C/C++ Program for Longest Increasing Subsequence

Difficulty Level : Medium
Last Updated : 04 Dec, 2018

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}.
longest-increasing-subsequence

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;
}

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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