Consider an array $X$ that contains $n$ positive integers. A subarray of $X$ is defined to be a sequence of array locations with consecutive indices. The C code snippet given below has been written to compute the length of the longest subarray of $X$ that contains at most two distinct integers. The code has two missing expressions labelled $(P)$ and $(Q)$.int first=0, second=0, len1=0, len2=0, maxlen=0; for (int i=0; i < n; i++) { if (X[i] == first) { len2++; len1++; } else if (X[i] == second) { len2++; len1 = (P) ; second = first; } else { len2 = (Q) ; len1 = 1; second = first; } if (len2 > maxlen) { maxlen = len2; } first = X[i]; }Which one of the following options gives the CORRECT missing expressions? (Hint: At the end of the $i$-th iteration, the value of len1 is the length of the longest subarray ending with $X[i]$ that contains all equal values, and len2 is the length of the longest subarray ending with $X[i]$ that contains at most two distinct values.)

At each iteration $i$:

• first stores the value of the immediately preceding element, $X[i-1]$.

• second stores the other distinct integer present in the current two-distinct-element window.

• len1 tracks the length of the continuous run of identical elements ending at $X[i]$.

• len2 tracks the length of the longest contiguous subarray ending at $X[i]$ with at most 2 distinct values.

Let us analyze the three branches:

1. Case 1 (X[i] == first): $X[i]$ extends the current streak of identical elements as well as the 2-distinct window. Both len1 and len2 increment by 1.

2. Case 2 (X[i] == second): $X[i]$ is one of the two allowed values, so the 2-distinct window extends (len2++). However, since $X[i] \neq \text{first}$, a new continuous run of identical elements starts with this single element $X[i]$. Therefore, the new len1 must be initialized to $1$. Hence, $(P) = 1$.

3. Case 3 (else): A third distinct element is encountered. The new valid 2-distinct subarray ending at $X[i]$ can only consist of the trailing contiguous run of first elements (whose count is stored in len1 from the previous iteration) plus the new element $X[i]$. Thus, the new len2 becomes $\text{len1} + 1$. Hence, $(Q) = \text{len1} + 1$.

Therefore, $(P) = 1$ and $(Q) = \text{len1} + 1$, which corresponds to Option (B).