Given a large number, n (having number digits up to 10^6) and various queries of the form :
Query(l, r) : find if the sub-string between the indices l and r (Both inclusive) are divisible by 3.
Input: n = 12468236544 Queries: l=0 r=1 l=1 r=2 l=3 r=6 l=0 r=10 Output: Divisible by 3 Divisible by 3 Not divisible by 3 Divisible by 3 Explanation: In the first query, 12 is divisible by 3 In the second query, 24 is divisible by 3 and so on.
We know that any number is divisible by 3 if the sum of its digits is divisible by 3. Hence the idea is to pre-process an auxiliary array that would store the sum of digits.
Mathematically, sum = 0 and for i from 0 to number of digits of number: sum[i+1] = sum[i]+ toInt(n[i]) where toInt(n[i]) represents the integer value of i'th digit of n
Once our auxiliary array is processed, we can answer each query in O(1) time, because the substring from indices l to r would be divisible by 3 only if, (sum[r+1]-sum[l])%3 == 0
Below is a the implementation program for the same.
Divisible by 3 Divisible by 3 Not divisible by 3 Divisible by 3
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