Longest Consecutive Sequence

Longest Consecutive Sequence - Problem

Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence.

You must write an algorithm that runs in O(n) time.

A consecutive sequence is a sequence of integers where each number is exactly one more than the previous number.

Input & Output

Example 1 — Mixed Numbers

$ Input: nums = [100,4,200,1,3,2]

› Output: 4

💡 Note: The longest consecutive elements sequence is [1,2,3,4], which has length 4

Example 2 — Duplicates

$ Input: nums = [0,3,7,2,5,8,4,6,0,1]

› Output: 9

💡 Note: The longest consecutive sequence is [0,1,2,3,4,5,6,7,8] with length 9

Example 3 — Single Element

$ Input: nums = [9]

› Output: 1

💡 Note: Single element forms a sequence of length 1

Constraints

0 ≤ nums.length ≤ 10⁵
-10⁹ ≤ nums[i] ≤ 10⁹

Visualization

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Asked in

G Google 45 f Facebook 38 a Amazon 32 M Microsoft 28

The key insight is to use a hash set for O(1) lookups and only start counting from sequence beginnings (where num-1 doesn't exist). Best approach is Hash Set with Time: O(n), Space: O(n).

Common Approaches

✓ Brute Force

⏱️ Time: O(n³) Space: O(1)

For every number in the array, check if consecutive numbers exist by searching the entire array. Count how long each sequence extends.

Hash Set Optimal

⏱️ Time: O(n) Space: O(n)

Convert array to hash set for O(1) lookups. For each number that could start a sequence (no num-1 exists), count consecutive numbers. This achieves O(n) time.

Sort First

⏱️ Time: O(n log n) Space: O(1)

Sort the array to group consecutive numbers together, then scan through once to find the longest run of consecutive elements.

Brute Force — Algorithm Steps

For each number n in array
Count consecutive numbers n+1, n+2, n+3... by searching array
Track maximum sequence length found

Visualization

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Step-by-Step Walkthrough

Start

Pick a number and count consecutive sequence

Search entire array for next consecutive number

Repeat

Repeat for every number, track maximum length

Code -

solution.c — C

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

bool contains(int* nums, int size, int target) {
    for (int i = 0; i < size; i++) {
        if (nums[i] == target) return true;
    }
    return false;
}

int solution(int* nums, int numsSize) {
    if (numsSize == 0) return 0;
    
    int maxLength = 1;
    
    for (int i = 0; i < numsSize; i++) {
        int currentNum = nums[i];
        int currentLength = 1;
        
        // Count consecutive numbers starting from nums[i]
        while (contains(nums, numsSize, currentNum + 1)) {
            currentNum++;
            currentLength++;
        }
        
        if (currentLength > maxLength) {
            maxLength = currentLength;
        }
    }
    
    return maxLength;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int count = 0;
    
    char* ptr = strchr(line, '[');
    if (ptr) {
        ptr++;
        char* token = strtok(ptr, ",]");
        while (token != NULL) {
            nums[count++] = atoi(token);
            token = strtok(NULL, ",]");
        }
    }
    
    int result = solution(nums, count);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

For each of n numbers, we search the array (O(n)) to find consecutive numbers, potentially checking n consecutive numbers

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables for counting

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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