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 first4to1to 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;
}
}