C/C++ Program for Longest Increasing Subsequence
The Longest Increasing Subsequence (LIS) problem is to find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order. For example, the length of LIS for {10, 22, 9, 33, 21, 50, 41, 60, 80} is 6 and LIS is {10, 22, 33, 50, 60, 80}.
More Examples:
Input : arr[] = {3, 10, 2, 1, 20} Output : Length of LIS = 3 The longest increasing subsequence is 3, 10, 20 Input : arr[] = {3, 2} Output : Length of LIS = 1 The longest increasing subsequences are {3} and {2} Input : arr[] = {50, 3, 10, 7, 40, 80} Output : Length of LIS = 4 The longest increasing subsequence is {3, 7, 40, 80}
Optimal Substructure:
Let arr[0..n-1] be the input array and L(i) be the length of the LIS ending at index i such that arr[i] is the last element of the LIS.
Then, L(i) can be recursively written as:
L(i) = 1 + max( L(j) ) where 0 < j < i and arr[j] < arr[i]; or
L(i) = 1, if no such j exists.
To find the LIS for a given array, we need to return max(L(i)) where 0 < i < n.
Thus, we see the LIS problem satisfies the optimal substructure property as the main problem can be solved using solutions to subproblems.
Following is a simple recursive implementation of the LIS problem. It follows the recursive structure discussed above.
C++
/* A Naive C++ recursive implementation of LIS problem */ #include <iostream> using namespace std; /* To make use of recursive calls, this function must return two things: 1) Length of LIS ending with element arr[n-1]. We use max_ending_here for this purpose 2) Overall maximum as the LIS may end with an element before arr[n-1] max_ref is used this purpose. The value of LIS of full array of size n is stored in *max_ref which is our final result */ int _lis( int arr[], int n, int * max_ref) { /* Base case */ if (n == 1) return 1; // 'max_ending_here' is length of LIS ending with arr[n-1] int res, max_ending_here = 1; /* Recursively get all LIS ending with arr[0], arr[1] ... arr[n-2]. If arr[i-1] is smaller than arr[n-1], and max ending with arr[n-1] needs to be updated, then update it */ for ( int i = 1; i < n; i++) { res = _lis(arr, i, max_ref); if (arr[i - 1] < arr[n - 1] && res + 1 > max_ending_here) max_ending_here = res + 1; } // Compare max_ending_here with the overall max. And // update the overall max if needed if (*max_ref < max_ending_here) *max_ref = max_ending_here; // Return length of LIS ending with arr[n-1] return max_ending_here; } // The wrapper function for _lis() int lis( int arr[], int n) { // The max variable holds the result int max = 1; // The function _lis() stores its result in max _lis(arr, n, &max); // returns max return max; } // Driver code int main() { int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 }; int n = sizeof (arr) / sizeof (arr[0]); cout << "Length of lis is " << lis(arr, n) << "\n" ; return 0; } // This code is contributed by Shubhamsingh10 |
C
/* A Naive C recursive implementation of LIS problem */ #include <stdio.h> #include <stdlib.h> /* To make use of recursive calls, this function must return two things: 1) Length of LIS ending with element arr[n-1]. We use max_ending_here for this purpose 2) Overall maximum as the LIS may end with an element before arr[n-1] max_ref is used this purpose. The value of LIS of full array of size n is stored in *max_ref which is our final result */ int _lis( int arr[], int n, int * max_ref) { /* Base case */ if (n == 1) return 1; // 'max_ending_here' is length of LIS ending with arr[n-1] int res, max_ending_here = 1; /* Recursively get all LIS ending with arr[0], arr[1] ... arr[n-2]. If arr[i-1] is smaller than arr[n-1], and max ending with arr[n-1] needs to be updated, then update it */ for ( int i = 1; i < n; i++) { res = _lis(arr, i, max_ref); if (arr[i - 1] < arr[n - 1] && res + 1 > max_ending_here) max_ending_here = res + 1; } // Compare max_ending_here with the overall max. And // update the overall max if needed if (*max_ref < max_ending_here) *max_ref = max_ending_here; // Return length of LIS ending with arr[n-1] return max_ending_here; } // The wrapper function for _lis() int lis( int arr[], int n) { // The max variable holds the result int max = 1; // The function _lis() stores its result in max _lis(arr, n, &max); // returns max return max; } /* Driver program to test above function */ int main() { int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 }; int n = sizeof (arr) / sizeof (arr[0]); printf ( "Length of lis is %d\n" , lis(arr, n)); return 0; } |
Length of lis is 5
Please refer complete article on Dynamic Programming | Set 3 (Longest Increasing Subsequence) for more details!