One number in the array appears twice, while another number from the range [1, n] is missing.

Our task is to find both:

Repeating Number (A)

Missing Number (B)

Example

Input: nums = [ 1 , 2 , 2 , 4 ]

Output: [ 2 , 3 ]

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Here, 2 appears twice and 3 is missing.

Brute Force Approach

Intuition

For every number from 1 to n , count how many times it appears in the array.

If frequency is 2 , it is the repeating number.

If frequency is 0 , it is the missing number.

Time & Space Complexity

Time Complexity: O(N²)

Space Complexity: O(1)

Java Code

class Solution { public int [] findErrorNums ( int [] nums ) {

int repeating = - 1 ; int missing = - 1 ; int n = nums .…