Given a Large number N ( number of digits in N can be up to 105). The task is to find the cuts required of a number such that maximum parts are divisible by 3.
Input: N = 1269 Output: 3 Cut the number as 12|6|9. So, 12, 6, 9 are the three numbers which are divisible by 3. Input: N = 71 Output: 0 However, we make cuts there is no such number that is divisible by 3.
Let’s calculate the values of the array res[0…n], where res[i] is the answer for the prefix of the length i. Obviously, res:=0, since for the empty string (the prefix of the length 0) the answer is 0.
For i>0 one can find res[i] in the following way:
- Let’s look at the last digit of the prefix of length i. It has index i-1. Either it doesn’t belong to segment divisible by 3, or it belongs.
- If it doesn’t belong, it means last digit can’t be used, so res[i]=res[i-1]. If it belongs then find shortest s[j..i-1] that is divisible by 3 and try to update res[i] with the value res[j]+1.
- A number is divisible by 3, if and only if the sum of its digits is divisible by 3. So the task is to find the shortest suffix of s[0…i-1] with the sum of digits divisible by 3. If such suffix is s[j..i-1] then s[0..j-1] and s[0..i-1] have the same remainder of the sum of digits modulo 3.
- Let’s maintain remIndex[0..2]- an array of the length 3, where remIndex[r] is the length of the longest processed prefix with the sum of digits equal to r modulo 3. Use remIndex[r]= -1 if there is no such prefix. It is easy to see that j=remIndex[r] where r is the sum of digits on the ith prefix modulo 3.
- So to find the maximal j<=i-1 that substring s[j..i-1] is divisible by 3, just check that remIndex[r] not equals to -1 and use j=remIndex[r], where r is the sum of digits on the i-th prefix modulo 3.
- It means that to handle case that the last digit belongs to divisible by 3 segment, try to update res[i] with value res[remIndex[r]]+1. In other words, just do if (remIndex[r] != -1) => res[i] = max(res[i], res[remIndex[r]] + 1).
Below is the implementation of the above approach:
We can use running_sum which keeps the sum of all successive integers, where none of the individual integer is divisible by 3. we can pass through each integer one by one and do the following:
- If the integer is divisible by 3 or the running_sum is non-zero and divisible by 3, increment the counter and reset running_sum.
- In case the integer is not divisible by 3, keep a track of sum of all such successive integers.
- Split a number into 3 parts such that none of the parts is divisible by 3
- Minimum Cuts can be made in the Chessboard such that it is not divided into 2 parts
- Maximum number of pieces in N cuts
- Minimum cuts required to divide the Circle into equal parts
- Divide number into two parts divisible by given numbers
- Recursively break a number in 3 parts to get maximum sum
- Divide a number into two parts such that sum of digits is maximum
- Break a number such that sum of maximum divisors of all parts is minimum
- Minimum and Maximum element of an array which is divisible by a given number k
- Minimum number of cuts required to make circle segments equal sized
- Split a string in equal parts such that all parts are palindromes
- Find the number of ways to divide number into four parts such that a = c and b = d
- Count number of ways to divide a number in 4 parts
- Break the number into three parts
- Divide a big number into two parts that differ by k
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.