CODEWITHSUNDEEP
python
Shandeep Murugasamy
Shandeep Murugasamy

Reputation: 311

Return the number of times that the string appears using loops

Return the number of times that the string "hi" appears anywhere in the given string. 

count_hi('abc hi ho') # → 1

count_hi('ABChi hi') # → 2

count_hi('hihi') # → 2

I have this solution with me;

def count_hi(str):
    return str.count("hi")

But I am looking for a solution using a given hint: Use the for i in range(len(str)-1): loop to look at each index in the string, except the last. For each i, extract the string starting at i and up to but not including i+2. Check if that string is "hi", and count how many times that happens.

I even tried this solution, but doesn't pass all the test cases:

def count_hi(str):
    count = 0
    for char in str:
        if char == 'hi':
           count += 1
    return count

Upvotes: 1

Views: 11253

Answers (6)

Yash Marmat
Yash Marmat

Reputation: 1195

def count_hi(str):
  my_list = []
  for each in range(len(str) - 1):
    if str[each] == "h":
      next_item = str[each + 1]
      my_list.append(str[each] + next_item)

  total = my_list.count("hi")
  return total

Upvotes: 0

Farha Rubena
Farha Rubena

Reputation: 1

def count_hi(str1):

#use the replace method to replace all spaces
  str1 = list(str1.replace(' ',''))
  print(str1)
        
  '''define a counter variable that increments by one everytime the if cond 
  is met'''

  count = 0
    
  for i in range(len(str1)-1):
    if str1[i] == "h" and str1[i+1] == "i":
      count += 1
  return count

Upvotes: 0

Rafael Sartori
Rafael Sartori

Reputation: 11

def count_hi(str):
  len_w = len(str)
  txt = str.replace("hi", "")
  len_wo = len(txt)
  return (len_w- len_wo)/2

Upvotes: 0

M.zar
M.zar

Reputation: 9

public int countHi(String str) {
  int count = 0;
  for( int i = 0 ; i < str.length()-1 ; i++){
  if ( str.substring(i , i+2).equals("hi"))
  count = count + 1;
  }
  return count;
}

Upvotes: -1

Zach Gates
Zach Gates

Reputation: 4155

You can split the string:

string = 'hire test foo hi bar high'
split_string = [[item]+['hi'] for item in string.split('hi') if item != ""]
split_string = sum(split_string, [])

And use a for loop to count the matching strings:

string_count = 0
for item in range(len(split_string)):
    if split_string[item] == 'hi':
        string_count += 1

Or, you can skip the for loop and directly count the list:

split_string.count('hi') # returns 3

Upvotes: 1

John1024
John1024

Reputation: 113864

Here is one version:

def count_hi(s):
    count = 0
    for i in range(len(s)-1):
        count += s[i]=='h' and s[i+1]=='i'
    return count

Here is another:

def count_hi2(s):
    count = 0
    for i in range(len(s)-1):
        count += s[i:i+2] == 'hi'
    return count

Discussion

Consider this code fragment:

for char in str:
    if char == 'hi':

This loops through the individual characters in the string str. Thus, in this loop, char is always one single character. Consequently, it will never be equal to two characters.

Also, it best practice to use a different name for strings: str is a builtin. Python will let you freely overwrite builtins but a consequence is that you will lose easy access them.

Upvotes: 4

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