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Formatted question description: https://leetcode.ca/all/665.html

# 665. Non-decreasing Array (Easy)

Given an array nums with n integers, your task is to check if it could become non-decreasing by modifying at most 1 element.

We define an array is non-decreasing if nums[i] <= nums[i + 1] holds for every i (0-based) such that (0 <= i <= n - 2).

Example 1:

Input: nums = [4,2,3]
Output: true
Explanation: You could modify the first 4 to 1 to get a non-decreasing array.


Example 2:

Input: nums = [4,2,1]
Output: false
Explanation: You can't get a non-decreasing array by modify at most one element.


Constraints:

• 1 <= n <= 10 ^ 4
• - 10 ^ 5 <= nums[i] <= 10 ^ 5

Related Topics:
Array

## Solution 1.

• class Solution {
public boolean checkPossibility(int[] nums) {
int length = nums.length;
boolean flag = false;
for (int i = 1; i < length; i++) {
int prevNum = nums[i - 1], curNum = nums[i];
if (prevNum > curNum) {
if (flag)
return false;
else {
if (i > 1 && i < length - 1) {
int prev2Num = nums[i - 2], nextNum = nums[i + 1];
if (curNum < prev2Num && nextNum < prevNum)
return false;
}
flag = true;
}
}
}
return true;
}
}

############

class Solution {
public boolean checkPossibility(int[] nums) {
for (int i = 0; i < nums.length - 1; ++i) {
int a = nums[i], b = nums[i + 1];
if (a > b) {
nums[i] = b;
if (isSorted(nums)) {
return true;
}
nums[i] = a;
nums[i + 1] = a;
return isSorted(nums);
}
}
return true;
}

private boolean isSorted(int[] nums) {
for (int i = 0; i < nums.length - 1; ++i) {
if (nums[i] > nums[i + 1]) {
return false;
}
}
return true;
}
}

• // OJ: https://leetcode.com/problems/non-decreasing-array/
// Time: O(N)
// Space: O(1)
class Solution {
public:
bool checkPossibility(vector<int>& A) {
int N = A.size(), R = N - 2;
for (; R >= 0 && A[R] <= A[R + 1]; --R);
for (int i = 0; i < N; ++i) {
if ((i == 0 || i == N - 1 || A[i + 1] >= A[i - 1]) && i >= R) return true;
if (i - 1 >= 0 && A[i] < A[i - 1]) break;
}
return false;
}
};

• class Solution:
def checkPossibility(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
if len(nums) < 2:
return True
count = 0
for i in range(1, len(nums)):
if nums[i] < nums[i - 1]:
if count == 1:
return False
if not (
i + 1 == len(nums)
or nums[i + 1] >= nums[i - 1]
or i - 2 < 0
or nums[i - 2] < nums[i]
):
return False
else:
count = 1
return True

############

class Solution(object):
def checkPossibility(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
flag = False
pre = float("-inf")
for i in range(len(nums) - 1):
if nums[i] < pre:
if nums[i + 1] >= nums[i - 1]:
nums[i] = nums[i + 1]
else:
nums[i - 1] = nums[i]
flag = True
break
pre = nums[i]
if not flag and len(nums) > 1 and nums[-1] < nums[-2]:
nums[-1] = nums[-2]
pre = float("-inf")
for num in nums:
if num < pre:
return False
pre = num
return True


• func checkPossibility(nums []int) bool {
isSorted := func(nums []int) bool {
for i, b := range nums[1:] {
if nums[i] > b {
return false
}
}
return true
}
for i := 0; i < len(nums)-1; i++ {
a, b := nums[i], nums[i+1]
if a > b {
nums[i] = b
if isSorted(nums) {
return true
}
nums[i] = a
nums[i+1] = a
return isSorted(nums)
}
}
return true
}

• function checkPossibility(nums: number[]): boolean {
const isSorted = (nums: number[]) => {
for (let i = 0; i < nums.length - 1; ++i) {
if (nums[i] > nums[i + 1]) {
return false;
}
}
return true;
};
for (let i = 0; i < nums.length - 1; ++i) {
const a = nums[i],
b = nums[i + 1];
if (a > b) {
nums[i] = b;
if (isSorted(nums)) {
return true;
}
nums[i] = a;
nums[i + 1] = a;
return isSorted(nums);
}
}
return true;
}