# Maximize count of Decreasing Consecutive Subsequences from an Array

• Last Updated : 07 Oct, 2021

Given an array arr[] consisting of N integers, the task is to find the maximum count of decreasing subsequences possible from an array which satisfies the following conditions:

• Each subsequence is in its longest possible form.
• The difference between adjacent elements of the subsequence is always 1.

Examples:

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Input: arr[] = {2, 1, 5, 4, 3}
Output:
Explanation:
Possible decreasing subsequences are { 5, 4, 3 } and { 2, 1 }.
Input: arr[] = {4, 5, 2, 1, 4}
Output:
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 ``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 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

 ``
Output
`3`

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