Editorial of Bangladesh 2.0 Round
“spoiler” text in github wiki pages String Algorithm Tutorial Only an odd position is available for the for the string S, when the length of the string is 1. Because if length is greater than 1, then always even positions exist. You need to just print the first character of the string S. Time Complexity : O(1) Code(C++) solution Pair sum Tutorial No matter what is N, there is always N/2 pair exist. But when N is even, we should decrease (N/2) by 1. Because there is a pair (i == j) that exists, which reflects our condition (i < j). Time Complexity : O(1) Code(C++) solution KuZ the Position Tutorial Soon we will update Code(C++) solution Number concatenation Tutorial