# Remove One Element to Make the Array Strictly Increasing Solution

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## Remove One Element to Make the Array Strictly Increasing

Given a 0-indexed integer array `nums`, return `true` if it can be made strictly increasing after removing exactly one element, or `false` otherwise. If the array is already strictly increasing, return `true`.

The array `nums` is strictly increasing if `nums[i - 1] < nums[i]` for each index `(1 <= i < nums.length).`

Example 1:

```Input: nums = [1,2,10,5,7]
Output: true
Explanation: By removing 10 at index 2 from nums, it becomes [1,2,5,7].
[1,2,5,7] is strictly increasing, so return true.
```

Example 2:

```Input: nums = [2,3,1,2]
Output: false
Explanation:
[3,1,2] is the result of removing the element at index 0.
[2,1,2] is the result of removing the element at index 1.
[2,3,2] is the result of removing the element at index 2.
[2,3,1] is the result of removing the element at index 3.
No resulting array is strictly increasing, so return false.```

Example 3:

```Input: nums = [1,1,1]
Output: false
Explanation: The result of removing any element is [1,1].
[1,1] is not strictly increasing, so return false.
```

Example 4:

```Input: nums = [1,2,3]
Output: true
Explanation: [1,2,3] is already strictly increasing, so return true.
```

Constraints:

• `2 <= nums.length <= 1000`
• `1 <= nums[i] <= 1000`

Program:

``````class Solution {
public:
bool canBeIncreasing(vector<int>& nums) {
int n=nums.size();
int chances=1; //  i have  only one chance to remove a number
int i=0;
while(i<nums.size()-1){
if(nums[i]>=nums[i+1]){
if(chances==0)return false;
if(i==0){
nums[i]=nums[i+1]-1;
}
else{
if(nums[i+1]>nums[i-1])nums[i]=nums[i-1];
else nums[i+1]=nums[i];
}
chances--;
}
i++;
}
return true;
}
};``````

Approach:

``````While travesring through the array Ill keep checking the strictly increasing  sequence for the condition
and when the condition doesn't satisfy the condition A[i] < A[i+1]:
Then we know for sure that we have to remove either of these two
here comes the main part i.e, which one to remove when you have a chance to remove both:
then it's always better to remove the first one.
You'll better understand it when you draw them as a graph on the paper``````

credit : here