Explanation :

First create an array A of length 26, representing the counts of each letter of the alphabet, with each value initialized to 0. Iterate through each character in S1 and add 1 to the corresponding entry in A. Once this iteration is complete, A will contain the counts for the letters in S1. Then, iterate through each character in S2, and subtract 1 from each corresponding entry in A. Now, if the each entry in A is 0, then S1 and S2 are anagrams; otherwise, S1 and S2 aren't anagrams.

Here is pseudocode for the procedure that was described:

    def areAnagrams(S1, S2)
      A = new Array(26)
      A.initializeValues(0)    
      for each character in S1
        arrayIndex = mapCharacterToNumber(character) //maps "a" to 0, "b" to 1, "c" to 2, etc...
        A[arrayIndex] += 1
      end

      for each character in S2
        arrayIndex = mapCharacterToNumber(character)
        A[arrayIndex] -= 1
      end
    
      for (i = 0; i < 26; i++)
        if A[i] != 0
          return false
        end
      end
      return true
    end
  

Explanation :

The first thought most people have it to try to get each robot to go back and forth, centered on their original spot, each time going a little further out in each direction. This won't work.

Another insight is that since you don't know which robot will be on the left or right, it probably doesn't make sense to give them different programs.

The solution to this problem is to just have both robots move consistently in one direction, but not at full speed. Then, have either robot pick up to full speed when it sees oil again (which will only happen for one robot). This robot will eventually catch up to the other robot, causing a collision.