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Leetcode 1550. Three Consecutive Odds

Problem Statement

Given an integer array arr, return true if there are three consecutive odd numbers in the array. Otherwise, return false.

Example

• Example 1:

  Input: arr = [2,6,4,1]
  Output: false
  Explanation: There are no three consecutive odds.
  

• Example 2:

  Input: arr = [1,2,34,3,4,5,7,23,12]
  Output: true
  Explanation: [5, 7, 23] are three consecutive odd numbers.
  

Constraints

• 1 <= arr.length <= 1000
• 1 <= arr[i] <= 1000

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Clarifying Questions

1. Q: Can the input array contain negative numbers? A: No, based on the constraints, the values range from 1 to 1000.
2. Q: Is the array guaranteed to have at least three elements? A: No, the length of the array can be as short as 1.

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Strategy

1. Iterate through the array: We will use a loop to go through each element of the array.
2. Check for three consecutive odd numbers: Use a sliding window of three elements to check if they are all odd. If they are, return true.
3. Return false if no such triplet is found: If the loop completes without finding three consecutive odd numbers, return false.

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Code

#include <vector>
using namespace std;

class Solution {
public:
    bool threeConsecutiveOdds(vector<int>& arr) {
        // We need to check if there are at least three elements
        if (arr.size() < 3) {
            return false;
        }
        
        // Iterate through the array with a window of size 3
        for (int i = 0; i < arr.size() - 2; ++i) {
            if (arr[i] % 2 != 0 && arr[i + 1] % 2 != 0 && arr[i + 2] % 2 != 0) {
                return true;
            }
        }
        
        // If no triplet of consecutive odd numbers is found, return false
        return false;
    }
};

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Time Complexity

• Time Complexity: O(n), where n is the number of elements in the array. We iterate through the array once, using a window of size 3.
• Space Complexity: O(1), because we are using a constant amount of extra space regardless of the input size.

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