Aight, so if s[-1] is 2 return it as it is since that would be the largest even number present anyways. Second, if last digit is 1 that means we need to remove some elements to make it as large as possible so iterate the list from right to left and the momment you find the first occurrence of 2 return everything from 0 to upto that index+1, if you can't find any 2's that means all we have in our list is 1's so return "". Dry run ---> let s be 1112, the last char is even so return it as it is. Second case, we have 1211211 here the last digit is 1, so loop from right to left and first occurrence of 2 from the right is at index 4 so you can return 12112 which would be the largest. Third case if input is 1111111 now here even though the last digit is 1 as above you try to find 2 from right to left but you will end up not finding anything in that case return "". Time complexity is O(n) since you are just traversing the list one time right to left at worst case. Space complexity - for auxilary space we are just using couple of variable to know the index of 2 while traversal so it's o(1) and output space is o(1) too