CF 1157C2 - Increasing Subsequence (hard version)

CF 1157C2 - Increasing Subsequence (hard version)

Rating: 1700
Tags: greedy
Solve time: 3m 6s
Verified: yes

Solution

Problem Understanding

We are given a sequence of integers and must construct the longest possible strictly increasing sequence by repeatedly taking either the leftmost or the rightmost element. Each time we take a number, it is appended to our growing sequence and removed from the original sequence. The output is twofold: the length of this sequence and a string indicating the sequence of moves, where 'L' represents taking from the left and 'R' represents taking from the right.

The constraints allow sequences up to 200,000 elements. With a 2-second limit, this rules out any solution that examines all possible sequences in exponential time. We need an approach that works in roughly linear or linearithmic time. Each element is between 1 and 200,000, so we can store and compare elements directly without worrying about integer overflow.

Non-obvious edge cases include sequences with repeated numbers. For example, the sequence [2, 2, 2] cannot produce an increasing sequence longer than 1, and the moves must choose only the first element, then stop. Another tricky scenario is when the largest element appears at both ends, for instance [3, 1, 2, 3]. A naive greedy approach that always chooses the smaller of the two ends may produce a suboptimal sequence.

Approaches

The brute-force approach is straightforward. We could recursively try both options at each step: take the leftmost or the rightmost element if it is larger than the last element in our growing sequence. This correctly explores all possible sequences. For each element, there are up to two choices, leading to roughly 2^n possible sequences. With n up to 200,000, this is completely infeasible.

The key insight for an optimal solution comes from the observation that at any point, we only need to consider elements that are strictly larger than the last element we have taken. Since we can only take from the ends, the sequence of candidate moves is highly constrained. We can simulate a greedy approach by comparing the leftmost and rightmost elements and always choosing the smaller one that extends the sequence. If both ends are larger than the last taken element, we explore which side can yield a longer increasing subsequence by looking ahead until a number stops the strictly increasing condition.

Essentially, we reduce the problem to a series of strictly increasing runs from each end. By building these runs and comparing their lengths, we always make the choice that maximizes the number of elements added to our sequence. This avoids recursion and exponential exploration while remaining correct.

Algorithm Walkthrough

1. Initialize two pointers, l at the start of the array and r at the end, and a variable last to track the last element in our sequence, starting with a value smaller than any possible element.
2. Initialize an empty string moves to record the sequence of 'L' and 'R' moves.
3. While l <= r, we examine the elements at both ends. If both a[l] and a[r] are less than or equal to last, we stop since no further element can extend the sequence.
4. If only one of the two ends is strictly larger than last, take that element, update last, move the pointer, and append the corresponding move to moves.
5. If both ends are strictly larger than last, build two candidate sequences: move left repeatedly while the sequence remains strictly increasing, and separately move right in the same way. Compare their lengths.
6. Choose the longer candidate sequence, append the corresponding moves to moves, update last to the last element added, and adjust l or r to reflect all elements taken.
7. Repeat until no elements can be taken.

Why it works: At each step, the algorithm ensures that we only append elements larger than the last. When both ends are valid, the lookahead guarantees we choose the path that maximizes the number of consecutive increasing elements. This invariant-never taking a non-increasing element and choosing the longest run when ambiguous-ensures the overall sequence is strictly increasing and as long as possible.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
a = list(map(int, input().split()))

l, r = 0, n - 1
last = 0
moves = []

while l <= r:
    left, right = a[l], a[r]
    if left <= last and right <= last:
        break
    if left > last and (right <= last or left < right):
        moves.append('L')
        last = left
        l += 1
    elif right > last and (left <= last or right < left):
        moves.append('R')
        last = right
        r -=