check if a string contains determinate sequence from last element to first-CodePudding

how can i check if a string contains determinate sequence from last element to first?

Example:

I have this sequence:

A, C, C, A, B, C

If sequence contains C [index:5], B[index:4], A[index:3] return true.

In this example the expected out is true

I have this sequence:

P, R, P, R, R

If sequence contains P [index:5], R[index:4], P[index:3] return true.

In this example the expected out is false

CodePudding user response：

Rabinzel's solution is good. If you want a solution using a loop you can do something like this:

Given seqeunce = 'ACCABC' and target = 'CBA'

def checkSeq(sequence, target):
    # This assumes target is the characters to be checked in reverse order from end of seqeuence
    if len(target) > len(sequence):
        return False

    pos = -1
    for ch in target:
        if sequence[pos] != ch:
            return False
        pos -= 1
    
    return True

Or this solution which ignores checking lengths and indexes entirely:

def checkSeq(sequence, target):
    # target is list of elements to check for at the end of sequence in the reverse order
    target = list(reversed(target))
    sequence = list(sequence)
    while sequence and target:
        if target.pop() != sequence.pop():
            return False
    return not (target and not sequence)

CodePudding user response：

If you have your target sequence as list, you could slice your input in reversed order and check if equal:

# 1st
#input list
lst = ['A', 'C', 'C', 'A', 'B', 'C']

#target list
target = ['C', 'B', 'A']
len_target = len(target)

# get the ending index dependent on length of input and target list
idx = len(lst) - len_target - 1 # 5 - 3 - 1 = 2
print(lst[:idx:-1] == target) # lst[:2:-1] slice starting from the back till idx 2 (2 not included)
# ['C', 'B', 'A'] == ['C', 'B', 'A'] --> True

# 2nd
lst = ['P', 'R', 'P', 'R', 'R']
target = ['P', 'R', 'P']
len_target = len(target)
idx = len(lst) - len_target - 1

print(lst[:idx:-1] == target)
False

If you want to read more about how to slice lists, this question has very detailed explanation.