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