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.

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

Java

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;
    }
}

// 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(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

665 - Non-decreasing Array

665. Non-decreasing Array (Easy)

Solution 1.

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