LeetCode #811 — MEDIUM

Subdomain Visit Count

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

A website domain "discuss.leetcode.com" consists of various subdomains. At the top level, we have "com", at the next level, we have "leetcode.com" and at the lowest level, "discuss.leetcode.com". When we visit a domain like "discuss.leetcode.com", we will also visit the parent domains "leetcode.com" and "com" implicitly.

A count-paired domain is a domain that has one of the two formats "rep d1.d2.d3" or "rep d1.d2" where rep is the number of visits to the domain and d1.d2.d3 is the domain itself.

For example, "9001 discuss.leetcode.com" is a count-paired domain that indicates that discuss.leetcode.com was visited 9001 times.

Given an array of count-paired domains cpdomains, return an array of the count-paired domains of each subdomain in the input. You may return the answer in any order.

Example 1:

Input: cpdomains = ["9001 discuss.leetcode.com"]
Output: ["9001 leetcode.com","9001 discuss.leetcode.com","9001 com"]
Explanation: We only have one website domain: "discuss.leetcode.com".
As discussed above, the subdomain "leetcode.com" and "com" will also be visited. So they will all be visited 9001 times.

Example 2:

Input: cpdomains = ["900 google.mail.com", "50 yahoo.com", "1 intel.mail.com", "5 wiki.org"]
Output: ["901 mail.com","50 yahoo.com","900 google.mail.com","5 wiki.org","5 org","1 intel.mail.com","951 com"]
Explanation: We will visit "google.mail.com" 900 times, "yahoo.com" 50 times, "intel.mail.com" once and "wiki.org" 5 times.
For the subdomains, we will visit "mail.com" 900 + 1 = 901 times, "com" 900 + 50 + 1 = 951 times, and "org" 5 times.

Constraints:

1 <= cpdomain.length <= 100
1 <= cpdomain[i].length <= 100
cpdomain[i] follows either the "rep_i d1_i.d2_i.d3_i" format or the "rep_i d1_i.d2_i" format.
rep_i is an integer in the range [1, 10⁴].
d1_i, d2_i, and d3_i consist of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: A website domain "discuss.leetcode.com" consists of various subdomains. At the top level, we have "com", at the next level, we have "leetcode.com" and at the lowest level, "discuss.leetcode.com". When we visit a domain like "discuss.leetcode.com", we will also visit the parent domains "leetcode.com" and "com" implicitly. A count-paired domain is a domain that has one of the two formats "rep d1.d2.d3" or "rep d1.d2" where rep is the number of visits to the domain and d1.d2.d3 is the domain itself. For example, "9001 discuss.leetcode.com" is a count-paired domain that indicates that discuss.leetcode.com was visited 9001 times. Given an array of count-paired domains cpdomains, return an array of the count-paired domains of each subdomain in the input. You may return the answer in any order.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

["9001 discuss.leetcode.com"]

Example 2

["900 google.mail.com", "50 yahoo.com", "1 intel.mail.com", "5 wiki.org"]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #811: Subdomain Visit Count
class Solution {
    public List<String> subdomainVisits(String[] cpdomains) {
        Map<String, Integer> cnt = new HashMap<>();
        for (String s : cpdomains) {
            int i = s.indexOf(" ");
            int v = Integer.parseInt(s.substring(0, i));
            for (; i < s.length(); ++i) {
                if (s.charAt(i) == ' ' || s.charAt(i) == '.') {
                    String t = s.substring(i + 1);
                    cnt.put(t, cnt.getOrDefault(t, 0) + v);
                }
            }
        }
        List<String> ans = new ArrayList<>();
        for (var e : cnt.entrySet()) {
            ans.add(e.getValue() + " " + e.getKey());
        }
        return ans;
    }
}

// Accepted solution for LeetCode #811: Subdomain Visit Count
func subdomainVisits(cpdomains []string) []string {
	cnt := map[string]int{}
	for _, s := range cpdomains {
		i := strings.IndexByte(s, ' ')
		v, _ := strconv.Atoi(s[:i])
		for ; i < len(s); i++ {
			if s[i] == ' ' || s[i] == '.' {
				cnt[s[i+1:]] += v
			}
		}
	}
	ans := make([]string, 0, len(cnt))
	for s, v := range cnt {
		ans = append(ans, strconv.Itoa(v)+" "+s)
	}
	return ans
}

# Accepted solution for LeetCode #811: Subdomain Visit Count
class Solution:
    def subdomainVisits(self, cpdomains: List[str]) -> List[str]:
        cnt = Counter()
        for s in cpdomains:
            v = int(s[: s.index(' ')])
            for i, c in enumerate(s):
                if c in ' .':
                    cnt[s[i + 1 :]] += v
        return [f'{v} {s}' for s, v in cnt.items()]

// Accepted solution for LeetCode #811: Subdomain Visit Count
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn subdomain_visits(cpdomains: Vec<String>) -> Vec<String> {
        let mut res: Vec<String> = vec![];
        let mut hm: HashMap<String, usize> = HashMap::new();
        for s in cpdomains {
            let (domains, count) = Self::parse(s);
            for domain in domains {
                let e = hm.entry(domain).or_default();
                *e += count;
            }
        }
        for (domain, count) in hm {
            res.push(format!("{} {}", count, domain));
        }
        res
    }

    fn parse(s: String) -> (Vec<String>, usize) {
        let mut domains: Vec<String> = vec![];
        let mut iter = s.split_whitespace();
        let count: usize = iter.next().unwrap().parse::<usize>().unwrap();
        let domain: String = iter.next().unwrap().parse::<String>().unwrap();
        for (i, c) in domain.chars().enumerate() {
            if c == '.' {
                let subdomain = &domain[i + 1..];
                domains.push(subdomain.to_string());
            }
        }
        domains.push(domain);
        (domains, count)
    }
}

#[test]
fn test() {
    let input: Vec<String> = vec_string!["9001 discuss.leetcode.com"];
    let mut answer: Vec<String> =
        vec_string!["9001 discuss.leetcode.com", "9001 leetcode.com", "9001 com"];
    let mut output = Solution::subdomain_visits(input);
    answer.sort();
    output.sort();
    assert_eq!(answer, output);

    let input: Vec<String> = vec_string![
        "900 google.mail.com",
        "50 yahoo.com",
        "1 intel.mail.com",
        "5 wiki.org"
    ];
    let mut answer: Vec<String> = vec_string![
        "901 mail.com",
        "50 yahoo.com",
        "900 google.mail.com",
        "5 wiki.org",
        "5 org",
        "1 intel.mail.com",
        "951 com"
    ];
    let mut output = Solution::subdomain_visits(input);
    answer.sort();
    output.sort();
    assert_eq!(answer, output);
}

// Accepted solution for LeetCode #811: Subdomain Visit Count
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #811: Subdomain Visit Count
// class Solution {
//     public List<String> subdomainVisits(String[] cpdomains) {
//         Map<String, Integer> cnt = new HashMap<>();
//         for (String s : cpdomains) {
//             int i = s.indexOf(" ");
//             int v = Integer.parseInt(s.substring(0, i));
//             for (; i < s.length(); ++i) {
//                 if (s.charAt(i) == ' ' || s.charAt(i) == '.') {
//                     String t = s.substring(i + 1);
//                     cnt.put(t, cnt.getOrDefault(t, 0) + v);
//                 }
//             }
//         }
//         List<String> ans = new ArrayList<>();
//         for (var e : cnt.entrySet()) {
//             ans.add(e.getValue() + " " + e.getKey());
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

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.

OPTIMIZED

O(n) time

O(1) space

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.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.