Question

2892. Minimizing Array After Replacing Pairs With Their Product

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Description

Given an integer array nums and an integer k, you can perform the following operation on the array any number of times:

• Select two adjacent elements of the array like x and y, such that x * y <= k, and replace both of them with a single element with value x * y (e.g. in one operation the array [1, 2, 2, 3] with k = 5 can become [1, 4, 3] or [2, 2, 3], but can't become [1, 2, 6]).

Return _the minimum possible length of _nums_ after any number of operations_.

 

Example 1:

Input: nums = [2,3,3,7,3,5], k = 20
Output: 3
Explanation: We perform these operations:
1. [2,3,3,7,3,5] -> [6,3,7,3,5]
2. [6,3,7,3,5] -> [18,7,3,5]
3. [18,7,3,5] -> [18,7,15]
It can be shown that 3 is the minimum length possible to achieve with the given operation.

Example 2:

Input: nums = [3,3,3,3], k = 6
Output: 4
Explanation: We can't perform any operations since the product of every two adjacent elements is greater than 6.
Hence, the answer is 4.

 

Constraints:

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

Solutions

Solution 1: Greedy

We use a variable $ans$ to record the current length of the array, and a variable $y$ to record the current product of the array. Initially, $ans = 1$ and $y = nums[0]$.

We start traversing from the second element of the array. Let the current element be $x$:

• If $x = 0$, then the product of the entire array is $0 \le k$, so the minimum length of the answer array is $1$, and we can return directly.
• If $x \times y \le k$, then we can merge $x$ and $y$, that is, $y = x \times y$.
• If $x \times y \gt k$, then we cannot merge $x$ and $y$, so we need to treat $x$ as a separate element, that is, $ans = ans + 1$, and $y = x$.

The final answer is $ans$.

The time complexity is $O(n)$, where n is the length of the array. The space complexity is $O(1)$.

class Solution:
  def minArrayLength(self, nums: List[int], k: int) -> int:
    ans, y = 1, nums[0]
    for x in nums[1:]:
      if x == 0:
        return 1
      if x * y <= k:
        y *= x
      else:
        y = x
        ans += 1
    return ans

class Solution {
  public int minArrayLength(int[] nums, int k) {
    int ans = 1;
    long y = nums[0];
    for (int i = 1; i < nums.length; ++i) {
      int x = nums[i];
      if (x == 0) {
        return 1;
      }
      if (x * y <= k) {
        y *= x;
      } else {
        y = x;
        ++ans;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int minArrayLength(vector<int>& nums, int k) {
    int ans = 1;
    long long y = nums[0];
    for (int i = 1; i < nums.size(); ++i) {
      int x = nums[i];
      if (x == 0) {
        return 1;
      }
      if (x * y <= k) {
        y *= x;
      } else {
        y = x;
        ++ans;
      }
    }
    return ans;
  }
};

func minArrayLength(nums []int, k int) int {
  ans, y := 1, nums[0]
  for _, x := range nums[1:] {
    if x == 0 {
      return 1
    }
    if x*y <= k {
      y *= x
    } else {
      y = x
      ans++
    }
  }
  return ans
}

function minArrayLength(nums: number[], k: number): number {
  let [ans, y] = [1, nums[0]];
  for (const x of nums.slice(1)) {
    if (x === 0) {
      return 1;
    }
    if (x * y <= k) {
      y *= x;
    } else {
      y = x;
      ++ans;
    }
  }
  return ans;
}