Three Consecutive Odds

Three Consecutive Odds - Problem

Array Easy

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

An odd number is any integer that is not divisible by 2.

Input & Output

Example 1 — Basic Case

$ Input: arr = [2,1,3,5,8]

› Output: true

💡 Note: The elements at indices 1, 2, and 3 are [1,3,5] which are three consecutive odd numbers

Example 2 — No Three Consecutive

$ Input: arr = [1,2,3,4,5]

› Output: false

💡 Note: No three consecutive elements are all odd. We have [1,2,3] but 2 is even

Example 3 — All Odds

$ Input: arr = [1,3,5,7,9]

› Output: true

💡 Note: The first three elements [1,3,5] are all odd and consecutive

Constraints

1 ≤ arr.length ≤ 1000
-1000 ≤ arr[i] ≤ 1000

Visualization

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Asked in

a Amazon 25 🍎 Apple 15

The key insight is to use a counter that tracks consecutive odd numbers, resetting to 0 when an even number is encountered. The optimal approach uses a single pass counter method with early termination. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Check Every Triplet

⏱️ Time: O(n) Space: O(1)

Iterate through the array and for each position, check if the current element and the next two elements are all odd. This approach examines every possible triplet of consecutive elements.

Counter Approach

⏱️ Time: O(n) Space: O(1)

Maintain a counter that increments when we encounter an odd number and resets to 0 when we encounter an even number. Return true when counter reaches 3.

Brute Force - Check Every Triplet — Algorithm Steps

Step 1: Loop through array from index 0 to n-3
Step 2: For each position, check if arr[i], arr[i+1], arr[i+2] are all odd
Step 3: Return true immediately if found, false if no triplet found

Visualization

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Step-by-Step Walkthrough

Start

Check first triplet at indices [0,1,2]

Continue

Move to next triplet [1,2,3], then [2,3,4]

Found

Return true when 3 consecutive odds found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool solution(int* arr, int arrSize) {
    for (int i = 0; i < arrSize - 2; i++) {
        if (arr[i] % 2 != 0 && arr[i + 1] % 2 != 0 && arr[i + 2] % 2 != 0) {
            return true;
        }
    }
    return false;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    // Parse array manually
    int arr[1000];
    int size = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        arr[size++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    bool result = solution(arr, size);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single loop through array, checking up to n-2 triplets

✓ Linear Growth

Space Complexity

O(1)

Only uses a few variables, no extra data structures

✓ Linear Space

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Related Problems

Common Approaches

Brute Force - Check Every Triplet — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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