Check if Array is Good - Practice Coding Problems

Check if Array is Good - Problem

Array Pattern Recognition Challenge

You're given an integer array nums and need to determine if it follows a specific "good" pattern.

An array is considered good if it's a permutation of a base[n] array, where:
• base[n] = [1, 2, 3, ..., n-1, n, n]
• Contains numbers 1 through n-1 exactly once
• Contains the number n exactly twice
• Has a total length of n+1

Examples of base arrays:
• base[1] = [1, 1] (length 2)
• base[3] = [1, 2, 3, 3] (length 4)
• base[4] = [1, 2, 3, 4, 4] (length 5)

Return true if the given array matches this pattern (in any order), otherwise return false.

Input & Output

example_1.py — Basic Valid Case

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

› Output: false

💡 Note: Length is 3, so n would be 2. But base[2] = [1, 2, 2] requires length 3 with numbers [1, 2, 2]. Our array has [1, 2, 3], which doesn't match.

example_2.py — Perfect Match

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

› Output: true

💡 Note: Length is 4, so n = 3. base[3] = [1, 2, 3, 3]. Our array contains exactly one 1, one 2, and two 3's, which matches the required pattern.

example_3.py — Edge Case

$ Input: nums = [1, 1]

› Output: true

💡 Note: Length is 2, so n = 1. base[1] = [1, 1]. Our array has exactly two 1's, which matches perfectly.

Constraints

1 ≤ nums.length ≤ 100
1 ≤ nums[i] ≤ 200
The array represents a permutation test case

Visualization

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Understanding the Visualization

Count Attendance

Go through the roster once, counting how many times each student number appears

Find Maximum

The highest student number tells us what 'n' should be for this class

Validate Pattern

Check: Do we have exactly n different student numbers? Do numbers 1 through n-1 appear once, and does n appear twice?

Key Takeaway

🎯 Key Insight: Use a hash table to count frequencies in one pass. The maximum value determines n, and we validate that numbers 1 to n-1 appear once while n appears twice.

Asked in

G Google 15 a Amazon 12 ⊞ Microsoft 8 f Meta 6

The optimal solution uses a hash table to count frequencies in one pass through the array. Key insight: the maximum value determines n, and we need exactly n unique numbers where numbers 1 to n-1 appear once each and n appears twice. Time: O(n), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n²)	O(1)	Check each expected number by scanning the entire array multiple times
One-Pass Hash Table (Optimal)	O(n)	O(n)	Count frequencies and validate pattern simultaneously in one pass
Two Pointers (Sorted Array)	O(n log n)	O(1)	Sort array and verify pattern using two pointers technique

Brute Force (Nested Loops) — Algorithm Steps

Find the maximum number in the array to determine 'n'
Check if array length equals n+1
For each number from 1 to n-1, scan entire array to count occurrences (should be 1)
For number n, scan entire array to count occurrences (should be 2)
Return true if all counts are correct

Visualization

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Step-by-Step Walkthrough

Find Maximum

Scan array to find max value (determines n)

Count Each Number

For each expected number, scan entire array to count occurrences

Validate Counts

Check if counts match expected pattern

Code -

solution.c — C

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

bool isArrayGood(int* nums, int numsSize) {
    if (numsSize == 0) return false;
    
    // Find maximum to determine n
    int maxNum = nums[0];
    for (int i = 1; i < numsSize; i++) {
        if (nums[i] > maxNum) {
            maxNum = nums[i];
        }
    }
    int n = maxNum;
    
    // Check if length is correct
    if (numsSize != n + 1) return false;
    
    // Check each number from 1 to n-1 appears exactly once
    for (int i = 1; i < n; i++) {
        int count = 0;
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] == i) count++;
        }
        if (count != 1) return false;
    }
    
    // Check that n appears exactly twice
    int countN = 0;
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] == n) countN++;
    }
    
    return countN == 2;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n numbers, we scan the entire array of length n

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables for counting

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

847 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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