Program to check three consecutive odds are present or not in Python

Suppose we have an array called nums, we have to check whether there are three consecutive odd numbers in nums or not.

So, if the input is like nums = [18,15,2,19,3,11,17,25,20], then the output will be True as there are three consecutive odds [3,11,17].

Algorithm

To solve this, we will follow these steps −

• Get the length of nums

• If length is 1 or 2, return False (impossible to have 3 consecutive elements)

• Otherwise, iterate through the array checking each group of 3 consecutive elements

• If all three elements in any group are odd, return True

• If no group of 3 consecutive odds is found, return False

Method 1: Using Modulo Operator

Check if each number is odd using the modulo operator ?

def solve(nums):
    length = len(nums)
    if length == 1 or length == 2:
        return False
    else:
        for i in range(len(nums) - 2):
            if nums[i] % 2 != 0 and nums[i+1] % 2 != 0 and nums[i+2] % 2 != 0:
                return True
        return False

nums = [18, 15, 2, 19, 3, 11, 17, 25, 20]
print(solve(nums))

True

Method 2: Using Bitwise AND

An alternative approach using bitwise AND operation to check if numbers are odd ?

def check_three_consecutive_odds(nums):
    if len(nums) < 3:
        return False
    
    for i in range(len(nums) - 2):
        if (nums[i] & 1) and (nums[i+1] & 1) and (nums[i+2] & 1):
            return True
    return False

# Test with different examples
test_cases = [
    [18, 15, 2, 19, 3, 11, 17, 25, 20],  # True
    [2, 4, 6, 8],                        # False (all even)
    [1, 3, 5],                           # True (all odd)
    [1, 2, 3, 4, 5],                     # False (no 3 consecutive odds)
    [1, 2]                               # False (less than 3 elements)
]

for i, nums in enumerate(test_cases):
    result = check_three_consecutive_odds(nums)
    print(f"Test {i+1}: {nums} ? {result}")

Test 1: [18, 15, 2, 19, 3, 11, 17, 25, 20] ? True
Test 2: [2, 4, 6, 8] ? False
Test 3: [1, 3, 5] ? True
Test 4: [1, 2, 3, 4, 5] ? False
Test 5: [1, 2] ? False

Comparison

Key Points

• Arrays with fewer than 3 elements cannot have 3 consecutive odds

• We only need to check up to len(nums) - 2 to avoid index errors

• Both modulo and bitwise methods work equally well for checking odd numbers

• The function returns True as soon as the first group of 3 consecutive odds is found

Conclusion

Use the modulo operator method for better readability. The bitwise AND approach offers a slight performance advantage but is less intuitive. Both methods efficiently solve the problem in O(n) time complexity.