Given an array arr[] consisting of N integers, the task is to find the maximum count of decreasing subsequences possible from an array that satisfies the following conditions:
- Each subsequence is in its longest possible form.
- The difference between adjacent elements of the subsequence is always 1.
Examples:
Input: arr[] = {2, 1, 5, 4, 3}
Output: 2
Explanation:
Possible decreasing subsequences are { 5, 4, 3 } and { 2, 1 }.
Input: arr[] = {4, 5, 2, 1, 4}
Output: 3
Explanation:
Possible decreasing subsequences are { 4 }, { 5, 4} and { 2, 1}.
Approach:
The idea is to use a HashMap to solve the problem. Follow the steps below:
- Maintain a HashMap to store the count of subsequences possible for an array element and maxSubsequences to count the total number of possible subsequences.
- Traverse the array, and for each element arr[i], check if any subsequence exists which can have arr[i] as the next element, by the count assigned to arr[i] in the HashMap.
- If exists, do the following:
- Assign arr[i] as the next element of the subsequence.
- Decrease count assigned to arr[i] in the HashMap, as the number of possible subsequences with arr[i] as the next element has decreased by 1.
- Similarly, increase count assigned to arr[i] – 1 in the HashMap, as the number of possible subsequences with arr[i] – 1 as the next element has increased by 1.
- Otherwise, increase maxCount, as a new subsequence is required and repeat the above step to modify the HashMap.
- After completing the traversal of the array, print the value of maxCount.
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum number// number of required subsequencesint maxSubsequences(int arr[], int n){ // HashMap to store number of // arrows available with // height of arrow as key unordered_map<int, int> m; // Stores the maximum count // of possible subsequences int maxCount = 0; // Stores the count of // possible subsequences int count; for (int i = 0; i < n; i++) { // Check if i-th element can be // part of any of the previous // subsequence if (m.find(arr[i]) != m.end()) { // Count of subsequences // possible with arr[i] as // the next element count = m[arr[i]]; // If more than one such // subsequence exists if (count > 1) { // Include arr[i] in a subsequence m[arr[i]] = count - 1; } // Otherwise else m.erase(arr[i]); // Increase count of subsequence possible // with arr[i] - 1 as the next element if (arr[i] - 1 > 0) m[arr[i] - 1] += 1; } else { // Start a new subsequence maxCount++; // Increase count of subsequence possible // with arr[i] - 1 as the next element if (arr[i] - 1 > 0) m[arr[i] - 1] += 1; } } // Return the answer return maxCount;}// Driver Codeint main(){ int n = 5; int arr[] = { 4, 5, 2, 1, 4 }; cout << maxSubsequences(arr, n) << endl; // This code is contributed by bolliranadheer} |
Java
// Java program to implement // the above approach import java.util.*; class GFG { // Function to find the maximum number // number of required subsequences static int maxSubsequences(int arr[], int n) { // HashMap to store number of // arrows available with // height of arrow as key HashMap<Integer, Integer> map = new HashMap<>(); // Stores the maximum count // of possible subsequences int maxCount = 0; // Stores the count of // possible subsequences int count; for (int i = 0; i < n; i++) { // Check if i-th element can be // part of any of the previous // subsequence if (map.containsKey(arr[i])) { // Count of subsequences // possible with arr[i] as // the next element count = map.get(arr[i]); // If more than one such // subsequence exists if (count > 1) { // Include arr[i] in a subsequence map.put(arr[i], count - 1); } // Otherwise else map.remove(arr[i]); // Increase count of subsequence possible // with arr[i] - 1 as the next element if (arr[i] - 1 > 0) map.put(arr[i] - 1, map.getOrDefault(arr[i] - 1, 0) + 1); } else { // Start a new subsequence maxCount++; // Increase count of subsequence possible // with arr[i] - 1 as the next element if (arr[i] - 1 > 0) map.put(arr[i] - 1, map.getOrDefault(arr[i] - 1, 0) + 1); } } // Return the answer return maxCount; } // Driver Code public static void main(String[] args) { int n = 5; int arr[] = { 4, 5, 2, 1, 4 }; System.out.println(maxSubsequences(arr, n)); } } |
C#
// C# program to implement // the above approach using System;using System.Collections.Generic; class GFG{ // Function to find the maximum number // number of required subsequences static int maxSubsequences(int []arr, int n) { // Dictionary to store number of // arrows available with // height of arrow as key Dictionary<int, int> map = new Dictionary<int, int>(); // Stores the maximum count // of possible subsequences int maxCount = 0; // Stores the count of // possible subsequences int count; for(int i = 0; i < n; i++) { // Check if i-th element can be // part of any of the previous // subsequence if (map.ContainsKey(arr[i])) { // Count of subsequences // possible with arr[i] as // the next element count = map[arr[i]]; // If more than one such // subsequence exists if (count > 1) { // Include arr[i] in a subsequence map.Add(arr[i], count - 1); } // Otherwise else map.Remove(arr[i]); // Increase count of subsequence possible // with arr[i] - 1 as the next element if (arr[i] - 1 > 0) if (map.ContainsKey(arr[i] - 1)) map[arr[i] - 1]++; else map.Add(arr[i] - 1, 1); } else { // Start a new subsequence maxCount++; // Increase count of subsequence possible // with arr[i] - 1 as the next element if (arr[i] - 1 > 0) if (map.ContainsKey(arr[i] - 1)) map[arr[i] - 1]++; else map.Add(arr[i] - 1, 1); } } // Return the answer return maxCount; } // Driver Code public static void Main(String[] args) { int n = 5; int []arr = { 4, 5, 2, 1, 4 }; Console.WriteLine(maxSubsequences(arr, n)); } } // This code is contributed by Amit Katiyar |
Python3
# Python program to implement# the above approachfrom collections import defaultdict# Function to find the maximum number# number of required subsequencesdef maxSubsequences(arr, n)->int: # Dictionary to store number of # arrows available with # height of arrow as key m = defaultdict(int) # Stores the maximum count # of possible subsequences maxCount = 0 # Stores the count # of possible subsequences count = 0 for i in range(0, n): # Check if i-th element can be # part of any of the previous # subsequence if arr[i] in m.keys(): # Count of subsequences # possible with arr[i] as # the next element count = m[arr[i]] # If more than one such # subsequence exists if count > 1: # Include arr[i] in a subsequence m[arr[i]] = count - 1 # Otherwise else: m.pop(arr[i]) # Increase count of subsequence possible # with arr[i] - 1 as the next element if arr[i] - 1 > 0: m[arr[i] - 1] += 1 else: maxCount += 1 # Increase count of subsequence possible # with arr[i] - 1 as the next element if arr[i] - 1 > 0: m[arr[i] - 1] += 1 # Return the answer return maxCount# Driver Codeif __name__ == '__main__': n = 5 arr = [4, 5, 2, 1, 4] print(maxSubsequences(arr, n))# This code is contributed by Riddhi Jaiswal. |
Javascript
<script>// Javascript program to implement// the above approach// Function to find the maximum number// number of required subsequencesfunction maxSubsequences(arr, n){ // Dictionary to store number of // arrows available with // height of arrow as key let map = new Map(); // Stores the maximum count // of possible subsequences let maxCount = 0; // Stores the count of // possible subsequences let count; for(let i = 0; i < n; i++) { // Check if i-th element can be // part of any of the previous // subsequence if (map.has(arr[i])) { // Count of subsequences // possible with arr[i] as // the next element count = map[arr[i]]; // If more than one such // subsequence exists if (count > 1) { // Include arr[i] in a subsequence map.add(arr[i], count - 1); } // Otherwise else map.delete(arr[i]); // Increase count of subsequence possible // with arr[i] - 1 as the next element if (arr[i] - 1 > 0) if (map.has(arr[i] - 1)) map[arr[i] - 1]++; else map.set(arr[i] - 1, 1); } else { // Start a new subsequence maxCount++; // Increase count of subsequence possible // with arr[i] - 1 as the next element if (arr[i] - 1 > 0) if (map.has(arr[i] - 1)) map[arr[i] - 1]++; else map.set(arr[i] - 1, 1); } } // Return the answer return maxCount;}// Driver code let n = 5; let arr = [ 4, 5, 2, 1, 4 ]; document.write(maxSubsequences(arr, n)); </script> |
Output
3
Time Complexity: O(n)
Auxiliary Space: O(n)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
