2731. Movement of Robots

Problem Description

In this problem, we are provided with an array nums which represents the initial positions of a group of robots on an infinite number line, where each element is a unique robot's starting position. Along with this, we have a string s that contains the direction each robot will head towards once they're commanded to move. The directions 'L' and 'R' stand for left (toward negative infinity) and right (toward positive infinity), respectively. Robots will move at a rate of one unit per second.

A key point in the problem is that if two robots meet at the same point at the same time, they instantaneously switch directions but continue moving without any delay. This collision means that we could consider the robots as if they pass through each other and continue in their initial directions for the purpose of calculation.

The task is to calculate the sum of all distances between pairs of robots after a given number of seconds d and to return this sum modulo 10^9 + 7 due to the potential for large numbers.

Intuition

At first glance, calculating the distances between robots after an arbitrary number of seconds seems complicated, especially when considering potential collisions. However, the key realization that simplifies this problem is to understand that we don't actually have to simulate the movement of robots and handle their collisions.

Because robots change direction instantaneously upon collision and there is no delay, we can imagine that the robots simply pass through each other and continue in their original direction. This means that after d seconds, regardless of any collisions that might have happened, we can calculate the new position of each robot by simply adding or subtracting d from their initial positions based on their intended direction.

After determining the theoretical new positions, we can sort these positions in ascending order. We then iterate through this sorted list to calculate the running sum of distances between each robot and all that come before it. This is effectively the cumulative distance from each of the robots to the start.

Summing up the distances while iterating through the sorted list provides the total sum of distances between all pairs of robots after d seconds. We take care to apply the modulo operation as required to handle the potential for very large numbers.

This approach negates the need to handle the complexity of collisions during the movement phase, which is what makes the problem solvable within the given constraints.

Learn more about Prefix Sum and Sorting patterns.

Solution Approach

The implementation of this problem's solution uses simple yet elegant logic and a few well-established Python techniques.

Firstly, we manipulate the initial positions of the robots according to the directions specified by the s string. This is carried out by a combination of enumeration and conditional addition or subtraction. Depending on whether the direction at index i in s is 'R' (right) or 'L' (left), we add or subtract the duration d from the corresponding starting position in nums.

The updated positions array structure is then sorted. Sorting here uses the default O(n log n) sorting algorithm, which is common practice in Python. Sorting is crucial as it prepares us to compute distances without having to track individual movements or collisions further.

Next, an iterative approach is used. The loop calculates the running sum of distances by traversing through the sorted array of new positions. To achieve this, two accumulator variables are used: ans, which accumulates the sum of distances, and s, which serves as the running sum of robot positions. The distance from a given robot to all robots before it in the sorted list is i * x - s for each robot x at index i. The product i * x gives the sum of distances from all previous robots to the current robot if they were i units apart, while subtracting s corrects for their actual positions.

Finally, the solution modulo 10^9 + 7 is applied to the cumulative distance sum to obtain the answer required by the problem statement.

The Python code implementation follows a clear sequence and is succinct, relying on a simple enumeration pattern and basic arithmetic to calculate the cumulative distances in the sorted array of robot positions. It does not employ complex data structures or algorithms but uses familiar constructs such as loops and sort functions effectively to derive the solution.

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Example Walkthrough

Let's illustrate the solution approach with a small example:

Suppose we have the array nums = [4, 1, 5] representing the starting positions of robots and a string s = "RLR" that indicates the direction each robot will move. We are also given d = 2 seconds to simulate the movement.

We process each position according to the direction provided. After d seconds, the new positions will be calculated as follows:
- For the first robot at nums[0] = 4 with direction s[0] = 'R', the new position will be 4 + 2 = 6.
- The second robot starts at nums[1] = 1 with direction s[1] = 'L', giving us a new position of 1 - 2 = -1.
- The third robot starts at nums[2] = 5 with direction s[2] = 'R', making the new position 5 + 2 = 7.
The updated array of positions then becomes [6, -1, 7].
We sort this array to get the order after d seconds: [-1, 6, 7].
To find the sum of all distances between pairs of robots after moving, we iterate through the sorted list and calculate the cumulative distance:
- The distance from -1 to 6 is 6 - (-1) = 7.
- The distance from 6 to 7 is 7 - 6 = 1.
Now we calculate the sum of all distances:
- For the first robot (-1), the cumulative sum is 0.
- For the second robot (6), we have (1 * 6) - (-1) = 6 + 1 = 7.
- For the final robot (7), we have (2 * 7) - (6 - 1) = 14 - 5 = 9.
The total sum of distances is therefore 0 + 7 + 9 = 16.

Applying the solution modulo 10^9 + 7, the final answer remains 16, since 16 is much smaller than 10^9 + 7.

Using this walkthrough, we gain a clear understanding of how to approach and solve the problem as outlined in the solution approach without directly considering robot collisions. This simplifies the calculations and allows for an efficient solution to the problem.

Solution Implementation

1class Solution:
2    def sumDistance(self, numbers: List[int], directions: str, distance: int) -> int:
3        # Define the modulo value for large numbers to prevent overflow
4        modulo_value = 10**9 + 7
5
6        # Update each number in the list according to the corresponding direction character
7        for index, direction_char in enumerate(directions):
8            if direction_char == "R":
9                # If character is "R", add the distance value
10                numbers[index] += distance
11            else:
12                # If character is not "R", subtract the distance value (implying "L" is encountered)
13                numbers[index] -= distance
14
15        # Sort the numbers to calculate total sum distance
16        numbers.sort()
17
18        # Initialize variables for the answer and the cumulative sum
19        total_sum_distance = 0
20        cumulative_sum = 0
21
22        # Calculate the sum-distance (sum of all |Pj - Pi|)
23        for index, number in enumerate(numbers):
24            # The current element's contribution is its value times its index
25            # minus the cumulative sum of all previous elements
26            total_sum_distance += index * number - cumulative_sum
27
28            # Update the cumulative sum with the current number
29            cumulative_sum += number
30
31        # Return the total sum distance modulo the defined modulo_value
32        return total_sum_distance % modulo_value
33

1class Solution {
2    public int sumDistance(int[] numbers, String direction, int distance) {
3        // Get the length of the input array
4        int n = numbers.length;
5      
6        // Create an array to store adjusted distances
7        long[] adjustedDistances = new long[n];
8      
9        // Calculate adjusted distances based on direction and store them
10        for (int i = 0; i < n; ++i) {
11            // Subtract or add the distance based on if the direction is 'L' or 'R'
12            adjustedDistances[i] = (long) numbers[i] + (direction.charAt(i) == 'L' ? -distance : distance);
13        }
14      
15        // Sort the adjusted distances
16        Arrays.sort(adjustedDistances);
17      
18        // Initialize variables for storing the result and the cumulative sum
19        long result = 0, cumulativeSum = 0;
20      
21        // Define modulo constant for large number handling
22        final int modulo = (int) 1e9 + 7;
23      
24        // Calculate the weighted sum of distances and update result
25        for (int i = 0; i < n; ++i) {
26            // Update the result with the current index times the element minus the cumulative sum so far
27            result = (result + i * adjustedDistances[i] - cumulativeSum) % modulo;
28            // Update cumulative sum with the current element's value
29            cumulativeSum += adjustedDistances[i];
30        }
31      
32        // Return the result cast back to integer
33        return (int) result;
34    }
35}
36

1#include <algorithm>  // Required for std::sort
2#include <vector>     // Required for std::vector
3#include <string>     // Required for std::string
4
5class Solution {
6public:
7    // Function to calculate the sum of distances based on conditions.
8    int sumDistance(vector<int>& nums, string s, int d) {
9        int n = nums.size(); // Get the size of the input vector 'nums'.
10        vector<long long> adjustedPositions(n); // Create a vector to store adjusted positions.
11
12        // Calculate adjusted positions based on 'L' or 'R' in string 's'.
13        for (int i = 0; i < n; ++i) {
14            adjustedPositions[i] = 1LL * nums[i] + (s[i] == 'L' ? -d : d);
15        }
16
17        // Sort the adjusted positions to access them in non-decreasing order.
18        sort(adjustedPositions.begin(), adjustedPositions.end());
19
20        long long answer = 0; // Initialize the sum of distances as 0.
21        long long prefixSum = 0; // Keep track of the sum of previous adjusted positions.
22        const int MOD = 1e9 + 7; // Define the modulus value for avoiding integer overflow.
23
24        // Calculate the sum of distances using prefix sums.
25        for (int i = 0; i < n; ++i) {
26            answer = (answer + i * adjustedPositions[i] - prefixSum) % MOD;
27            prefixSum = (prefixSum + adjustedPositions[i]) % MOD; // Update prefix sum.
28        }
29        return answer; // Return final answer.
30    }
31};
32

1function sumDistance(nums: number[], directions: string, distance: number): number {
2    // Define the length of the nums array
3    const length = nums.length;
4
5    // Update the nums array by adding or subtracting the distance based on the direction
6    for (let i = 0; i < length; ++i) {
7        nums[i] += directions[i] === 'L' ? -distance : distance;
8    }
9
10    // Sort the nums array in ascending order
11    nums.sort((a, b) => a - b);
12
13    // Initialize the answer and a variable to keep track of the cumulative sum
14    let answer = 0;
15    let cumulativeSum = 0;
16
17    // Define the modulus value to handle large numbers
18    const modulus = 1e9 + 7;
19
20    // Iterate over nums to calculate the final answer
21    for (let i = 0; i < length; ++i) {
22        // Update the answer according to the formula
23        answer = (answer + i * nums[i] - cumulativeSum) % modulus;
24
25        // Update the cumulative sum with the current number
26        cumulativeSum += nums[i];
27    }
28
29    // Return the answer
30    return answer;
31}
32

Time and Space Complexity

The given code's time complexity primarily comes from the sorting operation.

Sorting a list of n elements typically takes O(n log n) time. This is the dominating factor in the time complexity of this function.
The remaining part of the code iterates over the list once, which is an O(n) operation. However, since O(n log n) + O(n) simplifies to O(n log n), the overall time complexity remains O(n log n).

For space complexity:

The given code modifies the input list nums in-place and uses a fixed number of integer variables (mod, ans, s). Thus, apart from the input list, only constant extra space is used.
However, since the input list nums itself takes O(n) space, and we consider the space taken by inputs for space complexity analysis, the overall space complexity is O(n).

Therefore, we can conclude that:

The time complexity of the code is O(n log n).
The space complexity of the code is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.

Discover Your Strengths and Weaknesses: Take Our 2-Minute Quiz to Tailor Your Study Plan:

Question 1 out of 10

What's the output of running the following function using input [30, 20, 10, 100, 33, 12]?

1def fun(arr: List[int]) -> List[int]:
2    import heapq
3    heapq.heapify(arr)
4    res = []
5    for i in range(3):
6        res.append(heapq.heappop(arr))
7    return res
8

1public static int[] fun(int[] arr) {
2    int[] res = new int[3];
3    PriorityQueue<Integer> heap = new PriorityQueue<>();
4    for (int i = 0; i < arr.length; i++) {
5        heap.add(arr[i]);
6    }
7    for (int i = 0; i < 3; i++) {
8        res[i] = heap.poll();
9    }
10    return res;
11}
12

1class HeapItem {
2    constructor(item, priority = item) {
3        this.item = item;
4        this.priority = priority;
5    }
6}
7
8class MinHeap {
9    constructor() {
10        this.heap = [];
11    }
12
13    push(node) {
14        // insert the new node at the end of the heap array
15        this.heap.push(node);
16        // find the correct position for the new node
17        this.bubble_up();
18    }
19
20    bubble_up() {
21        let index = this.heap.length - 1;
22
23        while (index > 0) {
24            const element = this.heap[index];
25            const parentIndex = Math.floor((index - 1) / 2);
26            const parent = this.heap[parentIndex];
27
28            if (parent.priority <= element.priority) break;
29            // if the parent is bigger than the child then swap the parent and child
30            this.heap[index] = parent;
31            this.heap[parentIndex] = element;
32            index = parentIndex;
33        }
34    }
35
36    pop() {
37        const min = this.heap[0];
38        this.heap[0] = this.heap[this.size() - 1];
39        this.heap.pop();
40        this.bubble_down();
41        return min;
42    }
43
44    bubble_down() {
45        let index = 0;
46        let min = index;
47        const n = this.heap.length;
48
49        while (index < n) {
50            const left = 2 * index + 1;
51            const right = left + 1;
52
53            if (left < n && this.heap[left].priority < this.heap[min].priority) {
54                min = left;
55            }
56            if (right < n && this.heap[right].priority < this.heap[min].priority) {
57                min = right;
58            }
59            if (min === index) break;
60            [this.heap[min], this.heap[index]] = [this.heap[index], this.heap[min]];
61            index = min;
62        }
63    }
64
65    peek() {
66        return this.heap[0];
67    }
68
69    size() {
70        return this.heap.length;
71    }
72}
73
74function fun(arr) {
75    const heap = new MinHeap();
76    for (const x of arr) {
77        heap.push(new HeapItem(x));
78    }
79    const res = [];
80    for (let i = 0; i < 3; i++) {
81        res.push(heap.pop().item);
82    }
83    return res;
84}
85

[30, 20, 10]

[100, 33, 30]

[10, 12, 20]

I don’t know