The first one is d the second one is true the third one is false
Answer:
It can be a really good approach to use a local solver using the min conflicts heuristic in solving sudoku problems. It will work better actually. In this process, the value chosen is the value with the minimum conflicts. This is the general way a normal person would also tackle this problem. By this approach, if we keep taking the values with minimum conflicts the sudoku puzzle can be solved with a better performance.
Explanation:
Hi,
Answer => <span>Inode
Good Lessons </span>
Answer:
See explaination for the code
Explanation:
def wordsOfFrequency(words, freq):
d = {}
res = []
for i in range(len(words)):
if(words[i].lower() in d):
d[words[i].lower()] = d[words[i].lower()] + 1
else:
d[words[i].lower()] = 1
for word in words:
if d[word.lower()]==freq:
res.append(word)
return res
Note:
First a dictionary is created to keep the count of the lowercase form of each word.
Then, using another for loop, each word count is matched with the freq, if it matches, the word is appended to the result list res.
Finally the res list is appended.
