Three Consecutive Odds - Practice Coding Problems

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

Tap to expand

Understanding the Visualization

Input Array

Array with mixed odd and even numbers

Process

Find three consecutive odd numbers

Output

Return true if found, false otherwise

Key Takeaway

🎯 Key Insight: Use a counter to track consecutive odds, resetting when encountering an even number

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

Approach	Time	Space	Notes
✓ Counter Approach	O(n)	O(1)	Use a counter to track consecutive odd numbers
Brute Force - Check Every Triplet	O(n)	O(1)	Check every possible group of 3 consecutive elements

Counter Approach — Algorithm Steps

Step 1: Initialize counter to 0
Step 2: For each element, if odd increment counter, if even reset counter to 0
Step 3: Return true if counter reaches 3

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize

Start with counter = 0

Process

Increment for odd, reset for even

Success

Return true when counter reaches 3

Code -

solution.c — C

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

bool solution(int* arr, int arrSize) {
    int count = 0;
    for (int i = 0; i < arrSize; i++) {
        if (arr[i] % 2 == 1) {
            count++;
            if (count == 3) {
                return true;
            }
        } else {
            count = 0;
        }
    }
    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 pass through array with early termination possible

✓ Linear Growth

Space Complexity

O(1)

Only uses one counter variable

✓ Linear Space

23.5K Views

Medium Frequency

~10 min Avg. Time

842 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Counter Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler