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row1: abb
row2: a
______I dont think it will pass. Since you even count the letter in the same row. Actually to see if the nubmers of the letter equals the row, is not precise.
He is counting the letters occurrence in each string, the first condition ensure that same letters in the same string will only be counted once.
The second condition is just confirming the answer: if a letter have n counts accross n strings -> number of gem-element increases
r is 0 for first string , first condition
"if (letter[c - 'a'] == r) { letter[c - 'a']++;"
will increase the letter's index only once throught the whole string if a particular character occures even multiple times. and second condition
if (letter[c - 'a'] == rocks) { count++; } means if the particular index is equals to the number of string entered means it's present in every string so increase the counter.
I like this.
People have already explained how duplicates are avoided.
I particularly like the other feature (because I had to think about it to understand it),
if(letter[c-'a']==r){
means if up until now we have found this element in as many rocks as we have examined before this rock, it's still a gem-element, so it remains one after this rock as well.
But I don't like having
if(letter[c-'a']==rocks){count++;}
inside every every rock since it can only be true when r == rocks-1. I'd prefer to keep s[] outside the outer loop, and after all the rocks have been parsed, go through all the elements in the last rock again, counting if they have been found exactly rock times (and incrementing their count, so they aren't double-counted for duplicate elements in the last rock.)
Gemstones
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if (letter[c - 'a'] == r) { letter[c - 'a']++; if (letter[c - 'a'] == rocks) { count++; }
can you explain this concept??
Yeah please.
row1: abb row2: a ______I dont think it will pass. Since you even count the letter in the same row. Actually to see if the nubmers of the letter equals the row, is not precise.
He is counting the letters occurrence in each string, the first condition ensure that same letters in the same string will only be counted once. The second condition is just confirming the answer: if a letter have n counts accross n strings -> number of gem-element increases
r is 0 for first string , first condition "if (letter[c - 'a'] == r) { letter[c - 'a']++;" will increase the letter's index only once throught the whole string if a particular character occures even multiple times. and second condition if (letter[c - 'a'] == rocks) { count++; } means if the particular index is equals to the number of string entered means it's present in every string so increase the counter.
I like this. People have already explained how duplicates are avoided. I particularly like the other feature (because I had to think about it to understand it),
means if up until now we have found this element in as many rocks as we have examined before this rock, it's still a gem-element, so it remains one after this rock as well.
But I don't like having
inside every every rock since it can only be true when r == rocks-1. I'd prefer to keep s[] outside the outer loop, and after all the rocks have been parsed, go through all the elements in the last rock again, counting if they have been found exactly rock times (and incrementing their count, so they aren't double-counted for duplicate elements in the last rock.)