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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.
1import java.util.*;
2
3public class PrimeSubtraction {
4 private static boolean isPrime(int num) {
5 if (num <= 1) return false;
6 if (num <= 3) return true;
7 if (num % 2 == 0 || num % 3 == 0) return false;
8 for (int i = 5; i * i <= num; i += 6) {
9 if (num % i == 0 || num % (i + 2) == 0) return false;
10 }
11 return true;
12 }
13
14 public static boolean canBeStrictlyIncreasing(int[] nums) {
15 for (int i = 1; i < nums.length; i++) {
16 if (nums[i] <= nums[i - 1]) {
17 int diff = nums[i] - nums[i - 1];
18 for (int p = diff + 1; p < nums[i]; p++) {
19 if (isPrime(p)) {
20 nums[i] -= p;
21 break;
22 }
23 }
24 if (nums[i] <= nums[i - 1]) {
25 return false;
26 }
27 }
28 }
29 return true;
30 }
31
32 public static void main(String[] args) {
33 int[] nums = {4, 9, 6, 10};
34 System.out.println(canBeStrictlyIncreasing(nums));
35 }
36}In Java, this solution adheres to the same logical approach - checking the condition and modifying elements with the smallest prime necessary to meet the requirements for a strictly increasing order.
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.