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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 <stdio.h>
2#include <stdbool.h>
3#include <math.h>
4
5bool isPrime(int n) {
6 if (n <= 1) return false;
7 if (n <= 3) return true;
8 if (n % 2 == 0 || n % 3 == 0) return false;
9 for (int i = 5; i * i <= n; i += 6) {
10 if (n % i == 0 || n % (i + 2) == 0) return false;
11 }
12 return true;
13}
14
15bool canBeStrictlyIncreasing(int *nums, int n) {
16 for (int i = 1; i < n; i++) {
17 if (nums[i] <= nums[i - 1]) {
18 int diff = nums[i] - nums[i - 1];
19 for (int p = diff + 1; p < nums[i]; p++) {
20 if (isPrime(p)) {
21 nums[i] -= p;
22 break;
23 }
24 }
25 if (nums[i] <= nums[i - 1]) {
26 return false;
27 }
28 }
29 }
30 return true;
31}
32
33int main() {
34 int nums[] = {4, 9, 6, 10};
35 int n = sizeof(nums) / sizeof(nums[0]);
36 if (canBeStrictlyIncreasing(nums, n)) {
37 printf("true\n");
38 } else {
39 printf("false\n");
40 }
41 return 0;
42}This C solution iterates over the array elements and whenever a non-increasing pair is detected, it attempts to subtract the smallest prime number from the current element to make it bigger than the previous one. After adjusting each element, the code checks again if the array is strictly increasing.
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.
Using Java, we leverage the Sieve of Eratosthenes to store prime numbers so that during iteration, adjusting current elements are quicker. This produces a more efficient Alpha over a smaller computation differential.