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# Longest alternating subsequence

A sequence {X1, X2, .. Xn} is an alternating sequence if its elements satisfy one of the following relations :

X1 < X2 > X3 < X4 > X5 < …. xn or
X1 > X2 < X3 > X4 < X5 > …. xn

Examples:

Input: arr[] = {1, 5, 4}
Output: 3
Explanation: The whole arrays is of the form  x1 < x2 > x3

Input: arr[] = {10, 22, 9, 33, 49, 50, 31, 60}
Output: 6
Explanation: The subsequences {10, 22, 9, 33, 31, 60} or
{10, 22, 9, 49, 31, 60} or {10, 22, 9, 50, 31, 60}
are longest subsequence of length 6

Note: This problem is an extension of the longest increasing subsequence problem, but requires more thinking for finding optimal substructure property in this

## Longest alternating subsequence using dynamic programming:

To solve the problem follow the below idea:

We will solve this problem by dynamic Programming method, as it has optimal substructure and overlapping subproblems

Follow the below steps to solve the problem:

• Let A is given an array of length N
• We define a 2D array las[n][2] such that las[i][0] contains the longest alternating subsequence ending at index i and the last element is greater than its previous element
• las[i][1] contains the longest alternating subsequence ending at index i and the last element is smaller than its previous element, then we have the following recurrence relation between them,

las[i][0] = Length of the longest alternating subsequence
ending at index i and last element is greater
than its previous element

las[i][1] = Length of the longest alternating subsequence
ending at index i and last element is smaller
than its previous element

Recursive Formulation:

las[i][0] = max (las[i][0], las[j][1] + 1);
for all j < i and A[j] < A[i]

las[i][1] = max (las[i][1], las[j][0] + 1);
for all j < i and A[j] > A[i]

• The first recurrence relation is based on the fact that, If we are at position i and this element has to be bigger than its previous element then for this sequence (upto i) to be bigger we will try to choose an element j ( < i) such that A[j] < A[i] i.e. A[j] can become A[i]’s previous element and las[j][1] + 1 is bigger than las[i][0] then we will update las[i][0].
• Remember we have chosen las[j][1] + 1 not las[j][0] + 1 to satisfy the alternate property because in las[j][0] the last element is bigger than its previous one and A[i] is greater than A[j] which will break the alternating property if we update. So above fact derives the first recurrence relation, a similar argument can be made for the second recurrence relation also.

Below is the implementation of the above approach:

## C++

 `// C++ program to find longest alternating` `// subsequence in an array` `#include ` `using` `namespace` `std;`   `// Function to return max of two numbers` `int` `max(``int` `a, ``int` `b) { ``return` `(a > b) ? a : b; }`   `// Function to return longest alternating` `// subsequence length` `int` `zzis(``int` `arr[], ``int` `n)` `{`   `    ``/*las[i][0] = Length of the longest` `        ``alternating subsequence ending at` `        ``index i and last element is greater` `        ``than its previous element` `    ``las[i][1] = Length of the longest` `        ``alternating subsequence ending` `        ``at index i and last element is` `        ``smaller than its previous element */` `    ``int` `las[n][2];`   `    ``// Initialize all values from 1` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``las[i][0] = las[i][1] = 1;`   `    ``// Initialize result` `    ``int` `res = 1;`   `    ``// Compute values in bottom up manner` `    ``for` `(``int` `i = 1; i < n; i++) {`   `        ``// Consider all elements as` `        ``// previous of arr[i]` `        ``for` `(``int` `j = 0; j < i; j++) {`   `            ``// If arr[i] is greater, then` `            ``// check with las[j][1]` `            ``if` `(arr[j] < arr[i]` `                ``&& las[i][0] < las[j][1] + 1)` `                ``las[i][0] = las[j][1] + 1;`   `            ``// If arr[i] is smaller, then` `            ``// check with las[j][0]` `            ``if` `(arr[j] > arr[i]` `                ``&& las[i][1] < las[j][0] + 1)` `                ``las[i][1] = las[j][0] + 1;` `        ``}`   `        ``// Pick maximum of both values at index i` `        ``if` `(res < max(las[i][0], las[i][1]))` `            ``res = max(las[i][0], las[i][1]);` `    ``}` `    ``return` `res;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 10, 22, 9, 33, 49, 50, 31, 60 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``cout << ``"Length of Longest alternating "` `         ``<< ``"subsequence is "` `<< zzis(arr, n);`   `    ``return` `0;` `}`   `// This code is contributed by shivanisinghss2110`

## C

 `// C program to find longest alternating subsequence in` `// an array` `#include ` `#include `   `// function to return max of two numbers` `int` `max(``int` `a, ``int` `b) { ``return` `(a > b) ? a : b; }`   `// Function to return longest alternating subsequence length` `int` `zzis(``int` `arr[], ``int` `n)` `{` `    ``/*las[i][0] = Length of the longest alternating` `     ``subsequence ending at index i and last element is` `     ``greater than its previous element las[i][1] = Length of` `     ``the longest alternating subsequence ending at index i` `     ``and last element is smaller than its previous element` `   ``*/` `    ``int` `las[n][2];`   `    ``/* Initialize all values from 1  */` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``las[i][0] = las[i][1] = 1;`   `    ``int` `res = 1; ``// Initialize result`   `    ``/* Compute values in bottom up manner */` `    ``for` `(``int` `i = 1; i < n; i++) {` `        ``// Consider all elements as previous of arr[i]` `        ``for` `(``int` `j = 0; j < i; j++) {` `            ``// If arr[i] is greater, then check with` `            ``// las[j][1]` `            ``if` `(arr[j] < arr[i]` `                ``&& las[i][0] < las[j][1] + 1)` `                ``las[i][0] = las[j][1] + 1;`   `            ``// If arr[i] is smaller, then check with` `            ``// las[j][0]` `            ``if` `(arr[j] > arr[i]` `                ``&& las[i][1] < las[j][0] + 1)` `                ``las[i][1] = las[j][0] + 1;` `        ``}`   `        ``/* Pick maximum of both values at index i  */` `        ``if` `(res < max(las[i][0], las[i][1]))` `            ``res = max(las[i][0], las[i][1]);` `    ``}`   `    ``return` `res;` `}`   `/* Driver code */` `int` `main()` `{` `    ``int` `arr[] = { 10, 22, 9, 33, 49, 50, 31, 60 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``printf``(` `        ``"Length of Longest alternating subsequence is %d\n"``,` `        ``zzis(arr, n));` `    ``return` `0;` `}`

## Java

 `// Java program to find longest` `// alternating subsequence in an array` `import` `java.io.*;`   `class` `GFG {`   `    ``// Function to return longest` `    ``// alternating subsequence length` `    ``static` `int` `zzis(``int` `arr[], ``int` `n)` `    ``{` `        ``/*las[i][0] = Length of the longest` `            ``alternating subsequence ending at` `            ``index i and last element is` `            ``greater than its previous element` `        ``las[i][1] = Length of the longest` `            ``alternating subsequence ending at` `            ``index i and last element is` `            ``smaller than its previous` `            ``element */` `        ``int` `las[][] = ``new` `int``[n][``2``];`   `        ``/* Initialize all values from 1 */` `        ``for` `(``int` `i = ``0``; i < n; i++)` `            ``las[i][``0``] = las[i][``1``] = ``1``;`   `        ``int` `res = ``1``; ``// Initialize result`   `        ``/* Compute values in bottom up manner */` `        ``for` `(``int` `i = ``1``; i < n; i++) {` `            ``// Consider all elements as` `            ``// previous of arr[i]` `            ``for` `(``int` `j = ``0``; j < i; j++) {` `                ``// If arr[i] is greater, then` `                ``// check with las[j][1]` `                ``if` `(arr[j] < arr[i]` `                    ``&& las[i][``0``] < las[j][``1``] + ``1``)` `                    ``las[i][``0``] = las[j][``1``] + ``1``;`   `                ``// If arr[i] is smaller, then` `                ``// check with las[j][0]` `                ``if` `(arr[j] > arr[i]` `                    ``&& las[i][``1``] < las[j][``0``] + ``1``)` `                    ``las[i][``1``] = las[j][``0``] + ``1``;` `            ``}`   `            ``/* Pick maximum of both values at` `            ``index i */` `            ``if` `(res < Math.max(las[i][``0``], las[i][``1``]))` `                ``res = Math.max(las[i][``0``], las[i][``1``]);` `        ``}`   `        ``return` `res;` `    ``}`   `    ``/* Driver code*/` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `arr[] = { ``10``, ``22``, ``9``, ``33``, ``49``, ``50``, ``31``, ``60` `};` `        ``int` `n = arr.length;` `        ``System.out.println(``"Length of Longest "` `                           ``+ ``"alternating subsequence is "` `                           ``+ zzis(arr, n));` `    ``}` `}` `// This code is contributed by Prerna Saini`

## Python3

 `# Python3 program to find longest` `# alternating subsequence in an array`   `# Function to return max of two numbers`     `def` `Max``(a, b):`   `    ``if` `a > b:` `        ``return` `a` `    ``else``:` `        ``return` `b`   `# Function to return longest alternating` `# subsequence length`     `def` `zzis(arr, n):` `    ``"""las[i][0] = Length of the longest ` `        ``alternating subsequence ending at` `        ``index i and last element is greater` `        ``than its previous element` `    ``las[i][1] = Length of the longest ` `        ``alternating subsequence ending ` `        ``at index i and last element is` `        ``smaller than its previous element"""` `    ``las ``=` `[[``0` `for` `i ``in` `range``(``2``)]` `           ``for` `j ``in` `range``(n)]`   `    ``# Initialize all values from 1` `    ``for` `i ``in` `range``(n):` `        ``las[i][``0``], las[i][``1``] ``=` `1``, ``1`   `    ``# Initialize result` `    ``res ``=` `1`   `    ``# Compute values in bottom up manner` `    ``for` `i ``in` `range``(``1``, n):`   `        ``# Consider all elements as` `        ``# previous of arr[i]` `        ``for` `j ``in` `range``(``0``, i):`   `            ``# If arr[i] is greater, then` `            ``# check with las[j][1]` `            ``if` `(arr[j] < arr[i] ``and` `                    ``las[i][``0``] < las[j][``1``] ``+` `1``):` `                ``las[i][``0``] ``=` `las[j][``1``] ``+` `1`   `            ``# If arr[i] is smaller, then` `            ``# check with las[j][0]` `            ``if``(arr[j] > arr[i] ``and` `               ``las[i][``1``] < las[j][``0``] ``+` `1``):` `                ``las[i][``1``] ``=` `las[j][``0``] ``+` `1`   `        ``# Pick maximum of both values at index i` `        ``if` `(res < ``max``(las[i][``0``], las[i][``1``])):` `            ``res ``=` `max``(las[i][``0``], las[i][``1``])`   `    ``return` `res`     `# Driver Code` `arr ``=` `[``10``, ``22``, ``9``, ``33``, ``49``, ``50``, ``31``, ``60``]` `n ``=` `len``(arr)`   `print``(``"Length of Longest alternating subsequence is"``,` `      ``zzis(arr, n))`   `# This code is contributed by divyesh072019`

## C#

 `// C# program to find longest` `// alternating subsequence` `// in an array` `using` `System;`   `class` `GFG {`   `    ``// Function to return longest` `    ``// alternating subsequence length` `    ``static` `int` `zzis(``int``[] arr, ``int` `n)` `    ``{` `        ``/*las[i][0] = Length of the` `            ``longest alternating subsequence` `            ``ending at index i and last` `            ``element is greater than its` `            ``previous element` `        ``las[i][1] = Length of the longest` `            ``alternating subsequence ending at` `            ``index i and last element is` `            ``smaller than its previous` `            ``element */` `        ``int``[, ] las = ``new` `int``[n, 2];`   `        ``/* Initialize all values from 1 */` `        ``for` `(``int` `i = 0; i < n; i++)` `            ``las[i, 0] = las[i, 1] = 1;`   `        ``// Initialize result` `        ``int` `res = 1;`   `        ``/* Compute values in` `        ``bottom up manner */` `        ``for` `(``int` `i = 1; i < n; i++) {` `            ``// Consider all elements as` `            ``// previous of arr[i]` `            ``for` `(``int` `j = 0; j < i; j++) {` `                ``// If arr[i] is greater, then` `                ``// check with las[j][1]` `                ``if` `(arr[j] < arr[i]` `                    ``&& las[i, 0] < las[j, 1] + 1)` `                    ``las[i, 0] = las[j, 1] + 1;`   `                ``// If arr[i] is smaller, then` `                ``// check with las[j][0]` `                ``if` `(arr[j] > arr[i]` `                    ``&& las[i, 1] < las[j, 0] + 1)` `                    ``las[i, 1] = las[j, 0] + 1;` `            ``}`   `            ``/* Pick maximum of both` `            ``values at index i */` `            ``if` `(res < Math.Max(las[i, 0], las[i, 1]))` `                ``res = Math.Max(las[i, 0], las[i, 1]);` `        ``}`   `        ``return` `res;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr = { 10, 22, 9, 33, 49, 50, 31, 60 };` `        ``int` `n = arr.Length;` `        ``Console.WriteLine(``"Length of Longest "` `                          ``+ ``"alternating subsequence is "` `                          ``+ zzis(arr, n));` `    ``}` `}`   `// This code is contributed by anuj_67.`

## PHP

 ` ``\$arr``[``\$i``] ``and` `               ``\$las``[``\$i``][1] < ``\$las``[``\$j``][0] + 1)` `                ``\$las``[``\$i``][1] = ``\$las``[``\$j``][0] + 1;` `        ``}`   `        ``/* Pick maximum of both` `        ``values at index i */` `        ``if` `(``\$res` `< max(``\$las``[``\$i``][0], ``\$las``[``\$i``][1]))` `            ``\$res` `= max(``\$las``[``\$i``][0], ``\$las``[``\$i``][1]);` `    ``}`   `    ``return` `\$res``;` `}`   `// Driver Code` `\$arr` `= ``array``(10, 22, 9, 33, ` `             ``49, 50, 31, 60 );` `\$n` `= ``count``(``\$arr``);` `echo` `"Length of Longest alternating "` `.` `    ``"subsequence is "``, zzis(``\$arr``, ``\$n``) ;`   `// This code is contributed by anuj_67.` `?>`

## Javascript

 ``

Output

`Length of Longest alternating subsequence is 6`

Time Complexity: O(N2
Auxiliary Space: O(N), since N extra space has been taken

Efficient Approach: To solve the problem follow the below idea:

In the above approach, at any moment we are keeping track of two values (The length of the longest alternating subsequence ending at index i, and the last element is smaller than or greater than the previous element), for every element on the array. To optimize space, we only need to store two variables for element at any index i

inc = Length of longest alternative subsequence so far with current value being greater than it’s previous value.
dec = Length of longest alternative subsequence so far with current value being smaller than it’s previous value.
The tricky part of this approach is to update these two values.

“inc” should be increased, if and only if the last element in the alternative sequence was smaller than it’s previous element.
“dec” should be increased, if and only if the last element in the alternative sequence was greater than it’s previous element.

Follow the below steps to solve the problem:

• Declare two integers inc and dec equal to one
• Run a loop for i [1, N-1]
• If arr[i] is greater than the previous element then set inc equal to dec + 1
• Else if arr[i] is smaller than the previous element then set dec equal to inc + 1
• Return maximum of inc and dec

Below is the implementation of the above approach:

## C++

 `// C++ program for above approach` `#include ` `using` `namespace` `std;`   `// Function for finding` `// longest alternating` `// subsequence` `int` `LAS(``int` `arr[], ``int` `n)` `{`   `    ``// "inc" and "dec" initialized as 1` `    ``// as single element is still LAS` `    ``int` `inc = 1;` `    ``int` `dec = 1;`   `    ``// Iterate from second element` `    ``for` `(``int` `i = 1; i < n; i++) {`   `        ``if` `(arr[i] > arr[i - 1]) {`   `            ``// "inc" changes if "dec"` `            ``// changes` `            ``inc = dec + 1;` `        ``}`   `        ``else` `if` `(arr[i] < arr[i - 1]) {`   `            ``// "dec" changes if "inc"` `            ``// changes` `            ``dec = inc + 1;` `        ``}` `    ``}`   `    ``// Return the maximum length` `    ``return` `max(inc, dec);` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 10, 22, 9, 33, 49, 50, 31, 60 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``// Function Call` `    ``cout << LAS(arr, n) << endl;` `    ``return` `0;` `}`

## Java

 `// Java Program for above approach` `public` `class` `GFG {`   `    ``// Function for finding` `    ``// longest alternating` `    ``// subsequence` `    ``static` `int` `LAS(``int``[] arr, ``int` `n)` `    ``{`   `        ``// "inc" and "dec" initialized as 1,` `        ``// as single element is still LAS` `        ``int` `inc = ``1``;` `        ``int` `dec = ``1``;`   `        ``// Iterate from second element` `        ``for` `(``int` `i = ``1``; i < n; i++) {`   `            ``if` `(arr[i] > arr[i - ``1``]) {` `                ``// "inc" changes if "dec"` `                ``// changes` `                ``inc = dec + ``1``;` `            ``}` `            ``else` `if` `(arr[i] < arr[i - ``1``]) {`   `                ``// "dec" changes if "inc"` `                ``// changes` `                ``dec = inc + ``1``;` `            ``}` `        ``}`   `        ``// Return the maximum length` `        ``return` `Math.max(inc, dec);` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int``[] arr = { ``10``, ``22``, ``9``, ``33``, ``49``, ``50``, ``31``, ``60` `};` `        ``int` `n = arr.length;`   `        ``// Function Call` `        ``System.out.println(LAS(arr, n));` `    ``}` `}`

## Python3

 `# Python3 program for above approach` `def` `LAS(arr, n):`   `    ``# "inc" and "dec" initialized as 1` `    ``# as single element is still LAS` `    ``inc ``=` `1` `    ``dec ``=` `1`   `    ``# Iterate from second element` `    ``for` `i ``in` `range``(``1``, n):`   `        ``if` `(arr[i] > arr[i``-``1``]):`   `            ``# "inc" changes if "dec"` `            ``# changes` `            ``inc ``=` `dec ``+` `1` `        ``elif` `(arr[i] < arr[i``-``1``]):`   `            ``# "dec" changes if "inc"` `            ``# changes` `            ``dec ``=` `inc ``+` `1`   `    ``# Return the maximum length` `    ``return` `max``(inc, dec)`     `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[``10``, ``22``, ``9``, ``33``, ``49``, ``50``, ``31``, ``60``]` `    ``n ``=` `len``(arr)`   `    ``# Function Call` `    ``print``(LAS(arr, n))`

## C#

 `// C# program for above approach` `using` `System;`   `class` `GFG {`   `    ``// Function for finding` `    ``// longest alternating` `    ``// subsequence` `    ``static` `int` `LAS(``int``[] arr, ``int` `n)` `    ``{`   `        ``// "inc" and "dec" initialized as 1,` `        ``// as single element is still LAS` `        ``int` `inc = 1;` `        ``int` `dec = 1;`   `        ``// Iterate from second element` `        ``for` `(``int` `i = 1; i < n; i++) {` `            ``if` `(arr[i] > arr[i - 1]) {`   `                ``// "inc" changes if "dec"` `                ``// changes` `                ``inc = dec + 1;` `            ``}` `            ``else` `if` `(arr[i] < arr[i - 1]) {`   `                ``// "dec" changes if "inc"` `                ``// changes` `                ``dec = inc + 1;` `            ``}` `        ``}`   `        ``// Return the maximum length` `        ``return` `Math.Max(inc, dec);` `    ``}`   `    ``// Driver code` `    ``static` `void` `Main()` `    ``{` `        ``int``[] arr = { 10, 22, 9, 33, 49, 50, 31, 60 };` `        ``int` `n = arr.Length;`   `        ``// Function Call` `        ``Console.WriteLine(LAS(arr, n));` `    ``}` `}`   `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``

Output:

`6`

Time Complexity: O(N)
Auxiliary Space: O(1)