2892. Minimizing Array After Replacing Pairs With Their Product 🔒

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 $O(1)$.

Python3JavaC++GoTypeScript

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;
}

2892. Minimizing Array After Replacing Pairs With Their Product 🔒

Description

Solutions

Solution 1: Greedy

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