LeetCode //C - 1207. Unique Number of Occurrences

1207. Unique Number of Occurrences

Given an array of integers arr, return true if the number of occurrences of each value in the array is unique or false otherwise.
 

Example 1:

Input: arr = [1,2,2,1,1,3]
Output: true
Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two values have the same number of occurrences.

Example 2:

Input: arr = [1,2]
Output: false

Example 3:

Input: arr = [-3,0,1,-3,1,1,1,-3,10,0]
Output: true

Constraints:

• 1 <= arr.length <= 1000
• -1000 <= arr[i] <= 1000

From: LeetCode
Link: 1207. Unique Number of Occurrences

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Solution:

Ideas：

We can use a hash table (or an array, considering the constraints) to count occurrences. Then, we use another hash table to check for the uniqueness of these counts.

• occurrences is an array indexed from -1000 to 1000 (shifted by 1000 to accommodate negative numbers), used to count the occurrences of each integer in arr.
• seen is a boolean array used to track if a specific occurrence count has been seen before.
• We first iterate through the arr to count the occurrences of each number.
• Then, we iterate through the occurrences array to check if any count has been seen before. If a count is repeated, we return false. If we finish the loop without finding any duplicate counts, we return true.

Code:

bool uniqueOccurrences(int* arr, int arrSize) {
    
    
    int occurrences[2001] = {
    
    0}; // Array to count occurrences, offset by 1000 for negative numbers
    bool seen[1001] = {
    
    false}; // Array to track if an occurrence count has been seen

    // Count occurrences of each number
    for (int i = 0; i < arrSize; i++) {
    
    
        occurrences[arr[i] + 1000]++;
    }

    // Check for unique occurrences
    for (int i = 0; i < 2001; i++) {
    
    
        if (occurrences[i] > 0) {
    
    
            if (seen[occurrences[i]]) {
    
    
                // If we have already seen this count, it's not unique
                return false;
            }
            seen[occurrences[i]] = true;
        }
    }

    return true;
}