diff --git a/dynamic_programming/longest_increasing_subsequence.py b/dynamic_programming/longest_increasing_subsequence.py index 37ddef207..c04f6673f 100644 --- a/dynamic_programming/longest_increasing_subsequence.py +++ b/dynamic_programming/longest_increasing_subsequence.py @@ -1,12 +1,41 @@ -""" -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 -""" -def LIS(arr): - n= len(arr) - lis = [1]*n +''' +Author : Mehdi ALAOUI - for i in range(1, n): - for j in range(0, i): - if arr[i] > arr[j] and lis[i] <= lis[j]: - lis[i] = lis[j] + 1 - return max(lis) +This is a pure Python implementation of Dynamic Programming solution to the longest increasing subsequence of a given sequence. + +The problem is : +Given an ARRAY, to find the longest and increasing sub ARRAY in that given ARRAY and return it. +Example: [10, 22, 9, 33, 21, 50, 41, 60, 80] as input will return [10, 22, 33, 41, 60, 80] as output +''' + +def longestSub(ARRAY): #This function is recursive + + ARRAY_LENGTH = len(ARRAY) + if(ARRAY_LENGTH <= 1): #If the array contains only one element, we return it (it's the stop condition of recursion) + return ARRAY + #Else + PIVOT=ARRAY[0] + isFound=False + i=1 + LONGEST_SUB=[] + while(not isFound and i= ARRAY[i] ] + TEMPORARY_ARRAY = longestSub(TEMPORARY_ARRAY) + if ( len(TEMPORARY_ARRAY) > len(LONGEST_SUB) ): + LONGEST_SUB = TEMPORARY_ARRAY + else: + i+=1 + + TEMPORARY_ARRAY = [ element for element in ARRAY[1:] if element >= PIVOT ] + TEMPORARY_ARRAY = [PIVOT] + longestSub(TEMPORARY_ARRAY) + if ( len(TEMPORARY_ARRAY) > len(LONGEST_SUB) ): + return TEMPORARY_ARRAY + else: + return LONGEST_SUB + +#Some examples + +print(longestSub([4,8,7,5,1,12,2,3,9])) +print(longestSub([9,8,7,6,5,7])) \ No newline at end of file