Non-decreasing Array LT665

 Given an array 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 array[i] <= array[i + 1] holds for every i (1 <= i < n).

Example 1:

Input: [4,2,3]
Output: True
Explanation: You could modify the first 4 to 1 to get a non-decreasing array.

 

Example 2:

Input: [4,2,1]
Output: False
Explanation: You can't get a non-decreasing array by modify at most one element.

 

Note: The n belongs to [1, 10,000]. 

Idea 1. 慢慢分析不同情况，

false if more than 1 descending pair

if 1 descending pair, is it possible to do 1 midification? nums[i-1] > nums[i], can we modify nums[i-1] or nums[i]? if nums[i-2] <= nums[i], modifiy nums[i-1] = nums[i]; otherwise, modify nums[i] = nums[i-1], but will fail if nums[i-1] > nums[i+1].

Time complexity: O(n)

Space complexity: O(1)

modify the array while looping it

class Solution {
    public boolean checkPossibility(int[] nums) {
        boolean decreasing = false;
        for(int i = 1; i < nums.length; ++i) {
            if(nums[i-1] > nums[i]) {
                
                if(decreasing) {
                    return false;
                }
                if(i == 1 || nums[i-2] <= nums[i]) {
                  nums[i-1] = nums[i];
                }
                else {
                    nums[i] = nums[i-1];
                }
                decreasing = true;
            }
        }
        
        return true;
    }
}

用cnt可以更简洁

class Solution {
    public boolean checkPossibility(int[] nums) {
        int cnt = 0;
        for(int i = 1; cnt <=1 && i < nums.length; ++i) {
            if(nums[i-1] > nums[i]) {
                if(i == 1 || nums[i-2] <= nums[i]) {
                  nums[i-1] = nums[i];
                }
                else {
                    nums[i] = nums[i-1];
                }
                ++cnt;
            }
        }
        
        return cnt <= 1;
    }
}

Idea 1.b 不改变数组

class Solution {
    public boolean checkPossibility(int[] nums) {
        int cnt = 0;
        for(int i = 1; cnt <=1 && i < nums.length; ++i) {
            if(nums[i-1] > nums[i]) {
                if( (i>= 2 && nums[i-2] > nums[i])
                   && (i+1 < nums.length && nums[i-1] > nums[i+1])) {
                  return false;
                }
               
                ++cnt;
            }
        }
        
        return cnt <= 1;
    }
}

比较好理解的

class Solution {
    public boolean checkPossibility(int[] nums) {
       int pIndex = -1;
       for(int i = 1; i < nums.length; ++i) {
           if(nums[i-1] > nums[i]) {
               if(pIndex != -1) {
                   return false;
               }
               pIndex = i;
           }
       }
        
       return (pIndex == -1)
              || (pIndex == 1) || (pIndex == nums.length-1)
              || (nums[pIndex-2] <= nums[pIndex])
              || (nums[pIndex-1] <= nums[pIndex+1]);
    }
}