Given an array of n positive integers. We need to find the largest increasing sequence of consecutive positive integers.
Examples:
Input : arr[] = {5, 7, 6, 7, 8}
Output : Size of LIS = 4
LIS = 5, 6, 7, 8
Input : arr[] = {5, 7, 8, 7, 5}
Output : Size of LIS = 2
LIS = 7, 8
This problem can be solved easily by the concept of LIS where each next greater element differ from earlier one by 1. But this will take O(n^2) time complexity.
With the use of hashing we can finding the size of longest increasing sequence with consecutive integers in time complexity of O(n).
We create a hash table.. Now for each element arr[i], we perform hash[arr[i]] = hash[arr[i] – 1] + 1. So, for every element we know longest consecutive increasing subsequence ending with it. Finally we return maximum value from hash table.
Implementation:
C++
// C++ implementation of longest continuous increasing// subsequence#include <bits/stdc++.h>using namespace std;// Function for LISint findLIS(int A[], int n){ unordered_map<int, int> hash; // Initialize result int LIS_size = 1; int LIS_index = 0; hash[A[0]] = 1; // iterate through array and find // end index of LIS and its Size for (int i = 1; i < n; i++) { hash[A[i]] = hash[A[i] - 1] + 1; if (LIS_size < hash[A[i]]) { LIS_size = hash[A[i]]; LIS_index = A[i]; } } // print LIS size cout << "LIS_size = " << LIS_size << "\n"; // print LIS after setting start element cout << "LIS : "; int start = LIS_index - LIS_size + 1; while (start <= LIS_index) { cout << start << " "; start++; }}// driverint main(){ int A[] = { 2, 5, 3, 7, 4, 8, 5, 13, 6 }; int n = sizeof(A) / sizeof(A[0]); findLIS(A, n); return 0;} |
Java
// Java implementation of longest continuous increasing// subsequenceimport java.util.*;class GFG {// Function for LISstatic void findLIS(int A[], int n){ Map<Integer, Integer> hash = new HashMap<Integer, Integer>(); // Initialize result int LIS_size = 1; int LIS_index = 0; hash.put(A[0], 1); // iterate through array and find // end index of LIS and its Size for (int i = 1; i < n; i++) { hash.put(A[i], hash.get(A[i] - 1)==null? 1:hash.get(A[i] - 1)+1); if (LIS_size < hash.get(A[i])) { LIS_size = hash.get(A[i]); LIS_index = A[i]; } } // print LIS size System.out.println("LIS_size = " + LIS_size); // print LIS after setting start element System.out.print("LIS : "); int start = LIS_index - LIS_size + 1; while (start <= LIS_index) { System.out.print(start + " "); start++; }}// Driver codepublic static void main(String[] args){ int A[] = { 2, 5, 3, 7, 4, 8, 5, 13, 6 }; int n = A.length; findLIS(A, n);}}// This code is contributed by Princi Singh |
Python3
# Python3 implementation of longest # continuous increasing subsequence# Function for LISdef findLIS(A, n): hash = dict() # Initialize result LIS_size, LIS_index = 1, 0 hash[A[0]] = 1 # iterate through array and find # end index of LIS and its Size for i in range(1, n): # If the desired key is not present # in dictionary, it will throw key error, # to avoid this error this is necessary if A[i] - 1 not in hash: hash[A[i] - 1] = 0 hash[A[i]] = hash[A[i] - 1] + 1 if LIS_size < hash[A[i]]: LIS_size = hash[A[i]] LIS_index = A[i] # print LIS size print("LIS_size =", LIS_size) # print LIS after setting start element print("LIS : ", end = "") start = LIS_index - LIS_size + 1 while start <= LIS_index: print(start, end = " ") start += 1# Driver Codeif __name__ == "__main__": A = [ 2, 5, 3, 7, 4, 8, 5, 13, 6 ] n = len(A) findLIS(A, n)# This code is contributed by sanjeev2552 |
C#
// C# implementation of longest continuous increasing// subsequenceusing System;using System.Collections.Generic; class GFG {// Function for LISstatic void findLIS(int []A, int n){ Dictionary<int,int> hash = new Dictionary<int,int>(); // Initialize result int LIS_size = 1; int LIS_index = 0; hash.Add(A[0], 1); // iterate through array and find // end index of LIS and its Size for (int i = 1; i < n; i++) { if(hash.ContainsKey(A[i]-1)) { var val = hash[A[i]-1]; hash.Remove(A[i]); hash.Add(A[i], val + 1); } else { hash.Add(A[i], 1); } if (LIS_size < hash[A[i]]) { LIS_size = hash[A[i]]; LIS_index = A[i]; } } // print LIS size Console.WriteLine("LIS_size = " + LIS_size); // print LIS after setting start element Console.Write("LIS : "); int start = LIS_index - LIS_size + 1; while (start <= LIS_index) { Console.Write(start + " "); start++; }}// Driver codepublic static void Main(String[] args){ int []A = { 2, 5, 3, 7, 4, 8, 5, 13, 6 }; int n = A.Length; findLIS(A, n);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript implementation of longest continuous increasing// subsequence// Function for LISfunction findLIS(A, n) { let hash = new Map(); // Initialize result let LIS_size = 1; let LIS_index = 0; hash.set(A[0], 1); // iterate through array and find // end index of LIS and its Size for (let i = 1; i < n; i++) { hash.set(A[i], hash.get(A[i] - 1) == null ? 1 : hash.get(A[i] - 1) + 1); if (LIS_size < hash.get(A[i])) { LIS_size = hash.get(A[i]); LIS_index = A[i]; } } // print LIS size document.write("LIS_size = " + LIS_size + "<br>"); // print LIS after setting start element document.write("LIS : "); let start = LIS_index - LIS_size + 1; while (start <= LIS_index) { document.write(start + " "); start++; }}// Driver codelet A = [2, 5, 3, 7, 4, 8, 5, 13, 6];let n = A.length;findLIS(A, n);// This code is contributed by gfgking</script> |
Output:
LIS_size = 5 LIS : 2 3 4 5 6
Time Complexity : O(n)
Auxiliary Space: O(n)
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