CODEWITHSUNDEEP
pythonalgorithmfrequency
WOOSAL
WOOSAL

Reputation: 23

Starting Index of the Most Frequent Consecutive Number Algorithm

1) The purpose of the code: I have coded an algorithm, which is supposed to give me the exact number, which is the most frequent and consecutive at the same time number.

2) What I have tried: I have tried to write the whole code, and actually managed to get that exact number. I have also added the frequency of that number, which is the output.

3) What I need: I am looking for the algorithm, which will identify the first starting index of those consecutive numbers. For example, if the input is 123777321, as an index number 3 is needed, the reason being 777 is the most frequent consecutive number in this input, it should find its "index", and also print it out.

The Code I have written:

def maxRepeating(str):
    length = len(str)
    count = 0

    result = str[0]
    for i in range(length):

        current_count = 1
        for x in range(i + 1, length):

            if (str[i] != str[x]):
                break
            current_count += 1

        if current_count > count:
            count = current_count
            result = str[i]

    print("Longest same number sequence is of number {} by being repeated {} times in a row, with the first index starting at {}".format(result, count, i))



inputString = str(input("Please enter the string: "))

maxRepeating(inputString)

Example of an input: Please enter a string: 123777321

Example of an output: Longest same number sequence is of number 7 by being repeated 3 times in a row, with the first index starting at 3

Upvotes: 0

Views: 232

Answers (2)

chuck
chuck

Reputation: 1447

Just add a variable to track the starting index of the best sequence.

def maxRepeating(str):
    length = len(str)
    count = 0
    result = str[0]
    start_ind = None

    for i in range(length):

        current_count = 1
        for x in range(i + 1, length):

            if (str[i] != str[x]):
                break
            current_count += 1

        if current_count > count:
            count = current_count
            result = str[i]
            start_ind = i

    print("Longest same number sequence is of number {} by being repeated {} times in a row, with the first index starting at {}".format(result, count, start_ind))



inputString = str(input("Please enter the string: "))

maxRepeating(inputString)

Upvotes: 1

SomeDude
SomeDude

Reputation: 14238

From your comments, I assume you are trying to get the index at which the most frequently occurring element starts right? Declare another variable like max_index and update it each time you update count and use that to print the index.

.....
max_index = 0
for i in range(length):

        current_count = 1
        for x in range(i + 1, length):

            if (str[i] != str[x]):
                break
            current_count += 1

        if current_count > count:
            count = current_count
            result = str[i]
            max_index = i
print("Longest same number sequence is of number {} by being repeated {} times in a row, with the first index starting at {}".format(result, count, max_index))
......

Upvotes: 0

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