Find the longest sequence length whose sum is divisible by 3

Question

I have an exercise that needs to be done with O(n) time complexity, however, I can only solve it with an O(n^2) solution.

You have an array and you need to count the

Answer 1

Iterate through the array, summing the total as you go. Record the position of the first position where the modulo sum is 0. Also, record the position of he first position where the modulo sum is 1. And, finally, record the position of he first position where the modulo sum is 2.

Do the same thing backwards also, recording the last position where the modulo sum is 0, 1, and 2. That gives three possibilities for the longest sequence - you just check which pair are farthest apart.

Answer 2

As a non-CS person, this is interesting. First approach of mine was simply to calc the running sum mod 3. You'll get a sequence of {0,1,2}. Now look for the first and the last 0, the first and the last 1 and the first and the last 2, and compare their respective distances...

Answer 3

Let's take a look at prefix sums. A [L, R] subarray is divisble by 3 if and only if
prefixSum[L - 1] mod 3 = prefixSum[R] mod 3. This observation gives a very simple linear solution(because there are only 3 possible values of a prefix sum mod 3, we can simply find the first and the last one).

For example, if the input array is {1, 2, 3, -4, -1}, the prefix sums are {0, 1, 0, 0, 2, 1}. (there are n + 1 prefix sums because of an empty prefix). Now you can just take a look at the first and last occurrence of 0, 1 and 2.

Answer 4

You apply dynamic programming. For every position you compute 3 values:

• The longest sequence ending in that position which has sum s = 0 mod 3
• The longest sequence ending in that position which has sum s = 1 mod 3
• The longest sequence ending in that position which has sum s = 2 mod 3

So given this value for position i you can easily compute the new ones for position i+1.