Given a number of at most 100 digits. We have to check if it is possible, after removing certain digits, to obtain a number of at least one digit which is divisible by 8. We are forbidden to rearrange the digits.
Input : 1787075866 Output : Yes There exist more one or more subsequences divisible by 8. Example subsequences are 176, 16 and 8. Input : 6673177113 Output : No No subsequence is divisible by 8. Input : 3144 Output : Yes The subsequence 344 is divisible by 8.
Property of the divisibility by eight : number can be divided by eight if and only if its last three digits form a number that can be divided by eight. Thus, it is enough to test only numbers that can be obtained from the original one by crossing out and that contain at most three digits i.e we check all one digits, two digits and three digit number combinations.
Method 1 (Brute Force):
We apply the brute force approach. We permute all possible single digit, double digit and triple digit combinations using iterative ladder. If we encounter a single digit number divisible by 8 or a double digit number combination divisible by 8 or a triple digit number combination divisible by 8, then that will be the solution to our problem.
Method 2 (Dynamic Programming):
Though we have only 100 digit number, but for longer examples larger than that, our program might exceed the given time limit.
Thus, we optimize our code by using dynamic programming approach.
Let be the ith digit of the sample. We generate a matrix dp[i][j], 1<=i<=n and 0<=j<8. The value of dp is true if we can cross out some digits from the prefix of length i such that the remaining number gives j modulo eight, and false otherwise. For broad understanding of the concept, if at an index, we find element modulo 8 for that index we put the value of
For all other numbers, we build on a simple concept that either addition of that digit will contribute in formation of a number divisible by 8, or it shall be left out.
Note: We also have to keep it in mind that we cannot change the order
if we add the current digit to the previous result.
if we exclude the current digit in our formation.
Now, if such a number shall exist, we will get a “true” for any i in dp[i]
Using the dynamic approach, our time complexity cuts down to , where 8 is from which the number should be divisible and n is the length of our input. Therefore, the overall complexity is .
For this problem, we simply need to check if there exists a two-digit subsequence divisible by 8 (divisibility test for 8)
We first find all the 2 digit numbers divisible by 8 and map the tens place digit with unit place digit
i.e :- 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96
Ignore 48 as 8 is always divisible by 8 similarly 80 and 88 have 8 in them which make such subsequence always divisible by 8
So we map 1 to 6, 2 to 4, 3 to 2 and so on using map i.e stl map in C++.
After building the map we traverse string from the last index and check if the mapped value of present index number is visited or not hence we need a visited array for this which will store true if the number is visited, else false
first char from the last index is 9 so we check if 6 is visited (i.e 96 is subsequence or not), we mark 9 in visited array
next char is 6 so we check is 4 visited (i.e 64), we mark 6 in visited array
next char is 7 so we check is 2 visited (i.e 72), we mark 7 in visited array
next char is 3 so we check is 6 visited (i.e 36), yes 6 is marked hence we print Yes.
If you take a close look, visited array will always have 10 fields and map will always have same size, hence space complexity will , time complexity will be for traversing string.
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