0921MediumArrayDivide and ConquerSimulationPrefix Sum

Combining Meters After Stepped Scaling

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Trains the technique from

LeetCode 3653XOR After Range Multiplication Queries I

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Meter readings are given as nums, and a list of adjustments as queries. Each adjustment is [l, r, k, v] and multiplies, modulo 10^9 + 7, the readings at positions l, l + k, l + 2k and so on, taking every such position up to and including r.

Apply the adjustments in the order given, then return the bitwise exclusive-or of the whole list of readings.

Examples

Example 1

Input: nums = [13, 27, 41, 6, 19], queries = [[0, 4, 2, 3], [1, 3, 1, 7]]
Output: 980

The first adjustment strides by two from position 0, so it triples the readings at 0, 2 and 4. The second strides by one from position 1 through 3, so it multiplies those three by seven. Combining the finished readings bitwise gives this.

Example 2

Input: nums = [5, 5, 5, 5, 5, 5], queries = [[0, 5, 3, 2]]
Output: 0

Striding by three from position 0 reaches positions 0 and 3 only, doubling those two readings and leaving the other four alone.

Example 3

Input: nums = [2, 3, 5, 7, 11, 13, 17], queries = [[0, 6, 1, 1]]
Output: 20

Multiplying by one changes nothing, so the answer is the readings combined as they came.

Constraints

1 <= nums.length <= 1000
1 <= nums[i] <= 10^9
1 <= queries.length <= 1000
queries[i].length == 4
0 <= queries[i][0] <= queries[i][1] <= nums.length - 1
1 <= queries[i][2] <= nums.length
1 <= queries[i][3] <= 10^5

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def xor_after_queries(nums: list[int], queries: list[list[int]]) -> int:

Java

public int xorAfterQueries(int[] nums, int[][] queries)

The intended approach

Time O(n + sum over adjustments of the positions each touches), which is at most O(n * q) for n readings and q adjustments.Space O(1) beyond the readings themselves.

The adjustments cannot be reordered or merged, because multiplying under a remainder is applied to whatever the reading currently is, so the readings have to be carried forward as they change. What keeps this affordable is that an adjustment touches only the positions on its own step, so walking from `l` in strides of `k` visits exactly those and no others, costing the number of positions touched rather than the length of the list. Reduce after each multiplication so the readings stay inside the modulus, then exclusive-or the finished list in one pass. The exclusive-or is taken on the reduced readings, which is what the question asks for.

Pythonentry point: xor_after_queries

def xor_after_queries(nums, queries):
    MOD = 1000000007
    values = list(nums)
    n = len(values)
    for l, r, k, v in queries:
        at = l
        while at <= r:
            values[at] = values[at] * v % MOD
            at += k
    total = 0
    for x in values:
        total ^= x
    return total

The obvious version

For each adjustment, walk the whole list and pick out the positions it touches by testing every index against the range and the step, then multiply those. Costs the full length per adjustment and never strides.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_xor_after_queries(nums, queries):
    MOD = 1000000007
    values = list(nums)
    n = len(values)
    for query in queries:
        l = query[0]
        r = query[1]
        k = query[2]
        v = query[3]
        touched = []
        for i in range(n):
            if l <= i <= r and (i - l) % k == 0:
                touched.append(i)
        for i in touched:
            values[i] = values[i] * v % MOD
    total = 0
    for x in values:
        total ^= x
    return total

How this problem was checked

Reference solution executed against 11 test cases — 11 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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