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.

`/* A Naive C/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; `
`} ` |

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**Output:**

Length of lis is 5

Please refer complete article on Dynamic Programming | Set 3 (Longest Increasing Subsequence) for more details!

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