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LeetCode 2762: Continuous Subarrays Solution

Master LeetCode problem 2762 (Continuous Subarrays), a medium 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.

Problem Explanation

Explanation:

To solve this problem, we can iterate through the array and for each element, we can extend the subarray starting from that element as long as the condition |nums[i1] - nums[i2]| <= 2 is satisfied for all pairs of indices within that subarray.

We can keep track of the count of continuous subarrays that can be formed starting from each element in the array.

Here's the step-by-step algorithm:

  1. Initialize a variable result to store the total count of continuous subarrays.
  2. Iterate through the array from index 0 to n-1.
  3. For each element at index i, start a subarray with the element itself and check if the condition is satisfied for all pairs of indices within that subarray.
  4. While the condition is satisfied, increment the count of continuous subarrays and extend the subarray to the right.
  5. Repeat steps 3 and 4 for all elements in the array.
  6. Return the total count of continuous subarrays.

Time Complexity:

The time complexity of this solution is O(n) where n is the number of elements in the input array.

Space Complexity:

The space complexity is O(1) as we are using only a constant amount of extra space.

:

Solution Code

class Solution {
    public int countSubarrays(int[] nums) {
        int n = nums.length;
        int result = 0;
        
        for (int i = 0; i < n; i++) {
            int j = i;
            while (j < n) {
                if (isValidSubarray(nums, i, j)) {
                    result++;
                    j++;
                } else {
                    break;
                }
            }
        }
        
        return result;
    }
    
    private boolean isValidSubarray(int[] nums, int start, int end) {
        for (int i = start; i < end; i++) {
            for (int j = i + 1; j <= end; j++) {
                if (Math.abs(nums[i] - nums[j]) > 2) {
                    return false;
                }
            }
        }
        return true;
    }
}

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The time complexity for LeetCode 2762 (Continuous Subarrays) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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