Find the Longest Increasing Subsequence in Circular manner

Given an array, the task is to find to LIS (Longest Increasing Subsequence) in a circular way.

Examples :

Input : arr[] = {5, 4, 3, 2, 1}
Output : 2
Although there is no LIS in a given array
but in a circular form there can be
{1, 5}, {2, 5}, ......

Input : arr[]= {5, 6, 7, 1, 2, 3}
Output : 6
{1, 2, 3, 5, 6, 7} will be the LIS in the
circular manner.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

1. Append the same elements(i.e. whole array) with the given array.
2. For every window of size n(no. of elements in the given array), perform LIS.
3. Return maximum length.
For example : Given array is {1, 4, 6, 2, 3}

After appending elements resultant array
will be {1, 4, 6, 2, 3, 1, 4, 6, 2, 3}.

Now for every consecutive n elements perform LIS.
1- {1, 4, 6, 2, 3} --3 is length of LIS.
2- {4, 6, 2, 3, 1} --2 is length of LIS.
3- {6, 2, 3, 1, 4} --3
4- {2, 3, 1, 4, 6}-- 4 {2, 3, 4, 6}
5- {3, 1, 4, 6, 2} --3.
6- {1, 4, 6, 2, 3} Original list.

So, maximum length of LIS in circular manner is 4.

As in the last window we will have the same elements as in the given array which we don’t need to compute again, so we can append only n-1 elements to reduce the number of operations.

C++

 // C++ implementation to find LIS in circular way #include using namespace std;    // Utility function to find LIS using Dynamic programming int computeLIS(int circBuff[], int start, int end, int n) {     int LIS[end-start];        /* Initialize LIS values for all indexes */     for (int i = start; i < end; i++)         LIS[i] = 1;        /* Compute optimized LIS values in bottom up manner */     for (int i = start + 1; i < end; i++)            // Set j on the basis of current window         // i.e. first element of the current window         for (int j = start; j < i; j++ )             if (circBuff[i] > circBuff[j] && LIS[i] < LIS[j] + 1)                 LIS[i] = LIS[j] + 1;        /* Pick maximum of all LIS values */     int res = INT_MIN;     for (int i = start; i < end; i++)         res = max(res, LIS[i]);        return res; }    // Function to find Longest Increasing subsequence in // Circular manner int LICS(int arr[], int n) {     // Make a copy of given array by appending same     // array elements  to itself     int circBuff[2 * n];     for (int i = 0; i

Java

 // Java implementation to find LIS in circular way    class Test {     // Utility method to find LIS using Dynamic programming     static int computeLIS(int circBuff[], int start, int end, int n)     {         int LIS[] = new int[n+end-start];                 /* Initialize LIS values for all indexes */         for (int i = start; i < end; i++)             LIS[i] = 1;                 /* Compute optimized LIS values in bottom up manner */         for (int i = start + 1; i < end; i++)                     // Set j on the basis of current window             // i.e. first element of the current window             for (int j = start; j < i; j++ )                 if (circBuff[i] > circBuff[j] && LIS[i] < LIS[j] + 1)                     LIS[i] = LIS[j] + 1;                 /* Pick maximum of all LIS values */         int res = Integer.MIN_VALUE;         for (int i = start; i < end; i++)             res = Math.max(res, LIS[i]);                 return res;     }             // Function to find Longest Increasing subsequence in     // Circular manner     static int LICS(int arr[], int n)     {         // Make a copy of given array by appending same         // array elements  to itself         int circBuff[] = new int[2 * n];         for (int i = 0; i

Python3

 # Python3 implementation to find  # LIS in circular way Utility  # function to find LIS using  # Dynamic programmi def computeLIS(circBuff, start, end, n):     LIS = [0 for i in range(end)]            # Initialize LIS values      # for all indexes      for i in range(start, end):         LIS[i] = 1                # Compute optimized LIS values     # in bottom up manner     for i in range(start + 1, end):                    # Set j on the basis of current          # window i.e. first element of         # the current window          for j in range(start,i):             if (circBuff[i] > circBuff[j] and                  LIS[i] < LIS[j] + 1):                 LIS[i] = LIS[j] + 1                        # Pick maximum of all LIS values     res = -100000     for i in range(start, end):         res = max(res, LIS[i])     return res    # Function to find Longest Increasing  # subsequence in Circular manner def LICS(arr, n):            # Make a copy of given      # array by appending same      # array elements to itself      circBuff = [0 for i in range(2 * n)]     for i in range(n):         circBuff[i] = arr[i]     for i in range(n, 2 * n):         circBuff[i] = arr[i - n]                # Perform LIS for each     # window of size n      res = -100000     for i in range(n):         res = max(computeLIS(circBuff, i,                               i + n, n), res)     return res    # Driver code arr = [ 1, 4, 6, 2, 3 ] n = len(arr) print("Length of LICS is", LICS(arr, n))    # This code is contributed  # by sahilshelangia

C#

 // C# implementation to find  // LIS in circular way using System;    class Test {     // Utility method to find LIS     // using Dynamic programming     static int computeLIS(int []circBuff, int start,                                       int end, int n)     {         int []LIS = new int[n+end-start];                /* Initialize LIS values for all indexes */         for (int i = start; i < end; i++)             LIS[i] = 1;                /* Compute optimized LIS values             in bottom up manner */         for (int i = start + 1; i < end; i++)                    // Set j on the basis of current window             // i.e. first element of the current window             for (int j = start; j < i; j++ )                 if (circBuff[i] > circBuff[j] &&                              LIS[i] < LIS[j] + 1)                     LIS[i] = LIS[j] + 1;                /* Pick maximum of all LIS values */         int res = int.MinValue;         for (int i = start; i < end; i++)             res = Math.Max(res, LIS[i]);                return res;     }            // Function to find Longest Increasing      // subsequence in Circular manner     static int LICS(int []arr, int n)     {         // Make a copy of given array by          // appending same array elements to itself         int []circBuff = new int[2 * n];                    for (int i = 0; i

PHP

 \$circBuff[\$j] &&                       \$LIS[\$i] < \$LIS[\$j] + 1)                 \$LIS[\$i] = \$LIS[\$j] + 1;        /* Pick maximum of     all LIS values */     \$res = PHP_INT_MIN;     for (\$i = \$start; \$i < \$end; \$i++)         \$res = max(\$res, \$LIS[\$i]);        return \$res; }    // Function to find LIS  // in Circular manner function LICS(\$arr, \$n) {     // Make a copy of given array     // by appending same array     // elements to itself     for (\$i = 0; \$i < \$n; \$i++)         \$circBuff[\$i] = \$arr[\$i];     for (\$i = \$n; \$i < 2 * \$n; \$i++)         \$circBuff[\$i] = \$arr[\$i - \$n];        // Perform LIS for each      // window of size n     \$res = PHP_INT_MIN;     for (\$i = 0; \$i < \$n; \$i++)         \$res = max(computeLIS(\$circBuff, \$i,                                \$i + \$n, \$n),                                \$res);        return \$res; }    // Driver Code \$arr = array(1, 4, 6, 2, 3); \$n = sizeof(\$arr);    echo "Length of LICS is " ,              LICS(\$arr, \$n);        // This code is contributed by aj_36 ?>

Output :

Length of LICS is 4

Time complexity of above solution is O(n3). It can be reduced O(n2 Log n) using O(n Log n) algorithm to find LIS.

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