The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order.
Examples:
Input: [10, 22, 9, 33, 21, 50, 41, 60, 80]
Output: [10, 22, 33, 50, 60, 80] OR [10 22 33 41 60 80] or any other LIS of same length.
In the previous post, we have discussed The Longest Increasing Subsequence problem. However, the post only covered code related to the querying size of LIS, but not the construction of LIS. In this post, we will discuss how to print LIS using a similar DP solution discussed earlier.
Let arr[0..n-1] be the input array. We define vector L such that L[i] is itself is a vector that stores LIS of arr that ends with arr[i]. For example, for array [3, 2, 6, 4, 5, 1],
L[0]: 3 L[1]: 2 L[2]: 2 6 L[3]: 2 4 L[4]: 2 4 5 L[5]: 1
Therefore, for index i, L[i] can be recursively written as –
L[0] = {arr[O]}
L[i] = {Max(L[j])} + arr[i]
where j < i and arr[j] < arr[i] and if there is no such j then L[i] = arr[i]
Below is the implementation of the above idea –
C++
/* Dynamic Programming solution to construct Longest Increasing Subsequence */#include <iostream>#include <vector>using namespace std;// Utility function to print LISvoid printLIS(vector<int>& arr){ for (int x : arr) cout << x << " "; cout << endl;}// Function to construct and print Longest Increasing// Subsequencevoid constructPrintLIS(int arr[], int n){ // L[i] - The longest increasing sub-sequence // ends with arr[i] vector<vector<int> > L(n); // L[0] is equal to arr[0] L[0].push_back(arr[0]); // start from index 1 for (int i = 1; i < n; i++) { // do for every j less than i for (int j = 0; j < i; j++) { /* L[i] = {Max(L[j])} + arr[i] where j < i and arr[j] < arr[i] */ if ((arr[i] > arr[j]) && (L[i].size() < L[j].size() + 1)) L[i] = L[j]; } // L[i] ends with arr[i] L[i].push_back(arr[i]); } // L[i] now stores increasing sub-sequence of // arr[0..i] that ends with arr[i] vector<int> max = L[0]; // LIS will be max of all increasing sub- // sequences of arr for (vector<int> x : L) if (x.size() > max.size()) max = x; // max will contain LIS printLIS(max);}// Driver functionint main(){ int arr[] = { 3, 2, 6, 4, 5, 1 }; int n = sizeof(arr) / sizeof(arr[0]); // construct and print LIS of arr constructPrintLIS(arr, n); return 0;} |
Java
// Java program for // the above approach// Dynamic Programming // solution to construct Longest// Increasing Subsequenceimport java.util.*;class GFG{// Utility function to print LISstatic void printLIS(Vector<Integer> arr){ for (int x : arr) System.out.print(x + " "); System.out.println();}// Function to construct and print // Longest Increasing Subsequencestatic void constructPrintLIS(int arr[], int n){ // L[i] - The longest increasing // sub-sequence ends with arr[i] Vector<Integer> L[] = new Vector[n]; for (int i = 0; i < L.length; i++) L[i] = new Vector<Integer>(); // L[0] is equal to arr[0] L[0].add(arr[0]); // Start from index 1 for (int i = 1; i < n; i++) { // Do for every j less than i for (int j = 0; j < i; j++) { //L[i] = {Max(L[j])} + arr[i] // where j < i and arr[j] < arr[i] if ((arr[i] > arr[j]) && (L[i].size() < L[j].size() + 1)) L[i] = (Vector<Integer>) L[j].clone(); //deep copy } // L[i] ends with arr[i] L[i].add(arr[i]); } // L[i] now stores increasing sub-sequence of // arr[0..i] that ends with arr[i] Vector<Integer> max = L[0]; // LIS will be max of all increasing sub- // sequences of arr for (Vector<Integer> x : L) if (x.size() > max.size()) max = x; // max will contain LIS printLIS(max);}// Driver functionpublic static void main(String[] args){ int arr[] = {3, 2, 4, 5, 1}; int n = arr.length; // print LIS of arr constructPrintLIS(arr, n);}}// This code is contributed by gauravrajput1 |
Python3
# Dynamic Programming solution to construct Longest# Increasing Subsequence# Utility function to print LISdef printLIS(arr: list): for x in arr: print(x, end=" ") print()# Function to construct and print Longest Increasing# Subsequencedef constructPrintLIS(arr: list, n: int): # L[i] - The longest increasing sub-sequence # ends with arr[i] l = [[] for i in range(n)] # L[0] is equal to arr[0] l[0].append(arr[0]) # start from index 1 for i in range(1, n): # do for every j less than i for j in range(i): # L[i] = {Max(L[j])} + arr[i] # where j < i and arr[j] < arr[i] if arr[i] > arr[j] and (len(l[i]) < len(l[j]) + 1): l[i] = l[j].copy() # L[i] ends with arr[i] l[i].append(arr[i]) # L[i] now stores increasing sub-sequence of # arr[0..i] that ends with arr[i] maxx = l[0] # LIS will be max of all increasing sub- # sequences of arr for x in l: if len(x) > len(maxx): maxx = x # max will contain LIS printLIS(maxx)# Driver Codeif __name__ == "__main__": arr = [3, 2, 6, 4, 5, 1] n = len(arr) # construct and print LIS of arr constructPrintLIS(arr, n)# This code is contributed by# sanjeev2552 |
C#
// Dynamic Programming solution to construct Longest// Increasing Subsequenceusing System;using System.Collections.Generic; class GFG { // Utility function to print LIS static void printLIS(List<int> arr) { foreach(int x in arr) { Console.Write(x + " "); } Console.WriteLine(); } // Function to construct and print Longest Increasing // Subsequence static void constructPrintLIS(int[] arr, int n) { // L[i] - The longest increasing sub-sequence // ends with arr[i] List<List<int>> L = new List<List<int>>(); for(int i = 0; i < n; i++) { L.Add(new List<int>()); } // L[0] is equal to arr[0] L[0].Add(arr[0]); // start from index 1 for (int i = 1; i < n; i++) { // do for every j less than i for (int j = 0; j < i; j++) { /* L[i] = {Max(L[j])} + arr[i] where j < i and arr[j] < arr[i] */ if ((arr[i] > arr[j]) && (L[i].Count < L[j].Count + 1)) L[i] = L[j]; } // L[i] ends with arr[i] L[i].Add(arr[i]); } // L[i] now stores increasing sub-sequence of // arr[0..i] that ends with arr[i] List<int> max = L[0]; // LIS will be max of all increasing sub- // sequences of arr foreach(List<int> x in L) { if (x.Count > max.Count) { max = x; } } // max will contain LIS printLIS(max); } // Driver code static void Main() { int[] arr = { 3, 2, 4, 5, 1 }; int n = arr.Length; // construct and print LIS of arr constructPrintLIS(arr, n); }}// This code is contributed by divyesh072019 |
Javascript
<script>// JavaScript program Dynamic Programming solution to construct Longest// Increasing Subsequence function printLIS( x){ document.write(x+" ");}// Function to construct and print Longest Increasing// Subsequencefunction constructPrintLIS( arr, n){ // L[i] - The longest increasing sub-sequence // ends with arr[i] var L = new Array(n); for(var i = 0; i < n;i++){ L[i] = []; } // L[0] is equal to arr[0] L[0].push(arr[0]); // start from index 1 for (var i = 1; i < n; i++) { // do for every j less than i for (var j = 0; j < i; j++) { /* L[i] = {Max(L[j])} + arr[i] where j < i and arr[j] < arr[i] */ if ((arr[i] > arr[j]) && (L[i].length< L[j].length + 1)) L[i] = [...L[j]]; } // L[i] ends with arr[i] L[i].push(arr[i]); } // L[i] now stores increasing sub-sequence of // arr[0..i] that ends with arr[i] var max = 0; // LIS will be max of all increasing sub- // sequences of arr for (var x=0; x < L.length;x++){ if (L[x].length> L[max].length) max = x;} // L[max] will contain LIS L[max].forEach(printLIS); }var arr = [3, 2, 6, 4, 5, 1 ];var n = 6;// construct and print LIS of arrconstructPrintLIS(arr, n);</script> |
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Note that the time complexity of the above Dynamic Programming (DP) solution is O(n^3) (n^2 for two nested loops and n for copying another vector in a vector eg: L[i] = L[j] contributes O(n) also) and space complexity is O(n^2) as we are using 2d vector to store our LIS and there is a O(n Log n) non-DP solution for the LIS problem. See below post for O(n Log n) solution.
Construction of Longest Monotonically Increasing Subsequence (N log N)
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