LeetCode 3452: Sum of Good Numbers Solution

Master LeetCode problem 3452 (Sum of Good Numbers), 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.

3452. Sum of Good Numbers

Easy

Array

Problem Explanation

Explanation:

To solve this problem, we can iterate over the array nums and for each element at index i, check if it is strictly greater than the elements at indices i - k and i + k (if they exist). If either of these indices does not exist, we consider the element good as well. We can keep track of the sum of all good elements and return this sum as the final result.

Algorithm:

Initialize a variable sum to store the sum of good elements.
Iterate over the array nums from index 0 to n - 1 where n is the length of nums.
For each element nums[i], check if it is strictly greater than nums[i - k] and nums[i + k] (if they exist).
If the condition is satisfied or either of the indices does not exist, add nums[i] to sum.
Return the final sum as the result.

Time Complexity: O(n), where n is the length of the input array nums. Space Complexity: O(1).

Solution Code

class Solution {
    public int sumOfGoodNumbers(int[] nums, int k) {
        int sum = 0;
        for (int i = 0; i < nums.length; i++) {
            boolean good = true;
            if (i - k >= 0 && nums[i] <= nums[i - k]) {
                good = false;
            }
            if (i + k < nums.length && nums[i] <= nums[i + k]) {
                good = false;
            }
            if (good) {
                sum += nums[i];
            }
        }
        return sum;
    }
}

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