1550. Three Consecutive Odds

Description

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

Solutions

Solution 1

Python Code
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class Solution:
    def threeConsecutiveOdds(self, arr: List[int]) -> bool:
        cnt = 0
        for v in arr:
            if v & 1:
                cnt += 1
            else:
                cnt = 0
            if cnt == 3:
                return True
        return False

Java Code
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class Solution {
    public boolean threeConsecutiveOdds(int[] arr) {
        int cnt = 0;
        for (int v : arr) {
            if (v % 2 == 1) {
                ++cnt;
            } else {
                cnt = 0;
            }
            if (cnt == 3) {
                return true;
            }
        }
        return false;
    }
}

C++ Code
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class Solution {
public:
    bool threeConsecutiveOdds(vector<int>& arr) {
        int cnt = 0;
        for (int v : arr) {
            if (v & 1)
                ++cnt;
            else
                cnt = 0;
            if (cnt == 3) return true;
        }
        return false;
    }
};

Go Code
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func threeConsecutiveOdds(arr []int) bool {
	cnt := 0
	for _, v := range arr {
		if v%2 == 1 {
			cnt++
		} else {
			cnt = 0
		}
		if cnt == 3 {
			return true
		}
	}
	return false
}

TypeScript Code
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function threeConsecutiveOdds(arr: number[]): boolean {
    let cnt = 0;
    for (const v of arr) {
        if (v & 1) {
            ++cnt;
        } else {
            cnt = 0;
        }
        if (cnt == 3) {
            return true;
        }
    }
    return false;
}

Solution 2

Python Code
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class Solution:
    def threeConsecutiveOdds(self, arr: List[int]) -> bool:
        for i in range(len(arr) - 2):
            if arr[i] % 2 + arr[i + 1] % 2 + arr[i + 2] % 2 == 3:
                return True
        return False