CODEWITHSUNDEEP
python
Mist
Mist

Reputation: 57

Given 2 Strings str and word, you have to find how many words can you make from that given string

I am trying to solve some questions, even though the logic seems very straight forward, I am not able to construct a possible solution.

Given 2 Strings str and word, you have to find how many words can you make from that given string.

Input : str="This is a test string" word="tsit" , Output : 2

Explanation : there are 4 t's 4 s's 3 i's in the given str, by which you can only make 2 "tsit".

Input: str="Here is HashedIn Technologies" word="neurons" Output : 0

Explanation: since you do not have 'u' in str. thus u can't form word "neurons".

I was trying to use dictionary logic was increment count till any of the character count turns zero but how do i put it into code?

def freq(s, word):
s = s.lower()
aux_s = {}
aux_word = {}
for i in s:
    aux_s[i] = aux_s.get(i, 0) + 1
for i in word:
    aux_word[i] = aux_word.get(i, 0) + 1
count, ans = 0, float('inf')
for i,val in aux_s.items():
    if i in aux_word:
        pass

Upvotes: 1

Views: 994

Answers (2)

Aditya Sharma
Aditya Sharma

Reputation: 23

I have used Dictionary in the Solution , i have just changed a little bit on the logic which you have used.

import sys

test_str = "This is a test string"
test_str = test_str.lower()
word = "tsit"

res = {} 
word_count = {}

# loop to count frequency of all the characters in word
for keys in word:
  if keys != ' ':
    word_count[keys] = word_count.get(keys, 0) + 1

# loop to count frequency of all the characters in string
for keys in test_str: 
  if keys != ' ':   # if the char is space then ignore it
    res[keys] = res.get(keys, 0) + 1  # increment the count of the character if the character already exists

# assigning ans a max value
ans = sys.maxsize


for keys in word_count:
    if keys in res:
        a = res[keys]    #checking if the key exists in the res or not
    else:
        ans = 0   # if the char does not exist in the string then ans = 0
        break
    b = word_count[keys]
    n = a//b
    if n < ans:   # finding the lowest of the requirement
        ans = n   # assigning it to the answer
      
print(ans)   # print the answer

Upvotes: 0

Abhinav Mathur
Abhinav Mathur

Reputation: 8101

Since case doesn't matter as per your example,

str = "This is a test string".lower()
word = "tsit"
ans = len(str)
for i in word:
    ans = min(ans, str.count(i)//word.count(i))
print(ans)

Upvotes: 2

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