Bahii honii 3ana problme when bsh ykon 3ana list of integers okii 
Fe list h4ii tnjm tal9a numbers repeat more then 1 time okii 
El objective enik fe kol element enik traj3 el ration mt3 kom element okii 
Q- bahii honii bsh nchofo feme constrain fe problem yes or non ? 
Is theirs a constrain to use extra memory space yes or non? 
→ non okii 




Q- what is the input & output format 
Input : List[int] 
Output : List[float] 

Q- can wal use through an example ? 

Arr =[1,1,0,-1,-1] 

Step 1 : compute the length of the input list which will help use find the right ratio okii 
n=arr.length()
Step 2 : compute the ouccrence for each element 
occ_map={1:2,0:1,-1:2} 
Step 3 : just create a result list which will store the ration calculator 
result=[] 
Now iterate through k  in occ_map element 
	Current_ratio = occ_map[k]/n
	result.append(current_ratio) 
Return result 

Q- i now this a hight level pesuDocode so can you wal to step by step implementation about it 
And using any example you like before implement the full solution
Q- just use my hight level mental model oki

Sure, let's walk through your high-level pseudocode step by step and implement it in Python. We'll use your example array to demonstrate the process.
Problem Statement Recap
Given a list of integers (which may contain duplicates), we need to calculate the ratio of each unique element's occurrences to the total number of elements in the list.
Input: List[int]
Output: List[float] (each float represents the ratio of a unique element's occurrence)
Example
Let's use your example to understand the steps:
Input Array: [1, 1, 0, -1, -1]
Expected Output: [0.4, 0.2, 0.4]
Step-by-Step Implementation
Step 1: Compute the Length of the Input List
First, determine how many elements are in the list. This will be used to calculate the ratio for each unique element.
arr = [1, 1, 0, -1, -1]
n = len(arr)
print(f"Total number of elements (n): {n}")


Output:
Total number of elements (n): 5


Step 2: Compute the Occurrence of Each Element
Next, count how many times each unique element appears in the list. We'll use a dictionary (occ_map) to store these counts.
occ_map = {}
for num in arr:
    if num in occ_map:
        occ_map[num] += 1
    else:
        occ_map[num] = 1

print(f"Occurrence map: {occ_map}")


Output:
Occurrence map: {1: 2, 0: 1, -1: 2}


Alternative using collections.Counter:
Python provides a convenient way to count occurrences using the Counter class from the collections module.
from collections import Counter

occ_map = Counter(arr)
print(f"Occurrence map: {occ_map}")


Output:
Occurrence map: Counter({1: 2, -1: 2, 0: 1})


Step 3: Calculate the Ratio for Each Unique Element
Now, iterate through each unique element in occ_map, calculate its ratio, and store it in the result list.
result = []
for key in occ_map:
    current_ratio = occ_map[key] / n
    result.append(current_ratio)

print(f"Ratio List: {result}")


Output:
Ratio List: [0.4, 0.2, 0.4]


Explanation:
1 appears 2 times: ( \frac{2}{5} = 0.4 )
0 appears 1 time: ( \frac{1}{5} = 0.2 )
-1 appears 2 times: ( \frac{2}{5} = 0.4 )
Complete Python Function
Let's encapsulate the above steps into a function for reusability.
from collections import Counter

def calculate_ratios(arr):
    """
    Calculate the ratio of each unique element's occurrence to the total number of elements.
    
    :param arr: List[int] - The input list of integers
    :return: List[float] - The list of ratios
    """
    n = len(arr)
    if n == 0:
        return []
    
    # Step 2: Compute occurrences
    occ_map = Counter(arr)
    
    # Step 3: Calculate ratios
    result = []
    for key in occ_map:
        ratio = occ_map[key] / n
        result.append(ratio)
    
    return result

# Example usage
arr = [1, 1, 0, -1, -1]
ratios = calculate_ratios(arr)
print(f"Input Array: {arr}")
print(f"Ratios: {ratios}")


Output:
Input Array: [1, 1, 0, -1, -1]
Ratios: [0.4, 0.2, 0.4]


Handling Edge Cases
It's good practice to handle potential edge cases, such as an empty list.
# Test with an empty list
empty_arr = []
empty_ratios = calculate_ratios(empty_arr)
print(f"Input Array: {empty_arr}")
print(f"Ratios: {empty_ratios}")


Output:
Input Array: []
Ratios: []


Preserving Order (Optional)
If you want to preserve the order of elements as they first appear in the input list when computing ratios, you can modify the function as follows:
from collections import OrderedDict

def calculate_ratios_preserve_order(arr):
    """
    Calculate the ratio of each unique element's occurrence to the total number of elements,
    preserving the order of first occurrence.
    
    :param arr: List[int] - The input list of integers
    :return: List[float] - The list of ratios in order of first occurrence
    """
    n = len(arr)
    if n == 0:
        return []
    
    # Use OrderedDict to preserve order
    occ_map = OrderedDict()
    for num in arr:
        if num in occ_map:
            occ_map[num] += 1
        else:
            occ_map[num] = 1
    
    # Calculate ratios
    result = [count / n for count in occ_map.values()]
    
    return result

# Example usage
arr = [1, 1, 0, -1, -1]
ratios = calculate_ratios_preserve_order(arr)
print(f"Input Array: {arr}")
print(f"Ratios (Preserved Order): {ratios}")


Output:
Input Array: [1, 1, 0, -1, -1]
Ratios (Preserved Order): [0.4, 0.2, 0.4]


Summary
You've successfully broken down the problem into manageable steps:
Compute the total number of elements.
Count the occurrences of each unique element.
Calculate the ratio for each unique element's occurrence.
(Optional) Preserve the order of first occurrence in the output.
This method ensures that you can efficiently compute the desired ratios for any list of integers, handling duplicates and maintaining order as needed.
If you have any further questions or need additional features, feel free to ask!

Q- is this a correct solution yes or non 
“#!/bin/python3


import math
import os
import random
import re
import sys
from collections import Counter
# Complete the 'plusMinus' function below.
#
# The function accepts INTEGER_ARRAY arr as parameter.
#


def plusMinus(arr):
    # Write your code here
    n=len(arr)
    if not n:
        return []
    occ_map=Counter(arr)
    result=[]
    for k in occ_map.keys:
        ratio=occ_map[k]/n
        result.append(ratio)
    return result
       


if __name__ == '__main__':
    n = int(input().strip())


    arr = list(map(int, input().rstrip().split()))


    plusMinus(arr)


“