Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
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:
i and j, pair (i,j) and pair (j,i) are considered the same pair.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 * 1090 <= d <= 109nums.length == s.length s consists of 'L' and 'R' onlynums[i] will be unique.Problem summary: 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
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[-2,0,2] "RLL" 3
[1,0] "RL" 2
last-moment-before-all-ants-fall-out-of-a-plank)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2731: Movement of Robots
class Solution {
public int sumDistance(int[] nums, String s, int d) {
int n = nums.length;
long[] arr = new long[n];
for (int i = 0; i < n; ++i) {
arr[i] = (long) nums[i] + (s.charAt(i) == 'L' ? -d : d);
}
Arrays.sort(arr);
long ans = 0, sum = 0;
final int mod = (int) 1e9 + 7;
for (int i = 0; i < n; ++i) {
ans = (ans + i * arr[i] - sum) % mod;
sum += arr[i];
}
return (int) ans;
}
}
// Accepted solution for LeetCode #2731: Movement of Robots
func sumDistance(nums []int, s string, d int) (ans int) {
for i, c := range s {
if c == 'R' {
nums[i] += d
} else {
nums[i] -= d
}
}
sort.Ints(nums)
sum := 0
const mod int = 1e9 + 7
for i, x := range nums {
ans = (ans + i*x - sum) % mod
sum += x
}
return
}
# Accepted solution for LeetCode #2731: Movement of Robots
class Solution:
def sumDistance(self, nums: List[int], s: str, d: int) -> int:
mod = 10**9 + 7
for i, c in enumerate(s):
nums[i] += d if c == "R" else -d
nums.sort()
ans = s = 0
for i, x in enumerate(nums):
ans += i * x - s
s += x
return ans % mod
// Accepted solution for LeetCode #2731: Movement of Robots
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #2731: Movement of Robots
// class Solution {
// public int sumDistance(int[] nums, String s, int d) {
// int n = nums.length;
// long[] arr = new long[n];
// for (int i = 0; i < n; ++i) {
// arr[i] = (long) nums[i] + (s.charAt(i) == 'L' ? -d : d);
// }
// Arrays.sort(arr);
// long ans = 0, sum = 0;
// final int mod = (int) 1e9 + 7;
// for (int i = 0; i < n; ++i) {
// ans = (ans + i * arr[i] - sum) % mod;
// sum += arr[i];
// }
// return (int) ans;
// }
// }
// Accepted solution for LeetCode #2731: Movement of Robots
function sumDistance(nums: number[], s: string, d: number): number {
const n = nums.length;
for (let i = 0; i < n; ++i) {
nums[i] += s[i] === 'L' ? -d : d;
}
nums.sort((a, b) => a - b);
let ans = 0;
let sum = 0;
const mod = 1e9 + 7;
for (let i = 0; i < n; ++i) {
ans = (ans + i * nums[i] - sum) % mod;
sum += nums[i];
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.