# 1909. Remove One Element to Make the Array Strictly Increasing

## Description

Given a 0-indexed integer array nums, return true if it can be made strictly increasing after removing exactly one element, or false otherwise. If the array is already strictly increasing, return true.

The array nums is strictly increasing if nums[i - 1] < nums[i] for each index (1 <= i < nums.length).

Example 1:

Input: nums = [1,2,10,5,7]
Output: true
Explanation: By removing 10 at index 2 from nums, it becomes [1,2,5,7].
[1,2,5,7] is strictly increasing, so return true.


Example 2:

Input: nums = [2,3,1,2]
Output: false
Explanation:
[3,1,2] is the result of removing the element at index 0.
[2,1,2] is the result of removing the element at index 1.
[2,3,2] is the result of removing the element at index 2.
[2,3,1] is the result of removing the element at index 3.
No resulting array is strictly increasing, so return false.

Example 3:

Input: nums = [1,1,1]
Output: false
Explanation: The result of removing any element is [1,1].
[1,1] is not strictly increasing, so return false.


Constraints:

• 2 <= nums.length <= 1000
• 1 <= nums[i] <= 1000

## Solutions

• class Solution {
public boolean canBeIncreasing(int[] nums) {
int i = 1, n = nums.length;
for (; i < n && nums[i - 1] < nums[i]; ++i)
;
return check(nums, i - 1) || check(nums, i);
}

private boolean check(int[] nums, int i) {
int prev = Integer.MIN_VALUE;
for (int j = 0; j < nums.length; ++j) {
if (i == j) {
continue;
}
if (prev >= nums[j]) {
return false;
}
prev = nums[j];
}
return true;
}
}

• class Solution {
public:
bool canBeIncreasing(vector<int>& nums) {
int i = 1, n = nums.size();
for (; i < n && nums[i - 1] < nums[i]; ++i)
;
return check(nums, i - 1) || check(nums, i);
}

bool check(vector<int>& nums, int i) {
int prev = 0;
for (int j = 0; j < nums.size(); ++j) {
if (i == j) continue;
if (prev >= nums[j]) return false;
prev = nums[j];
}
return true;
}
};

• class Solution:
def canBeIncreasing(self, nums: List[int]) -> bool:
def check(nums, i):
prev = -inf
for j, num in enumerate(nums):
if i == j:
continue
if prev >= nums[j]:
return False
prev = nums[j]
return True

i, n = 1, len(nums)
while i < n and nums[i - 1] < nums[i]:
i += 1
return check(nums, i - 1) or check(nums, i)


• func canBeIncreasing(nums []int) bool {
i, n := 1, len(nums)
for ; i < n && nums[i-1] < nums[i]; i++ {

}
return check(nums, i-1) || check(nums, i)
}

func check(nums []int, i int) bool {
prev := 0
for j := 0; j < len(nums); j++ {
if i == j {
continue
}
if prev >= nums[j] {
return false
}
prev = nums[j]
}
return true
}

• function canBeIncreasing(nums: number[]): boolean {
const check = (p: number) => {
let prev = undefined;
for (let j = 0; j < nums.length; j++) {
if (p != j) {
if (prev !== undefined && prev >= nums[j]) {
return false;
}
prev = nums[j];
}
}
return true;
};
for (let i = 0; i < nums.length; i++) {
if (nums[i - 1] >= nums[i]) {
return check(i - 1) || check(i);
}
}
return true;
}


• impl Solution {
pub fn can_be_increasing(nums: Vec<i32>) -> bool {
let check = |p: usize| -> bool {
let mut prev = None;
for j in 0..nums.len() {
if p != j {
if let Some(value) = prev {
if value >= nums[j] {
return false;
}
}
prev = Some(nums[j]);
}
}
true
};
for i in 1..nums.len() {
if nums[i - 1] >= nums[i] {
return check(i - 1) || check(i);
}
}
true
}
}