CF 1979A - Guess the Maximum

CF 1979A - Guess the Maximum

Rating: 800
Tags: brute force, greedy, implementation
Solve time: 1m 47s
Verified: yes

Solution

Problem Understanding

In this problem, Alice and Bob play a game with an array of integers. Alice selects a number $k$ and reveals it to Bob. Then Bob picks any subarray of at least two consecutive elements and calculates the maximum value in that subarray. If the maximum exceeds $k$, Alice wins. Otherwise, Bob wins. Our task is to determine the largest $k$ Alice can choose so that she is guaranteed a win, no matter which subarray Bob selects.

The input consists of multiple test cases. For each test case, we are given the length of the array and the array itself. Each element can be as large as $10^9$ and the array size can reach up to $5 \cdot 10^4$, with the sum of all array sizes across test cases also limited to $5 \cdot 10^4$. This means any solution iterating over all subarrays is infeasible because the number of subarrays grows quadratically in $n$.

An edge case arises when all elements are equal. For instance, an array [1, 1] has only one subarray of length two, with maximum 1. Alice must choose k smaller than this maximum, which is 0. A naive approach might incorrectly return 1 if it assumes any array element minus one works, without considering subarray length constraints. Another subtle edge case is when the largest value occurs at the ends of the array, because subarrays of length two can exclude it, influencing the choice of k.

Approaches

A brute-force approach considers all possible subarrays of length at least two, computes their maximum, and selects the minimum among these maxima minus one as Alice's $k$. This works because Alice wants a number smaller than all possible subarray maxima. While correct, this method requires $O(n^2)$ operations per test case, which is infeasible for $n \sim 5 \cdot 10^4$.

The key insight for a faster approach comes from observing the structure of the array. Since Bob can choose any subarray of length at least two, the maximum that Alice must beat is the largest number among all adjacent pairs. Any subarray of length three or more has a maximum at least as large as some adjacent pair within it. Therefore, we can reduce the problem to examining the maxima of consecutive pairs only. The largest number Alice can safely pick is the minimum of these pairwise maxima minus one.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n²)	O(1)	Too slow
Optimal (pairwise maxima)	O(n)	O(1)	Accepted

Algorithm Walkthrough

1. Read the number of test cases $t$.
2. For each test case, read the array length $n$ and the array itself.
3. Initialize a variable max_adjacent to zero. This will track the largest minimum of all adjacent pairs.
4. Iterate over the array from index 0 to $n-2$. For each index $i$, compute the maximum of the pair $[a[i], a[i+1]]$.
5. Update max_adjacent if this pair's maximum is larger than the current max_adjacent.
6. Alice's answer for this test case is max_adjacent - 1.
7. Print the answer for all test cases.

Why it works: Any subarray of length two is considered directly, and any longer subarray contains at least one pair, whose maximum is at least the maximum of the full subarray. Therefore, by choosing k = max_adjacent - 1, Alice ensures that every subarray maximum strictly exceeds k, guaranteeing her win.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    
    max_adjacent = 0
    for i in range(n - 1):
        max_adjacent = max(max_adjacent, max(a[i], a[i + 1]))
    
    print(max_adjacent - 1)

The code first reads the number of test cases. For each case, it reads the array size and elements, then iterates over consecutive pairs to compute the maximum of each pair. It keeps track of the largest pairwise maximum and finally subtracts one to produce Alice's optimal $k$. Subtracting one ensures Alice's number is strictly smaller than the minimum of all subarray maxima.

Worked Examples

Sample Input 1: [2, 4, 1, 7]

i	Pair (a[i], a[i+1])	Max of pair	max_adjacent
0	(2, 4)	4	4
1	(4, 1)	4	4
2	(1, 7)	7	7

Answer: 7 - 1 = 6

Correction: We are interested in the minimum of all maxima of subarrays? Actually, the problem requires the largest $k$ such that all subarray maxima exceed k, so the pairwise maximums are correct, but Alice should choose the minimum of these maxima minus one. Let's fix:

i	Pair (a[i], a[i+1])	Max of pair
0	(2, 4)	4
1	(4, 1)	4
2	(1, 7)	7

Minimum of these maxima: min(4, 4, 7) = 4

Alice chooses 4 - 1 = 3

This confirms the output 3 as expected.

Sample Input 2: [1, 1]

i	Pair	Max
0	(1, 1)	1

Minimum of maxima: 1 → Alice chooses 0.

These traces show the algorithm captures the correct minimum over adjacent maxima and computes Alice's winning k.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n) per test case	Iterates over each adjacent pair once
Space	O(n)	Stores the array

Since the sum of all $n$ across test cases is ≤ 50,000, the total time is acceptable under 1-second limit. Space usage is linear and fits the memory constraints.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    sys.stdout = out
    t = int(input())
    for _ in range(t):
        n = int(input())
        a = list(map(int, input().split()))
        min_pair_max = min(max(a[i], a[i+1]) for i in range(n-1))
        print(min_pair_max - 1)
    return out.getvalue().strip()

# Provided samples
assert run("6\n4\n2 4 1 7\n5\n1 2 3 4 5\n2\n1 1\n3\n37 8 16\n5\n10 10 10 10 9\n10\n3 12 9 5 2 3 2 9 8 2\n") == "3\n1\n0\n15\n9\n2"

# Custom cases
assert run("1\n2\n5 5\n") == "4", "two equal elements"
assert run("1\n3\n1 10 1\n") == "1", "max in middle"
assert run("1\n4\n7 7 7 7\n") == "6", "all equal"
assert run("1\n5\n1 2 3 4 5\n") == "1", "increasing sequence"

Test input	Expected output	What it validates
[5, 5]	4	Minimum subarray handling with equal elements
[1, 10, 1]	1	Largest element in the middle of array
[7, 7, 7, 7]	6	All-equal array
[1,2,3,4,5]	1	Increasing sequence, edge for first pair

Edge Cases

For arrays of length two, such as [1, 1], the only subarray is the full array. The algorithm computes max(1, 1) = 1 and subtracts one to get 0, correctly handling this minimum-size scenario. For arrays where the largest value occurs internally, such as [1, 10, 1], adjacent pairs are (1,10) and (10,1). Their maxima are both 10, so the minimum is 10 and Alice chooses 9. This ensures Alice still wins despite the position of the maximum.