You are given a binary array consists of 0's and 1's , and q queries . q can be large !
in each query you are given a certain range [L , R].
suppose a[] = {1,0,0,1}
L = 1 , R = 3 .
do toggling , resultant array = {0,1,1,1}
L = 1 , R =4
count all one's , ans = 3.
you have to either toggle bits in the given range i.e make 0 = 1 and 1 = 0 .
and on another query you are about to count all 1's in the range of [L,R]. **** The problem gives a feel of segment tree + lazy propogation . but how to do toggling in segment tree .
how should we update the lazy tree !
Any idea !
Just use the trick that inverting a bit x can be done by taking - x + 1.
hey .. sorry but i couldnt understand a single thing .. it would be very helpful if u elaborate it quite a bit.
What I am saying is that toggling a bit is the same thing as multiplying it with -1 and then adding 1. The nice thing about this is that there are lazy segment trees that support doing addition and multiplication and sum/max/min on intervals, so for me solving this problem would just be applying a very general lazy segment tree.
I think that if you want to learn lazy segment trees then first try to implement a general lazy segment tree that support addition, multiplication, sum/min/max. After that you can solve nearly all segment tree problems using that one tree, so it is quite useful.
Link
i have seen it before posting. but how to use segment tree with lazy is not explaimed clearly there. if u could understand please let me know .
Lazy tags store if each range should be toggled. When a range is toggled, set
tree[node] = en-st+1 - tree[node];which uses the fact that toggling a bit value x is the same as doingx=1-x.so segment tree is storing the count of 1's in the range [l,r] ? and when we do lazy propogation , what value will we propogate ?
yes
I wrote a code for a similar problem. https://pastebin.com/R7iqy1sq Reply if you don't understand.