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The idea is to ensure every element is greater than its previous one by subtracting a suitable prime. We iterate through each element and subtract the smallest possible prime until the condition is met.
For each element in the array, find if there's a prime that can be subtracted to make the current element greater than the previous one. Repeat this for all elements until the array becomes strictly increasing.
Time Complexity: O(n * log(n)), primarily dominated by the prime-checking function.
Space Complexity: O(1), since we are modifying the array in place and not using any additional data structures.
1#include <iostream>
2#include <vector>
3#include <cmath>
4using namespace std;
5
6bool is_prime(int number) {
7 if (number <= 1) return false;
8 for (int i = 2; i <= sqrt(number); i++) {
9 if (number % i == 0) return false;
10 }
11 return true;
12}
13
14bool canBeeStrictlyIncreasing(vector<int>& nums) {
15 for (int i = 1; i < nums.size(); ++i) {
16 if (nums[i] <= nums[i - 1]) {
17 int diff = nums[i] - nums[i - 1];
18 bool found = false;
19
20 for (int p = diff + 1; p < nums[i]; ++p) {
21 if (is_prime(p)) {
22 nums[i] -= p;
23 found = true;
24 break;
25 }
26 }
27
28 if (!found || nums[i] <= nums[i - 1]) {
29 return false;
30 }
31 }
32 }
33 return true;
34}
35
36int main() {
37 vector<int> nums = {4, 9, 6, 10};
38 if (canBeeStrictlyIncreasing(nums)) {
39 cout << "true" << endl;
40 } else {
41 cout << "false" << endl;
42 }
43 return 0;
44}This C++ solution follows the same logic as the C solution; it iterates through the array, identifies when the array fails the strictly increasing condition, and tries to apply the smallest prime subtraction to meet the condition criteria.
This approach involves using the Sieve of Eratosthenes to precompute primes up to the maximum value in the array. For each element, we maintain a list of unpicked primes smaller than the current number and dynamically reduce the current element with the largest viable prime, proceeding until a strictly increasing order is established.
We use this optimized list to attempt to produce a valid increasing order to minimize operation count and improve efficiency.
Time Complexity: O(n * sqrt(m)), but preprocessing the prime sieve to reduce runtime cost per loop.
Space Complexity: O(m), for storing the boolean list to evaluate primes.
This C solution uses a sieve to precompute all primes less than a given maximum value. This approach allows us to quickly identify potential subtraction values to transform the array into one that is strictly increasing.