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nordhealth_test/solution.py

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+#!/usr/bin/env python3
+import sys
+from typing import List
+from collections import defaultdict
+
+"""
+Given an unsorted array A[]. The task is to print all unique pairs in the unsorted array with equal sum.
+Note: Print the result in the format as shown in the below examples.
+
+Examples:
+
+Input: A[] = { 6, 4, 12, 10, 22, 54, 32, 42, 21, 11}
+Output:
+Pairs : ( 4, 12) ( 6, 10) have sum : 16
+Pairs : ( 10, 22) ( 21, 11) have sum : 32
+Pairs : ( 12, 21) ( 22, 11) have sum : 33
+Pairs : ( 22, 21) ( 32, 11) have sum : 43
+Pairs : ( 32, 21) ( 42, 11) have sum : 53
+Pairs : ( 12, 42) ( 22, 32) have sum : 54
+Pairs : ( 10, 54) ( 22, 42) have sum : 64
+
+Input:A[]= { 4, 23, 65, 67, 24, 12, 86}
+Output:
+Pairs : ( 4, 86) ( 23, 67) have sum : 90
+"""
+def main() -> None:
+    argc = len(sys.argv)
+    argv = sys.argv
+    if argc != 2:
+        raise ValueError("Invalid input. You need to provide a single path to the input file. \n Ex.: `./solution.py ./input1.txt`")
+    
+    # Read input from the file and process the data
+    try:
+        input_data = read_input(argv[1])
+        print_pairs_with_same_sum(input_data)
+    except ValueError as e:
+        print(f"Error: {e}")
+        sys.exit(1)
+    
+
+def read_input(input_file: str) -> List[int]:
+    """ Reads the input file and returns a list of integers. """
+    try:
+        with open(input_file, "r") as file:
+            lst = file.readline().split(',')
+            parsed = [int(x.strip()) for x in lst]
+            return parsed
+    except Exception as e:
+        raise ValueError(f"Failed to read or parse the file: {e}")
+
+def print_pairs_with_same_sum(arr: List[int]) -> None:
+    """
+    Solution to the problem. Since performance is required it's in one method, but could be separated:
+    1). Calculate the sums into a hashmap and add the pairs
+    2). print the result
+    """
+    
+    # Not a valid array
+    if len(arr) < 2:
+        print("No pairs can be formed from the input file.")
+
+    sum_map = defaultdict(list)
+    
+    # Check for duplicate pairs. 
+    seen_pairs = set()
+
+    n = len(arr)
+    
+    for i in range(n):
+        # Time Complexity (O(n^2)), as we have n(n-1)/2 for the second loop.
+        for j in range(i + 1, n):
+            current_sum = arr[i] + arr[j]
+            pair = tuple((arr[i], arr[j]))
+
+            if pair not in seen_pairs:
+                # This will output nothing for [2,2,2,2]. Depending on the requirements we can store the pair using indices,
+                # then the result would be: 
+                # (A[0],A[1]) (A[0],A[2]) (A[0],A[3]) (A[1],A[2]) (A[1],A[3]) (A[2],A[3]) - sum: 4
+                # And we will not need the seen_pairs set.
+                sum_map[current_sum].append(pair)
+                seen_pairs.add(pair)
+    
+    # This is required to follow the example output, as it is presented as sorted by sum in the task description. 
+    # We can skip the sorting if this is not really part of the requiremetns.
+    for sum_val in sorted(sum_map.keys()):
+        pairs = sum_map[sum_val]
+        
+        if len(pairs) > 1:
+            pairs.sort()
+
+            print(f"Pairs : ", end="")
+            for pair in pairs:
+                print(f"( {pair[0]}, {pair[1]})", end=" ")
+            print(f"have sum : {sum_val}")
+
+
+if __name__ == "__main__":
+    main()