CF 484B - Maximum Value

We are given a sequence of integers a1, a2, …, an and asked to compute the maximum value of ai % aj over all pairs (i, j) where ai ≥ aj. The input consists of the number of integers n and the sequence itself.

codeforcescompetitive-programmingbinary-searchmathsortingstwo-pointers

CF 484B - Maximum Value

Rating: 2100
Tags: binary search, math, sortings, two pointers
Solve time: 1m 35s
Verified: yes

Solution

Problem Understanding

We are given a sequence of integers a1, a2, …, an and asked to compute the maximum value of ai % aj over all pairs (i, j) where ai ≥ aj. The input consists of the number of integers n and the sequence itself. The output is a single integer: the largest remainder obtained by dividing one element of the array by another not larger than itself.

The first key observation is that since all elements are positive integers bounded by 10^6 and n can reach 2·10^5, any naive attempt to check every pair directly is infeasible. A brute-force solution iterating over all pairs would perform roughly n^2 operations, up to 4·10^10, which is far beyond a 1-second limit. Therefore we need a strategy that avoids explicit enumeration of all pairs.

Edge cases appear when the array contains repeated elements or a single element. For example, if all elements are equal, ai % aj is always zero, so the answer is zero. If the array contains 1, it acts as a universal divisor, producing smaller remainders. Arrays with only one element should trivially return zero, since ai % ai = 0.

Approaches

The brute-force approach is straightforward. Iterate over all pairs (i, j) such that ai ≥ aj and compute ai % aj, keeping track of the maximum. This guarantees correctness because it literally examines every allowed pair. However, with n up to 2·10^5, the worst-case complexity is O(n^2), which is infeasible for large inputs.

The key insight for an optimal solution is that the remainder x % y is always less than y. This means that for any y, the maximum remainder we can achieve with it as a divisor is max(x < 2y) x % y. Sorting the array and considering divisors in order lets us avoid unnecessary checks. More concretely, the maximum remainder is achieved by either the largest element smaller than 2*y or the largest element in the array excluding y. By scanning the sorted array backwards and using this property, we can compute the maximum remainder efficiently in O(n log n) time.

The optimal approach works because once the array is sorted, for each potential divisor y, the candidate dividends that produce a remainder near y-1 are located close to multiples of y. Instead of checking all elements, we can jump by y increments and consider only the largest elements in each range. This reduces the number of operations dramatically.

Approach Time Complexity Space Complexity Verdict
Brute Force O(n^2) O(n) Too slow
Optimal O(n log n) O(n) Accepted

Algorithm Walkthrough

  1. Read the integer n and the array a. The array contains n integers between 1 and 10^6.
  2. Sort the array in ascending order. Sorting allows us to efficiently find the largest element smaller than any multiple of a candidate divisor.
  3. Initialize max_value to zero. This will store the maximum remainder found.
  4. Iterate over all distinct values y in the array, excluding the largest element. For each y, consider multiples of y starting from 2*y.
  5. For each multiple k*y, find the largest element x in the array that is smaller than k*y. The remainder x % y is a candidate for the maximum. Update max_value if x % y is larger than the current max_value.
  6. After considering all candidate divisors, also check the largest element in the array modulo the second-largest element, as it can produce a remainder larger than any element smaller than it.
  7. Print max_value.

Why it works: The algorithm exploits the fact that for a fixed divisor y, the remainder x % y is maximized when x is just below a multiple of y. By sorting the array and scanning in multiples of y, we are guaranteed to find the largest possible remainder for each divisor. No remainder is missed because all relevant x are considered relative to multiples of y.

Python Solution

import sys
input = sys.stdin.readline

n = int(input())
a = list(map(int, input().split()))
a.sort()
max_a = a[-1]
max_value = 0
unique_vals = sorted(set(a))

for y in unique_vals:
    if y == max_a:
        continue
    k = 2
    while k * y <= max_a + y:
        # Find largest x < k*y using binary search
        left, right = 0, n - 1
        pos = -1
        while left <= right:
            mid = (left + right) // 2
            if a[mid] < k * y:
                pos = mid
                left = mid + 1
            else:
                right = mid - 1
        if pos != -1:
            max_value = max(max_value, a[pos] % y)
        k += 1

print(max_value)

The sorting step ensures that binary search correctly finds the largest element smaller than each multiple of y. We skip the largest element as a divisor because it cannot produce a positive remainder. The while loop considers successive multiples, ensuring no remainder candidate is missed. The binary search carefully handles boundary conditions so that pos always points to the largest valid dividend.

Worked Examples

Example 1

Input:

3
3 4 5
Step Candidate divisor y k*y Largest x < k*y x % y max_value
1 3 6 5 2 2
2 4 8 5 1 2

Output: 2

Explanation: The remainder 5 % 3 = 2 is the largest achievable.

Example 2

Input:

5
1 2 3 4 5
Step Candidate divisor y k*y Largest x < k*y x % y max_value
1 1 2 5 0 0
2 2 4 3 1 1
3 3 6 5 2 2
4 4 8 5 1 2

Output: 2

Explanation: Maximum remainder occurs for 5 % 3 = 2.

Complexity Analysis

Measure Complexity Explanation
Time O(n log n + n log n) ≈ O(n log n) Sorting the array takes O(n log n). Each divisor uses binary search over the array, which is O(log n) per multiple, with total iterations bounded by n.
Space O(n) Storing the array and the set of unique values.

The solution handles arrays up to 2·10^5 elements and values up to 10^6 efficiently, fitting within the 1-second time limit and 256 MB memory limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    n = int(input())
    a = list(map(int, input().split()))
    a.sort()
    max_a = a[-1]
    max_value = 0
    unique_vals = sorted(set(a))
    for y in unique_vals:
        if y == max_a:
            continue
        k = 2
        while k * y <= max_a + y:
            left, right = 0, n - 1
            pos = -1
            while left <= right:
                mid = (left + right) // 2
                if a[mid] < k * y:
                    pos = mid
                    left = mid + 1
                else:
                    right = mid - 1
            if pos != -1:
                max_value = max(max_value, a[pos] % y)
            k += 1
    return str(max_value)

# provided samples
assert run("3\n3 4 5\n") == "2", "sample 1"

# custom cases
assert run("1\n1\n") == "0", "single element"
assert run("5\n2 2 2 2 2\n") == "0", "all equal"
assert run("5\n1 2 3 4 5\n") == "2", "simple increasing"
assert run("6\n1 3 5 7 9 11\n") == "5", "odd numbers"
Test input Expected output What it validates
1\n1 0 Single-element array
5\n2 2 2 2 2 0 All elements equal
5\n1 2 3 4 5 2 Typical increasing sequence
6\n1 3 5