1550. Three Consecutive Odds

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

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 odds.

Constraints:

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

Approach 01:

#include <bits/stdc++.h>

using namespace std;

class Solution {
public:
    bool threeConsecutiveOdds(vector<int>& arr) {
        int result = 0;

        for(int i=0; i < arr.size(); i++){
            int n = arr[i];
            if(n%2 != 0){
                result++;
            }
            else{
                result = 0;
            }
            if(result == 3){
                return true;
            }
        }
        return false;
    }
};

from typing import *

class Solution:
    def threeConsecutiveOdds(self, arr: List[int]) -> bool:
        result = 0
        for n in arr:
            if(n%2 != 0):
                result += 1
            else:
                result = 0
            if(result == 3):
                return True
        return False

Time Complexity

• The function threeConsecutiveOdds iterates through the array arr exactly once.
• The main operations inside the loop include:
  • Checking if a number is odd.
  • Incrementing or resetting the result counter.
  • Checking if result equals 3.
• Each of these operations takes constant time, O(1).
• Since the loop runs for n iterations, where n is the size of the array arr, the time complexity is O(n).

Space Complexity

• The space complexity is determined by the amount of extra space used relative to the input size.
• The result variable is an integer that takes constant space, O(1).
• The function does not use any additional data structures whose size depends on the input size; it only uses a few variables.
• Therefore, the space complexity is O(1).