665

665. Non-decreasing Array (Medium)

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

• n == nums.length
• 1 <= n <= 104
• -105 <= nums[i] <= 105

Related Topics:
Array

Similar Questions:

• Make Array Non-decreasing or Non-increasing (Hard)
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Solution 1. Two Passes

This problem is asking whether the array is non-decreasing after deleting at most one element.

We can try deleting one element at each position, and return true if it satisfies the following condition:

A[i+1] >= A[i-1] && i >= R

where A[R] is the rightmost element that A[R] > A[R+1]. That is, for any i >= R, the right part A[i+1 .. N-1] is non-decreasing.

Once A[i] < A[i-1], we should stop scanning because the left part won't be non-decreasing.

// OJ: https://leetcode.com/problems/non-decreasing-array/
// Author: github.com/lzl124631x
// 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;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/non-decreasing-array/
// Author: github.com/lzl124631x
// 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;
    }
};