Find a String in Python | HackerRank Solution

Hello coders, today we are going to solve Find a String Hacker Rank Solution in Python.

Find a String in Python - Hacker Rank Solution


In this challenge, the user enters a string and a substring. You have to print the number of times that the substring occurs in the given string. String traversal will take place from left to right, not from right to left.

NOTE: String letters are case-sensitive.

Input Format

The first line of input contains the original string. The next line contains the substring.


1 <= len(string) <= 200
Each character in the string is an ascii character.

Output Format

Output the integer number indicating the total number of occurrences of the substring in the original string.

Sample Input


Sample Output



There are a couple of new concepts:
In Python, the length of a string is found by the function len(s), where s is the string.
To traverse through the length of a string, use a for loop:

for i in range(0, len(s)):
    print (s[i])

A range function is used to loop over some length:

range (0, 5)

Here, the range loops over 0 to 45 is excluded.

Solution – Find a String in Python – Hacker Rank Solution

def count_substring(string, sub_string):
    for i in range(0, len(string)-len(sub_string)+1):
        if string[i] == sub_string[0]:
            for j in range (0, len(sub_string)):
                if string[i+j] != sub_string[j]:
            if flag==1:
                count += 1
    return count

if __name__ == '__main__':
    string = input().strip()
    sub_string = input().strip()
    count = count_substring(string, sub_string)

Disclaimer: The above Problem (Find a String) is generated by Hacker Rank but the Solution is provided by CodingBroz. This tutorial is only for Educational and Learning purposes.

2 thoughts on “Find a String in Python | HackerRank Solution”

  1. def count_substring(string, sub_string):
    x = len(sub_string)
    ct = 0
    for i in range(len(string)-len(sub_string)+1):
    if string[i:x+i] == sub_string:
    ct +=1
    return ct

  2. a=”abcdcdc”
    for i in range(0,len(a)-len(b)+1):
    if b==(a[i:len(b)+i]):

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