# Simple C++ 18 lines withnot modifying input

• Use flag to see if you have modified before. The problem is when nums[i]<nums[i-1], you need to choose to modify nums[i-1] or nums[i] (preferably nums[i-1] since we are moving forward to i+1 in next loop).

If it is the first time violation, consider 4 cases in order

1. if i==1, then we definitely modify nums[i-1] since modifying nums[0] will not affect any prior cells.
2. else if i==n-1, then it does not matter if we modify nums[i-1] or nums[i]. Whichever we will return true.
3. else if nums[i] >= nums[i-2], modify nums[i-1] to be nums[i]. We do not need to really modify anything since we move on to i+1 next loop and nums[i-1] does not matter anymore
4. else if nums[i-1] <= nums[i+1], modify nums[i] to become nums[i-1]. We do not need to really modify anything here as well since the purpose of modifying nums[i] is to make nums[i+1]>=nums[i]. However, nums[i]<nums[i-1] already so we do not need to modify nums[i] to be nums[i-1].

Since we have nums[i]<nums[i-1] already, if the above 4 cases all fail, it means we get another violation immediately so we return false. Otherwise, we just count this one violation and move forward without the need to modify anything.

``````class Solution {
public:
bool checkPossibility(vector<int>& nums) {
if(nums.size()<=1) return true;
bool flag=false;
for(int i=1; i<nums.size(); i++) {
if(nums[i]<nums[i-1]) {
if(!flag) { // first violation
flag=true;
if(i!=1 && i!=nums.size()-1 && nums[i]<nums[i-2] && nums[i-1]>nums[i+1])
return false; // immediate violation after the first
}
else return false; // second violation
}
}
return true;
}
};
``````

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