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Occurence of Each Number ​

Problem Description ​

You are given an integer array A.

You have to find the number of occurences of each number.

Return an array containing only the occurences with the smallest value's occurence first.

For example, A = [4, 3, 3], you have to return an array [2, 1], where 2 is the number of occurences for element 3, and 1 is the number of occurences for element 4. But, 2 comes first because 3 is smaller than 4.

Problem Constraints

1 <= |A| <= 10^5
1 <= Ai <= 10^9
1 <= |A| <= 10^5
1 <= Ai <= 10^9

Input Format

The first argument is the integer array A.
The first argument is the integer array A.

Output Format

Return an integer array denoting the occurences of each number.
Return an integer array denoting the occurences of each number.

Example Input

Input 1:
A = [1, 2, 3]

Input 2:
A = [4, 3, 3]
Input 1:
A = [1, 2, 3]

Input 2:
A = [4, 3, 3]

Example Output

Output 1:
[1, 1, 1]

Output 2:
[2, 1]
Output 1:
[1, 1, 1]

Output 2:
[2, 1]

Example Explanation

Explanation 1:
All the elements occur once, so the resultant array should be [1, 1, 1].

Explanation 2:
Explained in the description above.
Explanation 1:
All the elements occur once, so the resultant array should be [1, 1, 1].

Explanation 2:
Explained in the description above.

Solution ​

swift
import Foundation

class Solution {
	func findOccurences(_ A: inout [Int]) -> [Int] {
        var B = [Int:Int]()
        for v in A {
            if let c = B[v] {
                B[v] = (c + 1)
            } else {
                B[v] = 1
            }
        }
        
        return B.sorted{$0.key < $1.key}.map{$0.value}
	}
}
import Foundation

class Solution {
	func findOccurences(_ A: inout [Int]) -> [Int] {
        var B = [Int:Int]()
        for v in A {
            if let c = B[v] {
                B[v] = (c + 1)
            } else {
                B[v] = 1
            }
        }
        
        return B.sorted{$0.key < $1.key}.map{$0.value}
	}
}

References ​