Find X Value of Array II - Problem
Find X Value of Array II
Imagine you're managing a dynamic product catalog where you need to analyze different product combinations and their modular arithmetic properties after various updates and filtering operations.
Given an array
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For each query, you must:
1. Update
2. Remove prefix from index 0 to
3. Calculate x-value: Count how many ways you can remove a suffix such that the product of remaining elements ≡
The x-value represents the number of valid suffix removals that result in the desired modular product. Return an array where each element is the x-value for the corresponding query.
Imagine you're managing a dynamic product catalog where you need to analyze different product combinations and their modular arithmetic properties after various updates and filtering operations.
Given an array
nums of positive integers and a positive integer k, you need to process multiple queries. Each query contains:•
index_i: Position to update•
value_i: New value at that position•
start_i: Starting position for analysis•
x_i: Target remainder valueFor each query, you must:
1. Update
nums[index_i] to value_i (persists for subsequent queries)2. Remove prefix from index 0 to
start_i - 13. Calculate x-value: Count how many ways you can remove a suffix such that the product of remaining elements ≡
x_i (mod k)The x-value represents the number of valid suffix removals that result in the desired modular product. Return an array where each element is the x-value for the corresponding query.
Input & Output
example_1.py — Basic Query Processing
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Input:
nums = [2, 3, 4], k = 6, queries = [[1, 5, 0, 2]]
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Output:
[1]
💡 Note:
After updating nums[1] = 5, array becomes [2, 5, 4]. Starting from index 0, we check suffixes: [2] has product 2 ≡ 2 (mod 6) ✓, [2,5] has product 10 ≡ 4 (mod 6), [2,5,4] has product 40 ≡ 4 (mod 6). Only one way gives remainder 2.
example_2.py — Multiple Queries
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Input:
nums = [1, 2, 3], k = 5, queries = [[0, 4, 1, 1], [2, 6, 0, 4]]
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Output:
[0, 1]
💡 Note:
Query 1: Update nums[0]=4, start from index 1, array becomes [2,3], check remainder 1. Products: 2≡2, 6≡1. Count=1. Wait, that's wrong. Let me recalculate: [2] gives 2 mod 5, [2,3] gives 6≡1 mod 5, so count=1. Query 2: nums[2]=6, start from 0, array [4,2,6]. Products: 4≡4✓. Count=1.
example_3.py — Edge Case
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Input:
nums = [1], k = 3, queries = [[0, 2, 0, 2]]
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Output:
[1]
💡 Note:
Single element array. Update nums[0]=2, starting from index 0. Only one suffix possible: [2], which gives product 2 ≡ 2 (mod 3). Matches target x=2, so count=1.
Constraints
- 1 ≤ nums.length ≤ 105
- 1 ≤ nums[i] ≤ 109
- 1 ≤ k ≤ 109
- 1 ≤ queries.length ≤ 105
- 0 ≤ index_i < nums.length
- 1 ≤ value_i ≤ 109
- 0 ≤ start_i ≤ nums.length
Visualization
Tap to expand
Understanding the Visualization
1
Update Equipment
Modify the specified station's output value
2
Skip Initial Stations
Remove the prefix - start production from a specific station
3
Test All Stop Points
For each possible stopping point, check if the total production meets quality target
4
Count Valid Runs
Sum up all configurations that achieve the desired quality score (modulo k)
Key Takeaway
🎯 Key Insight: Use prefix products to transform O(N²) repeated multiplications into O(N) direct lookups, dramatically improving performance for multiple queries.
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Explanation
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