LeetCode 3525 Elixir solution

LeetCode 3525. Find X Value of Array II solution in Elixir using Segment Tree, Prefix Product, and Modular Arithmetic patterns. Each tree node stores the segment product modulo k and prefix counts for every possible remainder. Point updates modify the array efficiently, while range queries over [start, n - 1] return the number of prefixes whose product modulo k equals x. Since k <= 5, each merge takes constant time. Time complexity is O(n + q * log n), Space complexity is O(n).

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3525 LeetCode problem solution

LeetCode 3525. Find X Value of Array II solution using Segment Tree, Prefix Product, and Modular Arithmetic patterns. Each segment tree node stores the segment product modulo k and counts of prefix products for every possible remainder. For each query, we perform a point update and query the range [start, n - 1], where cnt[x] gives the required x-value. The merge operation combines prefix-product counts while preserving their left-to-right order. Time complexity: O(n + q log n), Space complexity: O(n), since k <= 5.

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LeetCode 3524 Elixir solution

LeetCode 3524 Find X Value of Array I uses Dynamic Programming and Modular Arithmetic technique to count subarrays by their product remainder. Instead of calculating full products, the solution keeps only k remainder states and updates them for each element. With k <= 5, the algorithm runs in O(n * k) time and uses O(k) extra space.

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3524 LeetCode problem solution

LeetCode 3524. Find X Value of Array I uses the Dynamic Programming and Modular Arithmetic technique to count subarrays by their product remainder modulo k. Since k <= 5, we can maintain counts for all possible remainders while extending subarrays with each new element. The Time complexity is O(n * k), and the Space complexity is O(k).

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3498 LeetCode problem solution

LeetCode 3498 Reverse Degree of a String problem solution using String Traversal and Character Mapping to compute reversed alphabet values in O(n) time and O(1) space complexity.

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