Check if Array is Good

Check if Array is Good - Problem

You are given an integer array nums. We consider an array good if it is a permutation of an array base[n].

base[n] = [1, 2, ..., n - 1, n, n] (in other words, it is an array of length n + 1 which contains 1 to n - 1 exactly once, plus two occurrences of n). For example, base[1] = [1, 1] and base[3] = [1, 2, 3, 3].

Return true if the given array is good, otherwise return false.

Note: A permutation of integers represents an arrangement of these numbers.

Input & Output

Example 1 — Valid Good Array

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

› Output: true

💡 Note: This array has length 4, so we expect base[3] = [1,2,3,3]. The input is a permutation of base[3], containing: 1 once, 2 once, and 3 twice.

Example 2 — Invalid Array

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

› Output: true

💡 Note: This array has length 4, so we expect base[3] = [1,2,3,3]. The input is a permutation of [1,2,3,3], just in different order.

Example 3 — Wrong Pattern

$ Input: nums = [1,1]

› Output: true

💡 Note: This array has length 2, so we expect base[1] = [1,1]. The input matches exactly.

Constraints

1 ≤ nums.length ≤ 100
1 ≤ nums[i] ≤ 1000

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that a good array must be a permutation of base[n] = [1,2,...,n-1,n,n]. Best approach uses frequency counting to verify the pattern in O(n) time. Time: O(n), Space: O(n).

Common Approaches

✓ Brute Force - Count Each Number

⏱️ Time: O(n²) Space: O(1)

Check if array length is n+1, then verify each number 1 to n-1 appears exactly once and n appears exactly twice by counting occurrences.

Frequency Count with Hash Map

⏱️ Time: O(n) Space: O(n)

Count frequency of each number using hash map, then check if frequencies match the expected pattern for a good array.

Sort and Compare

⏱️ Time: O(n log n) Space: O(1)

Sort the input array and directly compare it with the sorted base[n] array to check if they match exactly.

Brute Force - Count Each Number — Algorithm Steps

Step 1: Check if array length is valid (n+1)
Step 2: For each expected number, count its occurrences
Step 3: Verify counts match the good array pattern

Visualization

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Step-by-Step Walkthrough

Input Array

Given array [2,1,3,3] with length 4

Count Pattern

Check 1→1 time, 2→1 time, 3→2 times

Verify

All counts match base[3] pattern

Code -

solution.c — C

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

bool solution(int* nums, int numsSize) {
    if (numsSize < 2) return false;
    
    int expectedMax = numsSize - 1;
    
    // Check each number from 1 to expectedMax-1 appears exactly once
    for (int i = 1; i < expectedMax; i++) {
        int count = 0;
        for (int j = 0; j < numsSize; j++) {
            if (nums[j] == i) count++;
        }
        if (count != 1) return false;
    }
    
    // Check that expectedMax appears exactly twice
    int count = 0;
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] == expectedMax) count++;
    }
    
    return count == 2;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n numbers, we scan the entire array to count occurrences

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for counting, no extra data structures

✓ Linear Space

18.5K Views

Medium Frequency

~15 min Avg. Time

340 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Count Each Number — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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