Three Consecutive Odds

easy array

Problem

Return true if there are three consecutive odd numbers in arr; else false.

Inputarr = [1,2,34,3,4,5,7,23,12]

Outputtrue

5, 7, 23 are three consecutive odds.

arr

def three_consecutive_odds(arr):
    run = 0
    for x in arr:
        run = run + 1 if x % 2 else 0
        if run == 3: return True
    return False

function threeConsecutiveOdds(arr) {
  let r = 0;
  for (const x of arr) { r = (x & 1) ? r + 1 : 0; if (r === 3) return true; }
  return false;
}

class Solution {
    public boolean threeConsecutiveOdds(int[] arr) {
        int r = 0;
        for (int x : arr) { r = (x & 1) == 1 ? r + 1 : 0; if (r == 3) return true; }
        return false;
    }
}

bool threeConsecutiveOdds(vector& arr) {
    int r = 0;
    for (int x : arr) { r = (x & 1) ? r + 1 : 0; if (r == 3) return true; }
    return false;
}

Explanation

We just need to know if three odd numbers in a row appear anywhere. The simplest way is to keep a running counter of how many odds we have seen back to back as we scan once.

The variable run counts the current streak of odds. For each value x, if x is odd we do run + 1; if it is even the streak is broken so we reset run to 0.

The moment run hits 3 we can return True immediately — we have found three consecutive odds and there is no reason to keep looking. If the loop finishes without that happening, we return False.

Resetting on every even is the key: it guarantees the count only ever reflects an unbroken run of odds, not odds scattered around the array.

Example: [1,2,34,3,4,5,7,23,12]. The counter keeps resetting until we reach 5,7,23, where run climbs 1, 2, 3 and we return true.

Time: O(n) Space: O(1)