Count and Toggle Queries on a Binary Array
Given a size n in which initially all elements are 0. The task is to perform multiple queries of following two types. The queries can appear in any order.
1. toggle(start, end) : Toggle (0 into 1 or 1 into 0) the values from range ‘start’ to ‘end’.
2. count(start, end) : Count the number of 1’s within given range from ‘start’ to ‘end’.
Input : n = 5 // we have n = 5 blocks toggle 1 2 // change 1 into 0 or 0 into 1 Toggle 2 4 Count 2 3 // count all 1's within the range Toggle 2 4 Count 1 4 // count all 1's within the range Output : Total number of 1's in range 2 to 3 is = 1 Total number of 1's in range 1 to 4 is = 2
A simple solutionfor this problem is to traverse the complete range for “Toggle” query and when you get “Count” query then count all the 1’s for given range. But the time complexity for this approach will be O(q*n) where q=total number of queries.
An efficient solution for this problem is to use Segment Tree with Lazy Propagation. Here we collect the updates until we get a query for “Count”. When we get the query for “Count”, we make all the previously collected Toggle updates in array and then count number of 1’s with in the given range.
Below is the implementation of above approach:
This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.