Question

2601. Prime Subtraction Operation

中文文档

Description

You are given a 0-indexed integer array nums of length n.

You can perform the following operation as many times as you want:

• Pick an index i that you haven’t picked before, and pick a prime p strictly less than nums[i], then subtract p from nums[i].

Return _true if you can make nums a strictly increasing array using the above operation and false otherwise._

A strictly increasing array is an array whose each element is strictly greater than its preceding element.

 

Example 1:

Input: nums = [4,9,6,10]
Output: true
Explanation: In the first operation: Pick i = 0 and p = 3, and then subtract 3 from nums[0], so that nums becomes [1,9,6,10].
In the second operation: i = 1, p = 7, subtract 7 from nums[1], so nums becomes equal to [1,2,6,10].
After the second operation, nums is sorted in strictly increasing order, so the answer is true.

Example 2:

Input: nums = [6,8,11,12]
Output: true
Explanation: Initially nums is sorted in strictly increasing order, so we don't need to make any operations.

Example 3:

Input: nums = [5,8,3]
Output: false
Explanation: It can be proven that there is no way to perform operations to make nums sorted in strictly increasing order, so the answer is false.

 

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 1000
• nums.length == n

Solutions

Solution 1: Preprocessing prime numbers + binary search

We first preprocess all the primes within $1000$ and record them in the array $p$.

For each element $nums[i]$ in the array $nums$, we need to find a prime $p[j]$ such that $p[j] \gt nums[i] - nums[i + 1]$ and $p[j]$ is as small as possible. If there is no such prime, it means that it cannot be strictly increased by subtraction operations, return false. If there is such a prime, we will subtract $p[j]$ from $nums[i]$ and continue to process the next element.

If all the elements in $nums$ are processed, it means that it can be strictly increased by subtraction operations, return true.

The time complexity is $O(n \log n)$ and the space complexity is $O(n)$. where $n$ is the length of the array $nums$.

class Solution:
  def primeSubOperation(self, nums: List[int]) -> bool:
    p = []
    for i in range(2, max(nums)):
      for j in p:
        if i % j == 0:
          break
      else:
        p.append(i)

    n = len(nums)
    for i in range(n - 2, -1, -1):
      if nums[i] < nums[i + 1]:
        continue
      j = bisect_right(p, nums[i] - nums[i + 1])
      if j == len(p) or p[j] >= nums[i]:
        return False
      nums[i] -= p[j]
    return True

class Solution {
  public boolean primeSubOperation(int[] nums) {
    List<Integer> p = new ArrayList<>();
    for (int i = 2; i <= 1000; ++i) {
      boolean ok = true;
      for (int j : p) {
        if (i % j == 0) {
          ok = false;
          break;
        }
      }
      if (ok) {
        p.add(i);
      }
    }
    int n = nums.length;
    for (int i = n - 2; i >= 0; --i) {
      if (nums[i] < nums[i + 1]) {
        continue;
      }
      int j = search(p, nums[i] - nums[i + 1]);
      if (j == p.size() || p.get(j) >= nums[i]) {
        return false;
      }
      nums[i] -= p.get(j);
    }
    return true;
  }

  private int search(List<Integer> nums, int x) {
    int l = 0, r = nums.size();
    while (l < r) {
      int mid = (l + r) >> 1;
      if (nums.get(mid) > x) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  }
}

class Solution {
public:
  bool primeSubOperation(vector<int>& nums) {
    vector<int> p;
    for (int i = 2; i <= 1000; ++i) {
      bool ok = true;
      for (int j : p) {
        if (i % j == 0) {
          ok = false;
          break;
        }
      }
      if (ok) {
        p.push_back(i);
      }
    }
    int n = nums.size();
    for (int i = n - 2; i >= 0; --i) {
      if (nums[i] < nums[i + 1]) {
        continue;
      }
      int j = upper_bound(p.begin(), p.end(), nums[i] - nums[i + 1]) - p.begin();
      if (j == p.size() || p[j] >= nums[i]) {
        return false;
      }
      nums[i] -= p[j];
    }
    return true;
  }
};

func primeSubOperation(nums []int) bool {
  p := []int{}
  for i := 2; i <= 1000; i++ {
    ok := true
    for _, j := range p {
      if i%j == 0 {
        ok = false
        break
      }
    }
    if ok {
      p = append(p, i)
    }
  }
  for i := len(nums) - 2; i >= 0; i-- {
    if nums[i] < nums[i+1] {
      continue
    }
    j := sort.SearchInts(p, nums[i]-nums[i+1]+1)
    if j == len(p) || p[j] >= nums[i] {
      return false
    }
    nums[i] -= p[j]
  }
  return true
}

function primeSubOperation(nums: number[]): boolean {
  const p: number[] = [];
  for (let i = 2; i <= 1000; ++i) {
    let ok = true;
    for (const j of p) {
      if (i % j === 0) {
        ok = false;
        break;
      }
    }
    if (ok) {
      p.push(i);
    }
  }
  const search = (x: number): number => {
    let l = 0;
    let r = p.length;
    while (l < r) {
      const mid = (l + r) >> 1;
      if (p[mid] > x) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  };
  const n = nums.length;
  for (let i = n - 2; i >= 0; --i) {
    if (nums[i] < nums[i + 1]) {
      continue;
    }
    const j = search(nums[i] - nums[i + 1]);
    if (j === p.length || p[j] >= nums[i]) {
      return false;
    }
    nums[i] -= p[j];
  }
  return true;
}