Check if an Array Is Consecutive - Practice Coding Problems

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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Understanding the Visualization

Input Array

Given array with potentially unordered elements

Find Range

Determine expected consecutive range [min, min+n-1]

Verify

Check if all expected numbers exist without duplicates

Key Takeaway

🎯 Key Insight: A consecutive array has exactly n unique numbers covering the range [min, min+n-1]

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

Approach	Time	Space	Notes
✓ Hash Set Approach	O(n)	O(n)	Use hash set for O(1) lookups to verify all expected numbers exist
Brute Force - Check Each Expected Number	O(n²)	O(1)	For each number in expected range, search entire array to verify it exists
Sort First Approach	O(n log n)	O(1)	Sort array then check if adjacent elements differ by exactly 1

Hash Set Approach — Algorithm Steps

Add all numbers to hash set
Find minimum value
Check if set contains each number in range [min, min+n-1]

Visualization

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Step-by-Step Walkthrough

Build Set

Add all array elements to hash set

Check Range

Verify each expected number exists in set

Result

Consecutive if all numbers found and no duplicates

Code -

solution.c — C

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

#define MAX_SIZE 10000

bool solution(int* nums, int numsSize) {
    if (numsSize == 0) return false;
    
    // Simple hash set using array (assuming reasonable range)
    bool seen[MAX_SIZE * 2] = {false};
    int offset = MAX_SIZE;  // Handle negative numbers
    
    int minVal = nums[0];
    int uniqueCount = 0;
    
    for (int i = 0; i < numsSize; i++) {
        if (!seen[nums[i] + offset]) {
            seen[nums[i] + offset] = true;
            uniqueCount++;
        }
        if (nums[i] < minVal) {
            minVal = nums[i];
        }
    }
    
    // Check if unique count equals array length (no duplicates)
    if (uniqueCount != numsSize) {
        return false;
    }
    
    // Check if each expected number exists
    for (int expected = minVal; expected < minVal + numsSize; expected++) {
        if (!seen[expected + offset]) {
            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(n)

O(n) to build set + O(n) to check range = O(n)

✓ Linear Growth

Space Complexity

O(n)

Hash set stores up to n unique elements

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Hash Set Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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