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
Solution 1: Traversal
We can traverse the array to find the first position $i$ where $\textit{nums}[i] < \textit{nums}[i+1]$ is not satisfied. Then, we check if the array is strictly increasing after removing either $i$ or $i+1$. If it is, we return $\textit{true}$; otherwise, we return $\textit{false}$.
The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.
Python3
class Solution:
def canBeIncreasing(self, nums: List[int]) -> bool:
def check(k: int) -> bool:
pre = -inf
for i, x in enumerate(nums):
if i == k:
continue
if pre >= x:
return False
pre = x
return True
i = 0
while i + 1 < len(nums) and nums[i] < nums[i + 1]:
i += 1
return check(i) or check(i + 1)
Java
class Solution {
public boolean canBeIncreasing(int[] nums) {
int i = 0;
while (i + 1 < nums.length && nums[i] < nums[i + 1]) {
++i;
}
return check(nums, i) || check(nums, i + 1);
}
private boolean check(int[] nums, int k) {
int pre = 0;
for (int i = 0; i < nums.length; ++i) {
if (i == k) {
continue;
}
if (pre >= nums[i]) {
return false;
}
pre = nums[i];
}
return true;
}
}
C++
class Solution {
public:
bool canBeIncreasing(vector<int>& nums) {
int n = nums.size();
auto check = [&](int k) -> bool {
int pre = 0;
for (int i = 0; i < n; ++i) {
if (i == k) {
continue;
}
if (pre >= nums[i]) {
return false;
}
pre = nums[i];
}
return true;
};
int i = 0;
while (i + 1 < n && nums[i] < nums[i + 1]) {
++i;
}
return check(i) || check(i + 1);
}
};
Go
func canBeIncreasing(nums []int) bool {
check := func(k int) bool {
pre := 0
for i, x := range nums {
if i == k {
continue
}
if pre >= x {
return false
}
pre = x
}
return true
}
i := 0
for i+1 < len(nums) && nums[i] < nums[i+1] {
i++
}
return check(i) || check(i+1)
}
TypeScript
function canBeIncreasing(nums: number[]): boolean {
const n = nums.length;
const check = (k: number): boolean => {
let pre = 0;
for (let i = 0; i < n; ++i) {
if (i === k) {
continue;
}
if (pre >= nums[i]) {
return false;
}
pre = nums[i];
}
return true;
};
let i = 0;
while (i + 1 < n && nums[i] < nums[i + 1]) {
++i;
}
return check(i) || check(i + 1);
}
Rust
impl Solution {
pub fn can_be_increasing(nums: Vec<i32>) -> bool {
let check = |k: usize| -> bool {
let mut pre = 0;
for (i, &x) in nums.iter().enumerate() {
if i == k {
continue;
}
if pre >= x {
return false;
}
pre = x;
}
true
};
let mut i = 0;
while i + 1 < nums.len() && nums[i] < nums[i + 1] {
i += 1;
}
check(i) || check(i + 1)
}
}
C#
public class Solution {
public bool CanBeIncreasing(int[] nums) {
int n = nums.Length;
bool check(int k) {
int pre = 0;
for (int i = 0; i < n; ++i) {
if (i == k) {
continue;
}
if (pre >= nums[i]) {
return false;
}
pre = nums[i];
}
return true;
}
int i = 0;
while (i + 1 < n && nums[i] < nums[i + 1]) {
++i;
}
return check(i) || check(i + 1);
}
}