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
• class Solution {
• public boolean primeSubOperation(int[] nums) {
• for (int i = 0; i < nums.length; i++) {
• int min = 0;
• if (i > 1) min = Math.max(nums[i-1], 0);
• int max = nums[i];
• int prime = getPrime(min, max);
• nums[i] -= prime;
• }
• return isIncreasing(nums);
• }
• public boolean isIncreasing(int[] nums){
• for (int i = 1; i < nums.length; i++){
• if (nums[i] <= nums[i-1]){
• return false;
• }
• }
• return true;
• }
• public int getPrime(int min, int max) {
• for (int i = max-1; i > min; i--){
• if (isPrime(i) && (max – i > min)){
• return i;
• }
• }
• return 0;
• }
• public boolean isPrime(int n){
• for (int i = 2; i < n; i++){
• if (n % i == 0) {
• return false;
• }
• }
• return true;
• }
• }
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