Day 76 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! 🚀

100DaysOfCode #CodingChallenge #ProblemSolving #GeeksforGeeks #DeveloperJourney

Problem:

https://www.geeksforgeeks.org/problems/bus-conductor--170647/1

Bus Conductor
Difficulty: Easy Accuracy: 75.3%
You are conductor of a bus. You are given two arrays chairs[] and passengers[] of equal length, where chairs[i] is the position of the ith chair and passengers[j] is the position of the jth passenger. You may perform the following move any number of times:
• Increase or decrease the position of the ith passenger by 1 (i.e., moving the ith passenger from position x to x+1 or x-1)
Return the minimum number of moves required to move each passenger to get a chair.
Note: Although multiple chairs can occupy the same position, each passenger must be assigned to exactly one unique chair.
Examples:
Input: chairs[] = [3, 1, 5], passengers[] = [2, 7, 4]
Output: 4
Explanation: The passengers are moved as follows:

• The first passenger is moved from position 2 to position 1 using 1 move.
• The second passenger is moved from position 7 to position 5 using 2 moves.
• The third passenger is moved from position 4 to position 3 using 1 move. In total, 1 + 2 + 1 = 4 moves were used. Input: chairs[] = [2, 2, 6, 6], passengers[] = [1, 3, 2, 6] Output: 4 Explanation: Note that there are two chairs at position 2 and two chairs at position 6. The passangers are moved as follows:
• The first passenger is moved from position 1 to position 2 using 1 move.
• The second passenger is moved from position 3 to position 6 using 3 moves.
• The third passenger is not moved.
• The fourth passenger is not moved. In total, 1 + 3 + 0 + 0 = 4 moves were used. Constraints: 1 ≤ n ≤ 105 1 ≤ chairs[i], passengers[j] ≤ 104

Solution:
class Solution:
def findMoves(self, chairs, passengers):
chairs.sort()
passengers.sort()
moves = 0
for c, p in zip(chairs, passengers):
moves += abs(c - p)
return moves