Algorithm for doing many substring reversals? - string

Question

Suppose I have a string S of length N, and I want to perform M of the following operations:
choose 1 <= L,R <= N and reverse the substring S[L..R]
I am interested in what the final string looks like after all M operations. The obvious approach is to do the actual swapping, which leads to O(MN) worst-case behavior. Is there a faster way? I'm trying to just keep track of where an index ends up, but I cannot find a way to reduce the running time (though I have a gut feeling O(M lg N + N) -- for the operations and the final reading -- is possible).

Answer 1

Yeah, it's possible. Make a binary tree structure like
struct node {
struct node *child[2];
struct node *parent;
char label;
bool subtree_flipped;
};
Then you can have a logical getter/setter for left/right child:
struct node *get_child(struct node *u, bool right) {
return u->child[u->subtree_flipped ^ right];
}
void set_child(struct node *u, bool right, struct node *c) {
u->child[u->subtree_flipped ^ right] = c;
if (c != NULL) { c->parent = u; }
}
Rotations have to preserve flipped bits:
struct node *detach(struct node *u, bool right) {
struct node *c = get_child(u, right);
if (c != NULL) { c->subtree_flipped ^= u->subtree_flipped; }
return c;
}
void attach(struct node *u, bool right, struct node *c) {
set_child(u, right, c);
if (c != NULL) { c->subtree_flipped ^= u->subtree_flipped; }
}
// rotates one of |p|'s child up.
// does not fix up the pointer to |p|.
void rotate(struct node *p, bool right) {
struct node *u = detach(p, right);
struct node *c = detach(u, !right);
attach(p, right, c);
attach(u, !right, p);
}
Implement splay with rotations. It should take a "guard" pointer that is treated as a NULL parent for the purpose of splaying, so that you can splay one node to the root and another to its right child. Do this and then you can splay both endpoints of the flipped region and then toggle the flip bits for the root and the two subtrees corresponding to segments left unaffected.
Traversal looks like this.
void traverse(struct node *u, bool flipped) {
if (u == NULL) { return; }
flipped ^= u->subtree_flipped;
traverse(u->child[flipped], flipped);
visit(u);
traverse(u->child[!flipped], flipped);
}

Answer 2

Splay tree may help you, it supports reverse operation in an array, with total complexity O(mlogn)

Answer 3

#F. Ju is right, splay trees are one of the best data structures to achieve your goal.
However, if you don't want to implement them, or a solution in O((N + M) * sqrt(M)) is good enough, you can do the following:
We will perform sqrt(M) consecutive queries and then rebuilt the array from the scratch in O(N) time.
In order to do that, for each query, we will store the information that the queried segment [a, b] is reversed or not (if you reverse some range of elements twice, they become unreversed).
The key here is to maintain the information for disjoint segments here. Notice that since we are performing at most sqrt(M) queries before rebuilding the array, we will have at most sqrt(M) disjoint segments and we can perform query operation on sqrt(M) segments in sqrt(M) time. Let me know if you need a detailed explanation on how to "reverse" these disjoint segments.
This trick is very useful while solving problems like that and it is worth to know it.
UPDATE:
I solved the problem exactly corresponding to yours on HackerRank, during their contest, using the method I described.
Here is the problem
Here is my solution in C++.
Here is the discussion about the problem and a brief description of my method, please check my 3rd message there.

Answer 4

I'm trying to just keep track of where an index ends up
If you're just trying to follow one entry of the starting array, it's easy to do that in O(M) time.
I was going to just write pseudocode, but no hand-waving was needed so I ended up with what's probably valid C++.
// untested C++, but it does compile to code that looks right.
struct swap {
int l, r;
// or make these non-member functions for C
bool covers(int pos) { return l <= pos && pos <= r; }
int apply_if_covering(int pos) {
// startpos - l = r - endpos;
// endpos = l - startpos + r
if(covers(pos))
pos = l - pos + r;
return pos;
}
};
int follow_swaps (int pos, int len, struct swap swaps[], int num_swaps)
{
// pos = starting position of the element we want to track
// return value = where it will be after all the swaps
for (int i = 0 ; i < num_swaps ; i++) {
pos = swaps[i].apply_if_covering(pos);
}
return pos;
}
This compiles to very efficient-looking code.