LeetCode 1550: Three Consecutive Odds Solution

Problem Explanation

Explanation:

To solve this problem, we can iterate through the array and check if there are three consecutive odd numbers. We can do this by maintaining a count of consecutive odd numbers encountered so far. If the count reaches 3, we return true. Otherwise, we reset the count if we encounter an even number.

Start by initializing a count variable to 0.
Iterate through the array elements.
If the current element is odd, increment the count.
If the current element is even, reset the count to 0.
If at any point the count reaches 3, return true.
If we reach the end of the array without finding three consecutive odds, return false.

Time Complexity:

The time complexity of this approach 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 a constant amount of extra space regardless of the input size.

Solution Code

class Solution {
    public boolean threeConsecutiveOdds(int[] arr) {
        int count = 0;
        for (int num : arr) {
            if (num % 2 != 0) {
                count++;
                if (count == 3) {
                    return true;
                }
            } else {
                count = 0;
            }
        }
        return false;
    }
}

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1550. Three Consecutive Odds