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;
    }
}

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