Maximize count of Decreasing Consecutive Subsequences from an Array

Namachila November 19, 2025

0 0 6 minutes read

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 subsequences
int 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 Code
int 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 approach
 
from collections import defaultdict
 
# Function to find the maximum number
# number of required subsequences
def 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 Code
if __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 subsequences
function 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

Time Complexity: O(n)
Auxiliary Space: O(n)

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