How do you solve XOR After Range Multiplication Queries I on LeetCode?

XOR After Range Multiplication Queries I (LeetCode #3653) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + q * m) time complexity and O(n) space complexity. Key insight: Understanding the impact of range updates

What is the time complexity of XOR After Range Multiplication Queries I?

The optimal solution for XOR After Range Multiplication Queries I runs in O(n + q * m) time and uses O(n) space. We process each query in O(m) time where m is the number of indices affected, and we traverse nums once, leading to a more efficient solution.

What is the brute force solution for XOR After Range Multiplication Queries I?

The brute force approach for XOR After Range Multiplication Queries I is Brute Force, with O(n²) time complexity and O(1) space complexity. We directly apply each query to the array by iterating through the specified indices and updating the values. This straightforward method is easy to understand but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n + q * m).

What algorithm or data structure is used to solve XOR After Range Multiplication Queries I?

XOR After Range Multiplication Queries I uses the following concepts: Array, Divide and Conquer, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for XOR After Range Multiplication Queries I?

Clarify the problem requirements Discuss edge cases like empty arrays Explain your thought process while coding

What are common mistakes when solving XOR After Range Multiplication Queries I?

Not considering the modulo operation Incorrectly updating indices in the loop

#3653

XOR After Range Multiplication Queries I

Medium

ArrayDivide and ConquerSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + q * m)
Space	O(1)	O(n)

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Intuition

Time O(n + q * m)Space O(n)

Instead of applying each query directly, we can keep track of the multipliers for each index and apply them in a single pass after all queries are processed. This reduces unnecessary repeated calculations.

⚙️

Algorithm

3 steps

1Step 1: Create an array to store multipliers initialized to 1.
2Step 2: For each query, update the multiplier for the indices specified by l, r, and k.
3Step 3: After processing all queries, multiply each element in nums by its corresponding multiplier and compute the XOR.

solution.py10 lines

1def xor_after_queries(nums, queries):
2    multipliers = [1] * len(nums)
3    for l, r, k, v in queries:
4        for idx in range(l, r + 1, k):
5            multipliers[idx] = (multipliers[idx] * v) % (10**9 + 7)
6    result = 0
7    for i in range(len(nums)):
8        nums[i] = (nums[i] * multipliers[i]) % (10**9 + 7)
9        result ^= nums[i]
10    return result

ℹ

Complexity note: We process each query in O(m) time where m is the number of indices affected, and we traverse nums once, leading to a more efficient solution.

1Understanding the impact of range updates
2Efficiently managing multipliers reduces complexity

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.