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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++

 // C++ program for the above approach#include using namespace std; // Function that returns the maximum// number of envelopes that can be// inserted into another envelopesint maxEnvelopes(vector > 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 Codeint main(){    // Given the envelopes    vector > envelopes        = { { 4, 3 }, { 5, 3 }, { 5, 6 }, { 1, 2 } };     // Function Call    cout << maxEnvelopes(envelopes);     return 0;}

Java

 // Java program for the above approachimport java.util.*;import java.lang.*; class GFG{     // Function that returns the maximum// number of envelopes that can be// inserted into another envelopesstatic 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 Codepublic 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

Python3

 # Python3 program for the above approach # Function that returns the maximum# number of envelopes that can be# inserted into another envelopesdef 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 Codeif __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

C#

 // C# program to implement above approachusing System;using System.Collections;using System.Collections.Generic; 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        Array.Sort(envelopes, new comp());                                     // 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 = new int[][]{            new int[]{ 4, 3 },            new int[]{ 5, 3 },            new int[]{ 5, 6 },            new int[]{ 1, 2 }        };                 // Function call        Console.WriteLine(maxEnvelopes(envelopes));             }} class comp : IComparer{    public int Compare(int[] a, int[] b)    {        if(a[0] != b[0]) return a[0] - b[0];        return a[1] - b[1];    }} // This code is contributed by entertain2022.

Javascript



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++

 // C++ program for the above approach#include using namespace std; // Function that returns the maximum// number of envelopes that can be// inserted into another envelopesint maxEnvelopes(vector >& 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& a, vector& 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 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 Codeint main(){    // Given the envelopes    vector > envelopes        = { { 4, 3 }, { 5, 3 }, { 5, 6 }, { 1, 2 } };     // Function Call    cout << maxEnvelopes(envelopes);    return 0;}

Java

 // Java program for the above approachimport java.io.*;import java.util.*;import java.util.Arrays;import java.util.Collections; 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 increasing    // order of width and decreasing order    // of height is width is same    Arrays.sort(envelopes,new Comparator() {      @Override      public int compare(int[] a,                         int[] b)      {        return a[0] == b[0] ? b[1] - a[1] : a[0] - b[0];;      }    });     // To store the longest increasing    // sequence of height    ArrayList dp = new ArrayList();     // Finding LIS of the heights    // of the envelopes    for (int i = 0; i < N; ++i) {      int iter = Collections.binarySearch(dp, envelopes[i][1]);      if (iter < 0)        iter=Math.abs(iter)-1;       if(iter == dp.size())        dp.add(envelopes[i][1]);      else if (envelopes[i][1] < dp.get(iter))        dp.set(iter,envelopes[i][1]);    }     // Return the result    return dp.size();  }   // 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 Aman Kumar

Python3

 # Python program for the above approachfrom bisect import bisect_left as lower_bound # Function that returns the maximum# number of envelopes that can be# inserted into another envelopesdef maxEnvelopes(envelopes):    # Number of envelopes    N = len(envelopes)         if(N == 0):        return N         # Sort the envelopes in increasing    # order of width and decreasing order    # of height is width is same    envelopes.sort()         # To store the longest increasing    # sequence of height    dp = []         # Finding LIS of the heights    # of the envelopes    for i in range(N):        iter = lower_bound(dp,envelopes[i][1])        if(iter == len(dp)):            dp.append(envelopes[i][1])        elif(envelopes[i][1] < dp[iter]):            dp[iter] = envelopes[i][1]         # Return the result    return len(dp) # Driver Code # Given the envelopesenvelopes = [[4, 3], [5, 3], [5, 6], [1, 2]] # Function Callprint(maxEnvelopes(envelopes)) # This code is contributed by Pushpesh Raj

C#

 using System;using System.Linq;using System.Collections.Generic; 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 increasing    // order of width and decreasing order    // of height is width is same    Array.Sort(envelopes, (a, b) = > a[0] - b[0]);     // To store the longest increasing    // sequence of height    List dp = new List();     // Finding LIS of the heights    // of the envelopes    for (int i = 0; i < N; ++i) {      int iter = dp.BinarySearch(envelopes[i][1]);      if (iter < 0)        iter = ~iter;       if (iter == dp.Count)        dp.Add(envelopes[i][1]);      else if (envelopes[i][1] < dp[iter])        dp[iter] = envelopes[i][1];    }     // Return the result    return dp.Count;  }   // Driver Code  static void Main(string[] args)  {    // Given the envelopes    int[][] envelopes = new int[4][];    envelopes[0] = new int[] { 4, 3 };    envelopes[1] = new int[] { 5, 3 };    envelopes[2] = new int[] { 5, 6 };    envelopes[3] = new int[] { 1, 2 };     // Function Call    Console.WriteLine(maxEnvelopes(envelopes));  }} // This code is contributed by lokeshpotta20.

Javascript

 // JavaScript program for the above approach // Function that returns the maximum// number of envelopes that can be// inserted into another envelopesfunction maxEnvelopes(envelopes) {  // Number of envelopes  let N = envelopes.length;   if (N === 0) return N;   // Sort the envelopes in increasing  // order of width and decreasing order  // of height is width is same  envelopes.sort((a, b) => {    if (a[0] === b[0]) {      return b[1] - a[1];    } else {      return a[0] - b[0];    }  });   // To store the longest increasing  // sequence of height  let dp = [];   // Finding LIS of the heights  // of the envelopes  for (let i = 0; i < N; i++) {    let iter = dp.findIndex(x => x >= envelopes[i][1]);    if (iter === -1) {      dp.push(envelopes[i][1]);    } else if (envelopes[i][1] < dp[iter]) {      dp[iter] = envelopes[i][1];    }  }   // Return the result  return dp.length;} // Driver Codelet envelopes = [[4, 3], [5, 3], [5, 6], [1, 2]]; // Function Callconsole.log(maxEnvelopes(envelopes));// this contributed by devendra

Output:

3

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