Given the following list of $m$ pairs of non decreasing numbers,
$$ L = [(a_0,b_0), (a_1,b_1), \dots, (a_i,b_i), (a_j,b_j), \dots, (a_{m-2},b_{m-2}), (a_{m-1},b_{m-1})]\quad \text{with } a_i \leq a_j \wedge b_i \leq b_j, \forall i < j $$ and the set $S = \{ [l_1,u_1], [l_2,u_2], \dots, [l_n,u_n]\}$, with $l_1 = 0$, $u_n = m - 1$, $l_i \leq u_i \forall i \in [1,n]$, and $u_i < l_{i+1} \forall i \in [1,n)$, whose generic element $[l_i, u_i]$ is associated to the sublist $[(a_{l_i},b_{l_i}), \dots, (a_{u_i},b_{u_i})] \subset L$, determine $S$ such that all the following holds:
- $(a_i = a_{i + 1} \vee b_i = b_{i + 1}) \wedge a_{i + 1} - a_i \leq 1 \wedge b_{i + 1} - b_i \leq 1 \quad \forall i \in [l_j,u_j) \quad \forall j \in [1, n]$, i.e. from a pair in a sublist to the successive pair in the same sublist an increase of at most 1 is permitted at most on one side of the pair;
- $a_{u_i} < a_{l_{i+1}} \wedge b_{u_i} < b_{l_{i + 1}}\quad \forall i \in [1,n)$, i.e. from the last pair of a sublist to the first pair of the following sublist both sides must increase by at least 1;
- $C_S = \sum_{i = 1}^n (u_i - l_i + 1)$ is maximum, i.e. the intervals in $S$ should cover as much of $L$ as possible ($C_S$ is the coverage of $L$ provided by $S$).
Notice that it's not required that $l_{i+1} = u_i + 1 \quad \forall i \in [1,n)$, i.e. it's not necessary that the sublists $S$ covers the whole $L$, they just need to cover as much of it as possible (point 3), as long as also points 1 and 2 hold. This should imply that the solution to the problem exists.
The following are a few examples, where I've numbered the pairs for convenience.
For the input list $$ L = [(2,3)_0,(3,3)_1,(3,5)_2,(3,6)_3,(4,6)_4] $$ The only solution is $S = \{[0,0],[2,4]\}$ which leaves uncovered only the pair $(3,3)_1$; notice that trying to put $(3,3)_1$ in the same sublist as $(2,3)_0$, would imply that $(3,5)_2$ and $(3,6)_3$ should be given up because they have $(3,\cdot)$ in common with $(3,3)_1$, thus resulting in $\bar{S} = \{[0,1], [4,4]\}$, which is not a solution because it has lower coverage than $S$, i.e. $C_\bar{S} = 3 < 4 = C_S$.
On the other hand, the solution is not necessarily unique. For instance, given the list $$ L = [(1,1)_0,(1,2)_1,(2,2)_2,(2,4)_3,(3,4)_4,(3,5)_5] $$ the two solutions $S_1 = \{[0,2],[4,5]\}$ and $S_2 = \{[0,1],[3,5]\}$ are equally valid, as requirements 1 and 2 hold, and the summation in 3 have a value of 5 for both.