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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 <= 10001 <= arr[i] <= 1000
Solutions
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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; } } -
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; } }; -
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 -
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 } -
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; }