Introduction

In this blog post, we will explore how to solve the LeetCode problem "2731" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Some robots are standing on an infinite number line with their initial coordinates given by a 0-indexed integer array nums and will start moving once given the command to move. The robots will move a unit distance each second. You are given a string s denoting the direction in which robots will move on command. 'L' means the robot will move towards the left side or negative side of the number line, whereas 'R' means the robot will move towards the right side or positive side of the number line. If two robots collide, they will start moving in opposite directions. Return the sum of distances between all the pairs of robots d seconds after the command. Since the sum can be very large, return it modulo 109 + 7. Note: For two robots at the index i and j, pair (i,j) and pair (j,i) are considered the same pair. When robots collide, they instantly change their directions without wasting any time. Collision happens when two robots share the same place in a moment. For example, if a robot is positioned in 0 going to the right and another is positioned in 2 going to the left, the next second they'll be both in 1 and they will change direction and the next second the first one will be in 0, heading left, and another will be in 2, heading right. For example, if a robot is positioned in 0 going to the right and another is positioned in 1 going to the left, the next second the first one will be in 0, heading left, and another will be in 1, heading right. Example 1: Input: nums = [-2,0,2], s = "RLL", d = 3 Output: 8 Explanation: After 1 second, the positions are [-1,-1,1]. Now, the robot at index 0 will move left, and the robot at index 1 will move right. After 2 seconds, the positions are [-2,0,0]. Now, the robot at index 1 will move left, and the robot at index 2 will move right. After 3 seconds, the positions are [-3,-1,1]. The distance between the robot at index 0 and 1 is abs(-3 - (-1)) = 2. The distance between the robot at index 0 and 2 is abs(-3 - 1) = 4. The distance between the robot at index 1 and 2 is abs(-1 - 1) = 2. The sum of the pairs of all distances = 2 + 4 + 2 = 8. Example 2: Input: nums = [1,0], s = "RL", d = 2 Output: 5 Explanation: After 1 second, the positions are [2,-1]. After 2 seconds, the positions are [3,-2]. The distance between the two robots is abs(-2 - 3) = 5. Constraints: 2 <= nums.length <= 105 -2 109 <= nums[i] <= 2 109 0 <= d <= 109 nums.length == s.length s consists of 'L' and 'R' only nums[i] will be unique.

Explanation

Here's the breakdown of the approach, complexity analysis, and the Python code:

High-Level Approach:

• Track Robot Identities: The critical insight is that collisions only swap robot directions, not their underlying identities. So, we can track each robot's original index and initial direction.
• Calculate Final Positions: Based on the initial positions, directions, and the time d, calculate the final position of each robot as if there were no collisions.
• Calculate Distances: Sort the final positions. Then, iterate through the sorted positions, calculating the sum of distances between each robot and all robots to its left using prefix sums to optimize calculation.

Complexity Analysis:

• Runtime Complexity: O(n log n) due to sorting the final positions.
• Storage Complexity: O(n) to store the final positions and related data.

Python Code:

Code

    def sum_of_distances(nums, s, d):
    """
    Calculates the sum of distances between all pairs of robots after d seconds.

    Args:
        nums: A list of initial robot positions.
        s: A string representing the directions of the robots ('L' or 'R').
        d: The time duration.

    Returns:
        The sum of distances modulo 10^9 + 7.
    """

    n = len(nums)
    positions = []
    for i in range(n):
        if s[i] == 'R':
            positions.append(nums[i] + d)
        else:
            positions.append(nums[i] - d)

    positions.sort()

    total_distance = 0
    prefix_sum = 0
    MOD = 10**9 + 7

    for i in range(n):
        total_distance = (total_distance + (positions[i] * i - prefix_sum)) % MOD
        prefix_sum = (prefix_sum + positions[i]) % MOD

    return total_distance