have been attempting to cope with this downside set, and I’ve discovered how one can remedy it as effectively. However there may be one downside that’s
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On this problem, a string and an inventory of intervals are given. The string consists of English letters solely and it may well include each lowercase and uppercase letters.
For 2 totally different letters, we are saying that the primary letter is bigger than the second letter when the primary letter comes later within the alphabet than the second letter ignoring the case of the letters. For instance, the letter ‘Z’ and ‘t’ are higher than the letters ‘b’ and ‘G’, whereas the letters ‘B’ andd ‘b’ are equal as case isn’t thought of.
The duty is the next. For every given interval, it is advisable discover the rely of the best letter occurring within the string in that interval, ignoring the case of the letters, so occurrences of, for instance, a and A are occurrences of the identical letter.
Take into account, for instance, for the string “AbaBacD”. Within the interval, [0, 4], the best letter is ‘b’ with rely 2.
The primary line accommodates integer N, denoting the size of the enter string.
The second line accommodates string S.
The third line accommodates an integer Q, denoting the variety of intervals. Every line of the Q subsequent strains accommodates two space-separated integers xi and yi, denoting the start and the top of ith interval.
For every interval, print the rely of the best letter occurring within the string in that interval.
1. To lowercase the string 2. Loop by way of the queries array 3. Get the string from begin interval to finish interval 4. Discover biggest char within the string 5. Rely the prevalence, and add it to the array 6. Return the array
def getMaxCharCount(s, queries): # queries is a n x 2 array the place queries[i] and queries[i] represents x[i] and y[i] for the ith question. maxCount =  finalWord = s.decrease() for interval in queries: if interval == interval: maxCount.append(1) else: string = finalWord[interval:interval+1] maxCount.append(string.rely(max(string))) return maxCount
Any assist could be appreciated