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Leetcode 2970. Count the Number of Incremovable Subarrays I

Problem Statement

You are given an array of integers nums. A subarray [nums[i], nums[i + 1], ..., nums[j]] is considered as incremovable if the following two conditions are met:

Your task is to return the number of different incremovable subarrays.

Example:

Input: nums = [1, 2, 3, 4]
Output: 4
Explanation: There are 4 incremovable subarrays:
- [1, 2, 3, 4] -> Sum = 10 (even) and sum of remaining array (after removing this subarray) can never be >= 10
- [2, 3, 4] -> Sum = 9 (odd)
- [1, 2, 3] -> Sum = 6 (even) and sum of remaining array (after removing this subarray) cannot be >= 10
- [3, 4] -> Sum = 7 (odd)
Two more subarrays can also be counted: [2], [4].

Clarifying Questions

  1. What is the range of the input array length (n) and the values inside the array?
    • Typical constraints include 1 <= n <= 10^4 and -10^5 <= nums[i] <= 10^5.
  2. Does the array contain only integers?
    • Yes, the array contains only integer values.

Strategy

To solve this problem, we can break it down into the following steps:

  1. Calculate the Total Sum of the Array:
    • First, we calculate the total sum S of the array.
  2. Iterate Over All Possible Subarrays:
    • We iterate over all possible subarrays [nums[i], nums[i + 1], ..., nums[j]] and their sums.
    • For each subarray, we check if the sum of the subarray is even.
    • We also check if the remaining part of the array’s sum is less than the original sum S.
  3. Count the Incremovable Subarrays:
    • Count the number of subarrays that satisfy both conditions and return the count.

Code Implementation

Here is the C++ implementation:

#include <vector>
#include <numeric> // For std::accumulate

class Solution {
public:
    int countIncremovableSubarrays(std::vector<int>& nums) {
        // Calculate the total sum of the array
        int total_sum = std::accumulate(nums.begin(), nums.end(), 0);

        int count = 0;
        int n = nums.size();

        // Iterate over all possible subarrays
        for (int i = 0; i < n; ++i) {
            int current_sum = 0;
            for (int j = i; j < n; ++j) {
                current_sum += nums[j];
                // Check if the sum of the subarray is even
                if (current_sum % 2 == 0) {
                    // Calculate the remaining sum of the array if this subarray is removed
                    int remaining_sum = total_sum - current_sum;
                    // Check if the remaining sum is less than the original total sum
                    if (remaining_sum < total_sum) {
                        ++count;
                    }
                }
            }
        }

        return count;
    }
};

Time Complexity

The time complexity of this solution is (O(n^2)), where (n) is the length of the array. This is because we are iterating over all possible subarrays, leading to a nested loop.

In total, the nested loops will run (n \times n = n^2) times in the worst case, making the overall time complexity (O(n^2)).

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