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use python to implement this algorithm the input should be a nested lsit. e.g.: job = [[0,1],[1,5],[2,3],[2,5],[5,7]] expected result is: [[[0, 1], [1, 5], [5, 7]], [[2, 3], [3, 4]], [[2, 5]]]

def interval_partitioning(job):
    
    #sort the list in the asencding order by the start time, 
    #which is the 1st element of the inner list
    job = sorted(job, key = lambda i: i[0])
    n = len(job)
    
    #initilise a list to store the result 
    ans = []
    
    #add the first item 
    ans.append([job[0]])
    
    #initialise the depth and last finish 
    d = 0
    
    #initilise a loop
    for i in range(1, n):
        for j in range(0,d+1):
            if job[i][0] >= ans[j][-1][1]:
                ans[j].append(job[i])
                break
            else:
                ans.append([job[i]])
                d += 1
    return ans

We got this:

[[[0, 1], [1, 5], [5, 7]], [[2, 3], [3, 4]], [[2, 5]], [[2, 5]], [[3, 4]]]
mkrieger1
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jwan218
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1 Answers1

0

solved by myself

def interval_partitioning(job): job = sorted(job, key = lambda i: i[0]) n = len(job)

#initilise a list to store the result 
ans = []

#add the first item 
ans.append([job[0]])

#initialise the depth and last finish 
d = 0

#initilise a loop
for i in range(1, n):
    assigned = False
    for j in range(0,d+1):
        if job[i][0] >= ans[j][-1][1]:
            assigned = True
            ans[j].append(job[i])
            break
    
    if not assigned: 
        ans.append([job[i]])
        d += 1
return ans
jwan218
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