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MD ARIFUL HAQUE
MD ARIFUL HAQUE

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3524. Find X Value of Array I

#php #leetcode #algorithms #programming

3524. Find X Value of Array I

Difficulty: Medium

Topics: Staff, Array, Math, Dynamic Programming, Weekly Contest 446

You are given an array of positive integers nums, and a positive integer k.

You are allowed to perform an operation once on nums, where in each operation you can remove any non-overlapping prefix and suffix from nums such that nums remains non-empty.

You need to find the x-value of nums, which is the number of ways to perform this operation so that the product of the remaining elements leaves a remainder of x when divided by k.

Return an array result of size k where result[x] is the x-value of nums for 0 <= x <= k - 1.

A prefix of an array is a subarray1 that starts from the beginning of the array and extends to any point within it.

A suffix of an array is a subarray1 that starts at any point within the array and extends to the end of the array.

Note that the prefix and suffix to be chosen for the operation can be empty.

Example 1:

  • Input: nums = [1,2,3,4,5], k = 3
  • Output: [9,2,4]
  • Explanation:
    • For x = 0, the possible operations include all possible ways to remove non-overlapping prefix/suffix that do not remove nums[2] == 3.
    • For x = 1, the possible operations are:
      • Remove the empty prefix and the suffix [2, 3, 4, 5]. nums becomes [1].
      • Remove the prefix [1, 2, 3] and the suffix [5]. nums becomes [4].
    • For x = 2, the possible operations are:
      • Remove the empty prefix and the suffix [3, 4, 5]. nums becomes [1, 2].
      • Remove the prefix [1] and the suffix [3, 4, 5]. nums becomes [2].
      • Remove the prefix [1, 2, 3] and the empty suffix. nums becomes [4, 5].
      • Remove the prefix [1, 2, 3, 4] and the empty suffix. nums becomes [5].

Example 2:

  • Input: nums = [1,2,4,8,16,32], k = 4
  • Output: [18,1,2,0]
  • Explanation:
    • For x = 0, the only operations that do not result in x = 0 are:
      • Remove the empty prefix and the suffix [4, 8, 16, 32]. nums becomes [1, 2].
      • Remove the empty prefix and the suffix [2, 4, 8, 16, 32]. nums becomes [1].
      • Remove the prefix [1] and the suffix [4, 8, 16, 32]. nums becomes [2].
    • For x = 1, the only possible operation is:
      • Remove the empty prefix and the suffix [2, 4, 8, 16, 32]. nums becomes [1].
    • For x = 2, the possible operations are:
      • Remove the empty prefix and the suffix [4, 8, 16, 32]. nums becomes [1, 2].
      • Remove the prefix [1] and the suffix [4, 8, 16, 32]. nums becomes [2].
    • For x = 3, there is no possible way to perform the operation.

Example 3:

  • Input: nums = [1,1,2,1,1], k = 2
  • Output: [9,6]

Example 4:

  • Input: nums = [5], k = 3
  • Output: [0,0,1]

Example 5:

  • Input: nums = [1,1,1], k = 5
  • Output: [0,6,0,0,0]

Example 6:

  • Input: nums = [2,3,4], k = 1
  • Output: [6]

Constraints:

  • 1 <= nums[i] <= 10⁹
  • 1 <= nums.length <= 10⁵
  • 1 <= k <= 5

Hint:

  1. Use dynamic programming.
  2. Define dp[i][r] as the count of subarrays ending at index i whose product modulo k equals r.
  3. Compute dp[i][r] for each index i in nums and sum over all indices to get the final counts for each remainder.

Solution:

We treat every valid operation as choosing exactly one non-empty contiguous subarray to remain after removing a prefix and suffix. We count all such subarrays whose product modulo k equals each remainder x. This is done with a rolling DP over remainders modulo k, since k <= 5.

Approach

  • A valid operation is equivalent to selecting a non-empty subarray nums[l..r] to keep.
  • So the task becomes: count all non-empty contiguous subarrays by (product of subarray) % k.
  • Process nums from left to right.
  • Maintain dp[r]: number of subarrays ending at the previous index whose product modulo k is r.
  • For the current number num, let m = num % k.
  • New subarrays ending here are:
    • The single-element subarray [num], giving remainder m.
    • Every previous subarray extended by num, changing remainder from r to (r * m) % k.
  • Accumulate these counts into the final answer array.

Let's implement this solution in PHP: 3524. Find X Value of Array I

<?php
/**
 * @param Integer[] $nums
 * @param Integer $k
 * @return Integer[]
 */
function resultArray(array $nums, int $k): array
{
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
echo resultArray([1,2,3,4,5], 3) .  "\n";           // Output: [9,2,4]
echo resultArray([1,2,4,8,16,32], 4) .  "\n";       // Output: [18,1,2,0]
echo resultArray([1,1,2,1,1], 2) .  "\n";           // Output: [9,6]
echo resultArray([5], 3) .  "\n";                   // Output: [0,0,1]
echo resultArray([1,1,1], 5) .  "\n";               // Output: [0,6,0,0,0]
echo resultArray([2,3,4], 1) .  "\n";               // Output: [6]
?>
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Explanation:

  • Initialize ans and dp as arrays of size k filled with 0.
  • For each num in nums:
    • Compute m = num % k.
    • Create next array of size k filled with 0.
    • Add the single-element subarray: next[m] += 1.
    • For every remainder r from 0 to k - 1:
      • If dp[r] > 0, extend those subarrays:
        • next[(r * m) % k] += dp[r]
    • Add all counts in next to ans.
    • Set dp = next.
  • Return ans, where ans[x] is the number of valid operations giving remainder x.

Complexity Analysis

  • Time Complexity: O(n * k) For each element, we iterate over k possible remainders. Since k <= 5, this is effectively O(n).
  • Space Complexity: O(k) We only store DP and answer arrays of size k.

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  1. Subarray: A subarray is a contiguous sequence of elements within an array. ↩

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