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Count the number of contiguous increasing and decreasing subsequences in a sequence
• Difficulty Level : Medium
• Last Updated : 14 Feb, 2020

For a given distinct integer sequence of size N, the task is to count the number of contiguous increasing subsequence and contiguous decreasing subsequence in this sequence.

Examples:

Input: arr[] = { 80, 50, 60, 70, 40 }
Output: 1 2
Explanation:
The only one increasing subsequence is (50, 60, 70) and
two decreasing subsequences are (80, 50) and (70, 40).

Input: arr[] = { 10, 20, 23, 12, 5, 4, 61, 67, 87, 9 }
Output: 2 2
Explanation:
The increasing subsequences are (10, 20, 23) and (4, 61, 67, 87)
whereas the decreasing subsequences are (23, 12, 5, 4) and (87, 9).

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea behind solving this problem is to use two arrays which keeps the track of increasing or decreasing indices based on next elements.

1. Define two arrays max and min, such that the index of an element of the array is stored. For the first element of the array, if it is greater than its next element, it’s index is stored in the max array, else it is stored in the min array and so on.
2. For the last element of the array, if it is greater than the previous element, it’s index is stored in the max array, else it is stored in min array.
3. After this, all the maximal contiguous increasing subsequences are matched from (min to max), while all the maximal contiguous decreasing subsequences are matched from (max to min).
4. Finally, if the index of the first element of array is stored in the first index of min array, then the number of maximal contiguous increasing subsequence possible is size of min array and maximal contiguous decreasing subsequences possible is (size of max array – 1).
5. If the index of the first element of the array is stored in the first index of the max array, then the number of maximal contiguous increasing subsequence possible is(size of the max array -1) and maximal contiguous decreasing subsequences possible is the size of min array.

Below is the implementation of the above approach.

## C++

 `// C++ implementation of the above approach.`` ` `#include ``using` `namespace` `std;`` ` `// Function to count the number of maximal contiguous``// increasing and decreasing subsequences``void` `numOfSubseq(``int` `arr[], ``int` `n)`` ` `{`` ` `    ``int` `i, inc_count, dec_count;``    ``int` `max[n], min[n];`` ` `    ``// k2, k1 are used to store the``    ``// count of max and min array`` ` `    ``int` `k1 = 0, k2 = 0;`` ` `    ``// Comparison to store the index of``    ``// first element of array``    ``if` `(arr[0] < arr[1])``        ``min[k1++] = 0;``    ``else``        ``max[k2++] = 0;`` ` `    ``// Comparison to store the index``    ``// from second to second last``    ``// index of array``    ``for` `(i = 1; i < n - 1; i++) {``        ``if` `(arr[i] < arr[i - 1]``            ``&& arr[i] < arr[i + 1])``            ``min[k1++] = i;`` ` `        ``if` `(arr[i] > arr[i - 1]``            ``&& arr[i] > arr[i + 1])``            ``max[k2++] = i;``    ``}`` ` `    ``// Comparison to store the index``    ``// of last element of array``    ``if` `(arr[n - 1] < arr[n - 2])``        ``min[k1++] = n - 1;``    ``else``        ``max[k2++] = n - 1;`` ` `    ``// Count of number of maximal contiguous``    ``// increasing and decreasing subsequences``    ``if` `(min[0] == 0) {``        ``inc_count = k2;``        ``dec_count = k1 - 1;``    ``}``    ``else` `{``        ``inc_count = k2 - 1;``        ``dec_count = k1;``    ``}`` ` `    ``cout << ``"Increasing Subsequence Count: "``         ``<< inc_count << ``"\n"``;``    ``cout << ``"Decreasing Subsequence Count: "``         ``<< dec_count << ``"\n"``;``}`` ` `// Driver program to test above``int` `main()``{``    ``int` `arr[] = { 12, 8, 11, 13, 10, 15, 14, 16, 20 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``numOfSubseq(arr, n);``    ``return` `0;``}`

## Java

 `// Java implementation of the above approach. ``class` `GFG ``{``     ` `    ``// Function to count the number of maximal contiguous ``    ``// increasing and decreasing subsequences ``    ``static` `void` `numOfSubseq(``int` `arr[], ``int` `n) ``     ` `    ``{ ``     ` `        ``int` `i, inc_count, dec_count; ``        ``int` `max[] = ``new` `int``[n];``        ``int` `min[] = ``new` `int``[n]; ``     ` `        ``// k2, k1 are used to store the ``        ``// count of max and min array ``     ` `        ``int` `k1 = ``0``, k2 = ``0``; ``     ` `        ``// Comparison to store the index of ``        ``// first element of array ``        ``if` `(arr[``0``] < arr[``1``]) ``            ``min[k1++] = ``0``; ``        ``else``            ``max[k2++] = ``0``; ``     ` `        ``// Comparison to store the index ``        ``// from second to second last ``        ``// index of array ``        ``for` `(i = ``1``; i < n - ``1``; i++) ``        ``{ ``            ``if` `(arr[i] < arr[i - ``1``] ``                ``&& arr[i] < arr[i + ``1``]) ``                ``min[k1++] = i; ``     ` `            ``if` `(arr[i] > arr[i - ``1``] ``                ``&& arr[i] > arr[i + ``1``]) ``                ``max[k2++] = i; ``        ``} ``     ` `        ``// Comparison to store the index ``        ``// of last element of array ``        ``if` `(arr[n - ``1``] < arr[n - ``2``]) ``            ``min[k1++] = n - ``1``; ``        ``else``            ``max[k2++] = n - ``1``; ``     ` `        ``// Count of number of maximal contiguous ``        ``// increasing and decreasing subsequences ``        ``if` `(min[``0``] == ``0``) ``        ``{ ``            ``inc_count = k2; ``            ``dec_count = k1 - ``1``; ``        ``} ``        ``else` `        ``{ ``            ``inc_count = k2 - ``1``; ``            ``dec_count = k1; ``        ``} ``     ` `        ``System.out.println(``"Increasing Subsequence"` `+ ``                            ``" Count: "` `+ inc_count) ;``        ``System.out.println(``"Decreasing Subsequence"` `+ ``                            ``" Count: "` `+ dec_count) ; ``    ``} ``     ` `    ``// Driver code ``    ``public` `static` `void` `main (String[] args) ``    ``{ ``        ``int` `arr[] = { ``12``, ``8``, ``11``, ``13``, ``10``, ``15``, ``14``, ``16``, ``20` `}; ``        ``int` `n = arr.length; ``        ``numOfSubseq(arr, n); ``    ``} ``}`` ` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the above approach. `` ` `# Function to count the number of maximal contiguous``# increasing and decreasing subsequences``def` `numOfSubseq(arr, n):``    ``i, inc_count, dec_count ``=` `0``, ``0``, ``0``;``    ``max` `=` `[``0``]``*``n;``    ``min` `=` `[``0``]``*``n;`` ` `    ``# k2, k1 are used to store the``    ``# count of max and min array``    ``k1 ``=` `0``;``    ``k2 ``=` `0``;`` ` `    ``# Comparison to store the index of``    ``# first element of array``    ``if` `(arr[``0``] < arr[``1``]):``        ``min``[k1] ``=` `0``;``        ``k1 ``+``=` `1``;``    ``else``:``        ``max``[k2] ``=` `0``;``        ``k2 ``+``=` `1``;``     ` `    ``# Comparison to store the index``    ``# from second to second last``    ``# index of array``    ``for` `i ``in` `range``(``1``, n``-``1``):``        ``if` `(arr[i] < arr[i ``-` `1``] ``and` `arr[i] < arr[i ``+` `1``]):``            ``min``[k1] ``=` `i;``            ``k1 ``+``=` `1``;``         ` `        ``if` `(arr[i] > arr[i ``-` `1``] ``and` `arr[i] > arr[i ``+` `1``]):``            ``max``[k2] ``=` `i;``            ``k2 ``+``=` `1``;`` ` `    ``# Comparison to store the index``    ``# of last element of array``    ``if` `(arr[n ``-` `1``] < arr[n ``-` `2``]):``        ``min``[k1] ``=` `n ``-` `1``;``        ``k1 ``+``=` `1``;``    ``else``:``        ``max``[k2] ``=` `n ``-` `1``;``        ``k2 ``+``=` `1``;``     ` `    ``# Count of number of maximal contiguous``    ``# increasing and decreasing subsequences``    ``if` `(``min``[``0``] ``=``=` `0``):``        ``inc_count ``=` `k2;``        ``dec_count ``=` `k1 ``-` `1``;``    ``else``:``        ``inc_count ``=` `k2 ``-` `1``;``        ``dec_count ``=` `k1;``     ` `    ``print``(``"Increasing Subsequence Count: "` `, inc_count);``    ``print``(``"Decreasing Subsequence Count: "` `, dec_count);`` ` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[ ``12``, ``8``, ``11``, ``13``, ``10``, ``15``, ``14``, ``16``, ``20` `];``    ``n ``=` `len``(arr);``    ``numOfSubseq(arr, n);`` ` `# This code is contributed by 29AjayKumar`

## C#

 `// C# implementation of the above approach. ``using` `System;`` ` `class` `GFG``{``         ` `// Function to count the number of maximal contiguous ``// increasing and decreasing subsequences ``static` `void` `numOfSubseq(``int` `[]arr, ``int` `n) `` ` `{ `` ` `    ``int` `i, inc_count, dec_count; ``    ``int` `[]max = ``new` `int``[n];``    ``int` `[]min = ``new` `int``[n]; `` ` `    ``// k2, k1 are used to store the ``    ``// count of max and min array ``    ``int` `k1 = 0, k2 = 0; `` ` `    ``// Comparison to store the index of ``    ``// first element of array ``    ``if` `(arr[0] < arr[1]) ``        ``min[k1++] = 0; ``    ``else``        ``max[k2++] = 0; `` ` `    ``// Comparison to store the index ``    ``// from second to second last ``    ``// index of array ``    ``for` `(i = 1; i < n - 1; i++) ``    ``{ ``        ``if` `(arr[i] < arr[i - 1] ``            ``&& arr[i] < arr[i + 1]) ``            ``min[k1++] = i; `` ` `        ``if` `(arr[i] > arr[i - 1] ``            ``&& arr[i] > arr[i + 1]) ``            ``max[k2++] = i; ``    ``} `` ` `    ``// Comparison to store the index ``    ``// of last element of array ``    ``if` `(arr[n - 1] < arr[n - 2]) ``        ``min[k1++] = n - 1; ``    ``else``        ``max[k2++] = n - 1; `` ` `    ``// Count of number of maximal contiguous ``    ``// increasing and decreasing subsequences ``    ``if` `(min[0] == 0) ``    ``{ ``        ``inc_count = k2; ``        ``dec_count = k1 - 1; ``    ``} ``    ``else``    ``{ ``        ``inc_count = k2 - 1; ``        ``dec_count = k1; ``    ``} `` ` `    ``Console.WriteLine(``"Increasing Subsequence"` `+ ``                        ``" Count: "` `+ inc_count) ;``    ``Console.WriteLine(``"Decreasing Subsequence"` `+ ``                        ``" Count: "` `+ dec_count) ; ``} ``     ` `// Driver code ``static` `public` `void` `Main ()``{``    ``int` `[]arr = { 12, 8, 11, 13, 10, 15, 14, 16, 20 }; ``    ``int` `n = arr.Length; ``    ``numOfSubseq(arr, n); ``} ``}`` ` `// This code is contributed by ajit.`
Output:
```Increasing Subsequence Count: 3
Decreasing Subsequence Count: 3
```

Time Complexity: O(N)

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