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.
Constraints:
2 <= nums.length <= 10001 <= nums[i] <= 1000
Solution 1: Brute Force
Enumerate the element to remove and check.
Time complexity: O(n2)
Space complexity: O(n) -> O(1)
C++
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// Author: Huahua class Solution { public: bool canBeIncreasing(vector<int>& nums) { const int n = nums.size(); auto check = [&](int k) { vector<int> arr(nums); arr.erase(begin(arr) + k); for (int i = 1; i < n - 1; ++i) if (arr[i] <= arr[i - 1]) return false; return true; }; for (int i = 0; i < n; ++i) if (check(i)) return true; return false; } }; |
C++/O(1) Space
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// Author: Huahua class Solution { public: bool canBeIncreasing(vector<int>& nums) { const int n = nums.size(); auto check = [&](int k) { for (int i = 1; i < n; ++i) { int j = (i - 1 == k) ? (i - 2) : (i - 1); if (i != k && j >= 0 && nums[i] <= nums[j]) return false; } return true; }; for (int i = 0; i < n; ++i) if (check(i)) return true; return false; } }; |
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