, 3}, {5, 6}, {1, 2}}
Output: 3
Explanation: 
The maximum number of envelopes that can be put into another envelope 
is 3.
({1, 2}, {4, 3}, {5, 6})

Input: envelope[] = {{3, 6}, {5, 4}, {4, 8}, {6, 9}, {10, 7}, {12, 12}}
Output: 4
Explanation:
The maximum number of envelopes that can be put into another envelope is 4.
 ({3, 6}, {4, 8}, {6, 9}, {12, 12})



Naive Approach: This problem is similar to the Longest Increasing Subsequence problem of Dynamic Programming. The idea is to sort the envelopes in non-decreasing order and for each envelope check the number of envelopes that can be put inside that envelope. Follow the steps below to solve the problem:

  • Sort the array in the non-decreasing order of width and height.
  • Initialize a dp[] array, where dp[i] stores the number of envelopes that can be put inside with envelope[i] as the largest envelope.
  • For each envelope[i], loop through the envelopes smaller than itself and check if the width and the height of the smaller envelope is strictly less than that of envelope[i]. If it is less, than the smaller envelope can be put inside envelope[i].
  • The maximum of the dp[] array gives the maximum number of envelopes that can be put inside one another.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns the maximum
// number of envelopes that can be
// inserted into another envelopes
int maxEnvelopes(vector<vector<int> > envelopes)
{
    // Number of envelopes
    int N = envelopes.size();
  
    if (N == 0)
        return N;
  
    // Sort the envelopes in
    // non-decreasing order
    sort(envelopes.begin(),
        envelopes.end());
  
    // Initialize dp[] array
    int dp[N];
  
    // To store the result
    int max_envelope = 1;
  
    dp[0] = 1;
  
    // Loop through the array
    for (int i = 1; i < N; ++i) {
        dp[i] = 1;
  
        // Find envelopes count for
        // each envelope
        for (int j = 0; j < i; ++j) {
  
            if (envelopes[i][0] > envelopes[j][0]
                && envelopes[i][1] > envelopes[j][1]
                && dp[i] < dp[j] + 1)
                dp[i] = dp[j] + 1;
        }
  
        // Store maximum envelopes count
        max_envelope = max(max_envelope,
                        dp[i]);
    }
  
    // Return the result
    return max_envelope;
}
  
// Driver Code
int main()
{
    // Given the envelopes
    vector<vector<int> > envelopes
        = { { 4, 3 }, { 5, 3 }, { 5, 6 }, { 1, 2 } };
  
    // Function Call
    cout << maxEnvelopes(envelopes);
  
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
import java.lang.*;
  
class GFG{
      
// Function that returns the maximum
// number of envelopes that can be
// inserted into another envelopes
static int maxEnvelopes(int[][] envelopes)
{
      
    // Number of envelopes
    int N = envelopes.length;
      
    if (N == 0)
        return N;
      
    // Sort the envelopes in
    // non-decreasing order
    Arrays.sort(envelopes,
               (a, b) -> (a[0] != b[0]) ? 
                          a[0] - b[0] :
                          a[1] - b[1]);
                            
    // Initialize dp[] array
    int[] dp = new int[N];
      
    // To store the result
    int max_envelope = 1;
      
    dp[0] = 1;
      
    // Loop through the array
    for(int i = 1; i < N; ++i)
    {
        dp[i] = 1;
          
        // Find envelopes count for
        // each envelope
        for(int j = 0; j < i; ++j) 
        {
              
            if (envelopes[i][0] > envelopes[j][0] &&
                envelopes[i][1] > envelopes[j][1] &&
                          dp[i] < dp[j] + 1)
                dp[i] = dp[j] + 1;
        }
          
        // Store maximum envelopes count
        max_envelope = Math.max(max_envelope, dp[i]);
    }
      
    // Return the result
    return max_envelope;
}
  
// Driver Code
public static void main (String[] args)
{
      
    // Given the envelopes
    int[][] envelopes = { { 4, 3 }, { 5, 3 }, 
                          { 5, 6 }, { 1, 2 } };
      
    // Function call
    System.out.println(maxEnvelopes(envelopes));
}
}
  
// This code is contributed by offbeat

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Python3

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# Python3 program for the above approach
  
# Function that returns the maximum
# number of envelopes that can be
# inserted into another envelopes
def maxEnvelopes(envelopes):
  
    # Number of envelopes
    N = len(envelopes)
  
    if (N == 0):
        return N
  
    # Sort the envelopes in
    # non-decreasing order
    envelopes = sorted(envelopes)
  
    # Initialize dp[] array
    dp = [0] * N
  
    # To store the result
    max_envelope = 1
  
    dp[0] = 1
  
    # Loop through the array
    for i in range(1, N):
        dp[i] = 1
  
        # Find envelopes count for
        # each envelope
        for j in range(i):
  
            if (envelopes[i][0] > envelopes[j][0]
                and envelopes[i][1] > envelopes[j][1]
                and dp[i] < dp[j] + 1):
                dp[i] = dp[j] + 1
  
        # Store maximum envelopes count
        max_envelope = max(max_envelope, dp[i])
  
    # Return the result
    return max_envelope
  
# Driver Code
if __name__ == '__main__':
  
    # Given the envelopes
    envelopes = [ [ 4, 3 ], [ 5, 3 ], 
                [ 5, 6 ], [ 1, 2 ] ]
  
    # Function Call
    print(maxEnvelopes(envelopes))
  
# This code is contributed by Mohit Kumar

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Output: 

3

Time Complexity: O(N2)
Auxiliary Space: O(N)

Efficient Approach:To optimize the naive approach the idea is to use the concept of Binary Search and Longest Increasing Subsequence. Sorting the envelopes in the increasing order of width and the decreasing order of height if width is same, reduces the problem to finding the longest increasing sequence of height of the envelope. This approach works as width is already sorted in increasing order and only maximum increasing sequence of height is sufficient to find the maximum number of envelopes. The efficient way to find the Longest Increasing Sequence in N×log(N) approach is discussed in this article.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns the maximum
// number of envelopes that can be
// inserted into another envelopes
int maxEnvelopes(vector<vector<int> >& envelopes)
{
    // Number of envelopes
    int N = envelopes.size();
  
    if (N == 0)
        return N;
  
    // Sort the envelopes in increasing
    // order of width and decreasing order
    // of height is width is same
    sort(envelopes.begin(), envelopes.end(),
        [](vector<int>& a, vector<int>& b) {
            return a[0] < b[0]
                    or (a[0] == b[0] and a[1] > b[1]);
        });
  
    // To store the longest increasing
    // sequence of height
    vector<int> dp;
  
    // Finding LIS of the heights
    // of the envelopes
    for (int i = 0; i < N; ++i) {
        auto iter = lower_bound(dp.begin(),
                                dp.end(),
                                envelopes[i][1]);
  
        if (iter == dp.end())
            dp.push_back(envelopes[i][1]);
        else if (envelopes[i][1] < *iter)
            *iter = envelopes[i][1];
    }
  
    // Return the result
    return dp.size();
}
  
// Driver Code
int main()
{
    // Given the envelopes
    vector<vector<int> > envelopes
        = { { 4, 3 }, { 5, 3 }, { 5, 6 }, { 1, 2 } };
  
    // Function Call
    cout << maxEnvelopes(envelopes);
    return 0;
}

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Output: 

3

Time Complexity: O(N*log(N))
Auxiliary Space: O(N)

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Improved By : mohit kumar 29, offbeat