LeetCode 2970: Count the Number of Incremovable Subarrays I Solution

Master LeetCode problem 2970 (Count the Number of Incremovable Subarrays I), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2970. Count the Number of Incremovable Subarrays I

Easy

Array Two Pointers Binary Search Enumeration

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible subarrays of the given array nums. For each subarray, we check if removing it results in a strictly increasing array. If it does, we increment a counter to keep track of the number of incremovable subarrays. We can check if an array is strictly increasing by comparing each element with the next one in the subarray.

Algorithm:

Initialize a counter variable result to keep track of the number of incremovable subarrays.
Iterate through all possible subarrays of nums.
For each subarray, check if removing it results in a strictly increasing array.
If yes, increment the result counter.
Return the final result count.

Time Complexity:

The time complexity of this approach is O(n^3) where n is the length of the input array nums. This is because we are iterating through all possible subarrays.

Space Complexity:

The space complexity of this approach is O(1) as we are not using any extra space that grows with the input size.

Solution Code

class Solution {
    public int countIncremovableSubarrays(int[] nums) {
        int result = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = i; j < nums.length; j++) {
                if (isIncremovable(nums, i, j)) {
                    result++;
                }
            }
        }
        return result;
    }
    
    private boolean isIncremovable(int[] nums, int start, int end) {
        for (int i = start; i < end; i++) {
            if (nums[i] >= nums[i + 1]) {
                return false;
            }
        }
        return true;
    }
}

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