Binary Strings and Blocks

First, kill the scary definition

Let a binary string have $k$ blocks.

If you remove an end block, the remaining string has $k-1$ blocks.

If you remove an internal block, its left and right neighboring blocks have the same character, because binary blocks alternate. So after removal those two neighbors merge, and the remaining string has $k-2$ blocks.

Now check when we can get an odd number of blocks:

• $k=1$: removing the only block gives $0$ blocks, so not beautiful.
• even $k \ge 2$: remove an end block, getting $k-1$, which is odd.
• odd $k \ge 3$: remove an internal block, getting $k-2$, which is odd.

So a binary string is beautiful exactly when it has at least two blocks. In normal human language: it must contain both 0 and 1. That whole block-removal circus boils down to “not constant”. Classic Codeforces nonsense, but useful nonsense.

Therefore, a constraint $s[l:r]$ is beautiful iff the substring is not constant.

That means inside positions $l,l+1,\dots,r$, there must be at least one place where adjacent characters differ.

Turn strings into change arrays

Define

$$x_i = [s_i \ne s_{i+1}] \quad \text{for } 1 \le i < n.$$

Then $x$ is a binary array of length $n-1$.

A substring $s[l:r]$ is constant iff all adjacent differences inside it are zero:

$$x_l=x_{l+1}=\dots=x_{r-1}=0.$$

So the constraint $[l,r]$ becomes:

$$\text{there is at least one } 1 \text{ in } x[l:r-1].$$

In other words, every given interval $[l,r-1]$ on the change array must be hit by at least one selected position where $x_i=1$.

Once we count valid change arrays $x$, each one corresponds to exactly two original strings: choose $s_1$ as 0 or 1, then every $x_i$ determines whether the next bit flips.

So the final answer is:

$$2 \cdot \#\{\text{valid } x\}.$$

Counting valid change arrays

Now the problem is:

Given intervals on positions $1..n-1$, count binary arrays where every interval contains at least one 1.

Let’s build the array from left to right and track the position of the last 1.

Suppose we are currently at position $i$ in the change array. For every constraint interval ending at $i$, say $[L,i]$, we need the last 1 position to be at least $L$.

If several intervals end at $i$, we only care about the maximum left endpoint among them:

$$need[i] = \max L \text{ over all intervals } [L,i].$$

While processing prefix $1..i$, after applying constraints ending at $i$, the last 1 must be at least:

$$bound_i = \max(bound_{i-1}, need[i]).$$

This is because old constraints still matter. Once the last 1 is too far left, you’re cooked.

DP state

Let $dp[j]$ mean:

After processing the current prefix, the last 1 is exactly at position $j$.

We also allow $j=0$, meaning there has been no 1 yet.

At position $i$, we choose $x_i$:

• If $x_i=0$, the last 1 stays the same.
• If $x_i=1$, the last 1 becomes $i$.

After that, states with last `1 < bound_i$ are invalid.

A direct DP would be too slow if we move all states every time. But notice the transition is simple:

• choosing 0 keeps all existing states;
• choosing 1 creates state $i$ from the sum of all valid previous states.

So we maintain an array of DP values by last-one position and a running total of all currently valid states.

At each position $i$:

1. Set $dp[i] = total$, because any valid prefix can append 1.
2. Add it to total.
3. Increase the required lower bound using constraints ending here.
4. Remove all states $j < bound$ from total.

Order matters a bit, but constraints ending at $i$ apply after choosing $x_i$, and $x_i=1$ at position $i$ is allowed to satisfy them. So adding $dp[i]$ before deleting invalid old states is perfectly fine.

Initially, before processing anything, there is one empty prefix with no 1:

$$dp[0]=1, \quad total=1.$$

At the end, total is the number of valid change arrays.

Multiply by $2$ for the initial bit of $s$.

Why this works

Every constraint $[l,r]$ in the original string becomes an interval $[l,r-1]$ in the change array.

For an interval ending at $i$, requiring at least one 1 inside it is exactly requiring:

$$lastOne \ge L.$$

Taking the max over all constraints seen so far gives one lower bound on lastOne. The DP stores exactly how many prefixes have each possible lastOne, and the running total stores only states that still satisfy all constraints processed so far.

So no constraint is missed, and no invalid array sneaks through. Very satisfying. Slightly annoying that the original statement made us do interpretive dance first, but whatever.

Complexity

For each test case, we process each constraint once and each change-array position once.

Time complexity:

$$O(n+m)$$

Memory complexity:

$$O(n)$$