Day 68 of 100 days dsa coding challenge

Taking on a new challenge: solving GeeksforGeeks POTD daily and sharing my solutions! 💻🔥
The goal: sharpen problem-solving skills, level up coding, and learn something new every day. Follow my journey! 🚀

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Problem:

https://www.geeksforgeeks.org/problems/construct-an-array-from-its-pair-sum-array/1

Construct an array from its pair-sum array

Difficulty: Easy Accuracy: 49.0%

Given a pair-sum array arr[] construct the original array. A pair-sum array for an array is the array that contains sum of all pairs in ordered form, i.e., arr[0] is sum of res[0] and res[1], arr[1] is sum of res[0] and res[2] and so on.
Note: If the size of original array res[] is n, then the size of pair-sum array arr[] would be n * (n -1) /2. We may assume that the pair-sum array arr[] is appropriate in size.
Note that, if the original array is correct then the driver code will print true, else false;

Examples:
Input: arr[] = [4, 5, 3]
Output: true
Explanation: A valid original array is [3, 1, 2], pairwise sums are (3 + 1), (3 + 2) and (1 + 2).
Input: arr[] = [3]
Output: true
Explanation: One of the valid original array is [1, 2].
Constraints:
1 ≤ n ≤ 103
1 ≤ arr[i] ≤ 109

Solution:
class Solution:
def constructArr(self, arr):
if len(arr) == 1:
return [1, arr[0]-1]

    s1, s2, s3 = arr[0], arr[1], arr[2]
    a = (s1 + s2 - s3) // 2
    res = [a]

    n = 2
    i = 1
    while len(res) < n:
        res.append(arr[i] - a)
        i += 1

    expected_pairs = []
    for i in range(len(res)):
        for j in range(i + 1, len(res)):
            expected_pairs.append(res[i] + res[j])

    expected_pairs.sort()
    return res if expected_pairs == arr else []