Check if an Array Is Consecutive

Check if an Array Is Consecutive - Problem

Given an integer array nums, return true if nums is consecutive, otherwise return false.

An array is consecutive if it contains every number in the range [x, x + n - 1] (inclusive), where x is the minimum number in the array and n is the length of the array.

Input & Output

Example 1 — Consecutive Array

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

› Output: true

💡 Note: The minimum is 1, array length is 5. Expected range [1,2,3,4,5] - all numbers present, so it's consecutive.

Example 2 — Missing Number

$ Input: nums = [1,0,3]

› Output: false

💡 Note: The minimum is 0, array length is 3. Expected range [0,1,2] but we have [0,1,3] - missing 2, so not consecutive.

Example 3 — Duplicate Numbers

$ Input: nums = [1,1,2,3]

› Output: false

💡 Note: Contains duplicates. For consecutive array [1,2,3,4], we need unique numbers 1,2,3,4 but have 1,1,2,3.

Constraints

1 ≤ nums.length ≤ 10⁵
-10⁶ ≤ nums[i] ≤ 10⁶

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that a consecutive array must contain exactly the numbers [min, min+n-1] with no duplicates or gaps. Best approach is Hash Set for O(1) lookups. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force - Check Each Expected Number

⏱️ Time: O(n²) Space: O(1)

Find the minimum value, then for each number from min to min+n-1, scan the entire array to check if it exists. This ensures all expected numbers are present.

Hash Set Approach

⏱️ Time: O(n) Space: O(n)

Store all numbers in a hash set for constant-time lookups, then check if every number in the expected range [min, min+n-1] exists in the set. Also handles duplicates automatically.

Sort First Approach

⏱️ Time: O(n log n) Space: O(1)

Sort the array first, then iterate through to verify each adjacent pair differs by exactly 1. This transforms the problem into a simple sequential check.

Brute Force - Check Each Expected Number — Algorithm Steps

Find minimum value in array
For each number in range [min, min+n-1], search array
Return false if any number is missing

Visualization

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Step-by-Step Walkthrough

Find Minimum

Identify starting point of consecutive range

Check Each Expected

Search array for each number in range [min, min+n-1]

Result

Return true if all numbers found, false otherwise

Code -

solution.c — C

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

bool solution(int* nums, int numsSize) {
    if (numsSize == 0) return false;
    
    int minVal = nums[0];
    for (int i = 1; i < numsSize; i++) {
        if (nums[i] < minVal) {
            minVal = nums[i];
        }
    }
    
    // Check if each expected number exists in array
    for (int expected = minVal; expected < minVal + numsSize; expected++) {
        bool found = false;
        for (int i = 0; i < numsSize; i++) {
            if (nums[i] == expected) {
                found = true;
                break;
            }
        }
        if (!found) {
            return false;
        }
    }
    
    return true;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n expected numbers, we search through n array elements

⚡ Linearithmic

Space Complexity

O(1)

Only using variables, no extra data structures

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check Each Expected Number — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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