CF 1903D1 - Maximum And Queries (easy version)

Rating: 1700
Tags: binary search, bitmasks, brute force, greedy
Solve time: 2m 15s
Verified: no

Solution

Problem Understanding

We are given an array of integers and a budget of operations, where each operation allows increasing a single element by 1. For each query, which provides a different budget, we want the largest value achievable by taking the bitwise AND of all array elements after using at most that many operations. The array itself does not change between queries, so each query is independent.

The input sizes are substantial: the array can have up to 100,000 elements, and there can be up to 100,000 queries. The total product of n * q is bounded at 100,000, which means that we cannot afford an algorithm that examines all elements for every query in a nested loop. Each element can be as large as 10^6, so the bitwise representation has at most 20 bits. This suggests that bit-level manipulation is feasible and likely necessary. The number of operations k can be enormous, up to 10^18, so we cannot simulate each increment individually.

A naive approach might attempt to try every possible way to spend the operations, or even try all possible resulting arrays. For example, if the array is [1, 2, 3] and k=3, one might try all combinations of adding up to 3 total units across the elements. This explodes combinatorially and is infeasible even for small n.

An important subtlety is that some elements may already have high bits set, while others need significant increases to match. For instance, consider [1, 2, 7] with k=1. The maximum AND is 0, because increasing any single element by 1 cannot make all elements share a common high bit. A careless solution that just sums the increments or tries greedy additions from left to right may report the wrong AND. We must carefully evaluate which bits can be “turned on” across all elements under the given budget.

Approaches

The brute-force solution would attempt every combination of distributing k operations to maximize the AND. This is correct in principle because if we could try all possibilities, we would eventually find the best. But for n=10^5 and k up to 10^18, this is impossible. Even iterating over all array elements per query without clever logic would give O(n*q) = 10^10 operations in the worst case.

The key insight is to think in terms of bits rather than absolute numbers. The bitwise AND of numbers is dominated by the positions of the highest bits that are set in all elements. If we want to make the AND as large as possible, we can check bits from the highest down and see whether we can increase elements to set that bit using the available operations. For each bit, we calculate the total cost to raise all numbers that do not already have the bit set to at least that value. If the cost is within the current query's k, we keep the bit in the final AND.

This observation reduces the problem from tracking exact numbers to a greedy, bit-by-bit check from high to low. The total number of bits is small (20 for numbers up to 10^6), so for each query, we only need to evaluate at most 20 bits across all elements. This yields a feasible O(n * 20) per query, which is acceptable given the n*q ≤ 10^5 constraint.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(k * n)	O(n)	Too slow
Bitwise Greedy	O(n * 20) per query	O(n)	Accepted

Algorithm Walkthrough

Sort the bits of interest in descending order, starting from the highest bit possible in any element (20th bit for numbers ≤10^6). We handle bits one at a time from the most significant to the least significant because higher bits contribute more to the AND.
Initialize the answer to 0. This variable will store the AND we can achieve with the current budget.
For each bit, compute the cost to set that bit in all elements. For each element that does not already have the bit set, calculate how much we need to add to turn on that bit without changing higher bits that are already set. This is (1 << bit) - (a[i] & (1 << bit - 1)) simplified to the amount needed to reach the next number where this bit is 1.
If the total cost for the current bit is less than or equal to k, deduct this cost from k and include the bit in the answer using bitwise OR.
Proceed to the next lower bit and repeat the process until all bits have been evaluated.
Output the final answer for the query.

Why it works: The algorithm maintains the invariant that at each step, the current answer represents the highest AND achievable using the operations spent so far. By starting from the highest bit and only committing to bits that can be turned on for all elements, we ensure no opportunity is missed, and no bit is set when it cannot be applied to all elements. This guarantees that the result is the true maximum AND achievable under the given budget.

Python Solution

import sys
input = sys.stdin.readline

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

queries = [int(input()) for _ in range(q)]

max_bit = 0
for num in a:
    max_bit = max(max_bit, num.bit_length())

for k in queries:
    ans = 0
    remaining = k
    for bit in reversed(range(max_bit + 1)):
        cost = 0
        mask = 1 << bit
        for num in a:
            if not (num & mask):
                increment = mask - (num & (mask - 1))
                cost += increment
        if cost <= remaining:
            remaining -= cost
            ans |= mask
    print(ans)

The solution first calculates the highest relevant bit across all elements. Then for each query, it iterates from high to low bits, computing how much it costs to turn that bit on for all elements. Only if the total cost fits in the query budget k do we update the answer. A subtle point is computing the increment correctly for numbers that already have lower bits set, which avoids accidentally overshooting.

Worked Examples

Sample 1

Input:

Step	Bit	Remaining k	Cost	Include?	Answer
1	2	2	2	Yes	2
2	1	0	1	No	2
3	0	0	0	Yes	2

Second query k=10:

Step	Bit	Remaining k	Cost	Include?	Answer
1	2	10	2	Yes	4
2	1	8	2	Yes	6
3	0	6	0	Yes	6

This demonstrates that higher bits are considered first, and we only include them if the budget allows, yielding the correct maximum AND.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * max_bit * q) ≈ O(n * 20 * q)	Each query iterates over all bits and all elements.
Space	O(n + q)	Storing the array and the queries.

Since n * q ≤ 10^5 and max_bit ≤ 20, this algorithm easily fits in the 2-second time limit and uses modest memory.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    # solution call
    n, q = map(int, input().split())
    a = list(map(int, input().split()))
    queries = [int(input()) for _ in range(q)]
    max_bit = max(num.bit_length() for num in a)
    for k in queries:
        ans = 0
        remaining = k
        for bit in reversed(range(max_bit + 1)):
            cost = 0
            mask = 1 << bit
            for num in a:
                if not (num & mask):
                    increment = mask - (num & (mask - 1))
                    cost += increment
            if cost <= remaining:
                remaining -= cost
                ans |= mask
        print(ans)
    return output.getvalue().strip()

# Provided sample
assert run("4 2\n1 3 7 5\n2\n10\n") == "2\n6"

# Custom: minimum input
assert run("1 1\n0\n0\n") == "0"

# All elements equal
assert run("3 1\n7 7 7\n5\n") == "7"

# Large k to max all bits
assert run("3 1\n1 2 3\n100\n") == "3"

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