Write a function solution that, given a string S consisting of N letters 'a' and/or 'b' returns

First, we need to check if the string contains both 'a' and 'b'. If it contains only 'a' or only

Write a function solution that, given a string S consisting...

Write a function solution that, given a string S consisting of N letters 'a' and/or 'b' returns true when all occurrences of letter 'a' are before all occurrences of letter 'b' and returns false otherwise. Examples: 1. Given S = "aabbb", the function should return true. 2. Given S = "ba", the function should return false. 3. Given S = "aaa", the function should return true. Note that 'b' does not need to occur in S. 4. Given S = "b", the function should return true. Note that 'a' does not need to occur in S. 5. Given S = "abba", the function should return false. Write an efficient algorithm for the following assumptions: N is an integer within the range [1..300,000]; string S consists only of the characters "a" and/or "b".

Transcript

00:01 Right it java method now that would let's a supply strength and if a country before before any of the two items that will pick for any positions any consecutive positions that would compare and we'll say true otherwise you say false all right so i'm going to create a method which will be bullion so it's a public static bullying i'll say a first right strength i mean you know that string by x then you need to say golden result which would be equals you true initially right so i'll simply return results all right so this is what i'm going to do i'm going to split the string into the individual car taxes so i'll say string i'm going to first of all the four trim it for all the spaces to go i'll say to lower case for them to be converted to your case then i'll say split so i split by white spaces right and then i would say if there is does do is equals one if there's only one item then we simply return true right which here the resource is true okay if that's not the case if we have more than one items in the list so the assumption is that we we only have a and b in the list say for a x i equals you one because i've already passes zero point now say i which is less than the list dot next right now have i plus plus so that it lives through them so say if the first position which is i minus one all right right is equals you now let me use equals right so equals to a and so for and i'm going to use this and the second position which is i all right is equals you use equals b okay that is the first position is a the second position is b for any two positions that we are considering them i'll simply retain the results alright so if that's not the case and it actually comes to the end of the list so here let me actually return false okay let me because i retain force let me let me let me let me come before so if b comes before a for any of the circumstances that i've mentioned before then i'm going to have because b comes before a i'm going to return force but if it comes to the end of the loop all right i'm simply going to return the results which is having true okay so i have a number of test cases in other to so let me run that at the last part and we'll see what we've had.

04:21 Okay, so b comes before a that is false.

04:24 A comes before b that is true.

04:26 We have only b which is true, a which is true...

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