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
    
    

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