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#
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| 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)
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| 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;
}
}
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| 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;
}
};
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| 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
}
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| 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;
}
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| 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
}
}
|