LeetCode #1487 — MEDIUM

Making File Names Unique

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

The Problem

Problem Statement

Given an array of strings names of size n. You will create n folders in your file system such that, at the i^th minute, you will create a folder with the name names[i].

Since two files cannot have the same name, if you enter a folder name that was previously used, the system will have a suffix addition to its name in the form of (k), where, k is the smallest positive integer such that the obtained name remains unique.

Return an array of strings of length n where ans[i] is the actual name the system will assign to the i^th folder when you create it.

Example 1:

Input: names = ["pes","fifa","gta","pes(2019)"]
Output: ["pes","fifa","gta","pes(2019)"]
Explanation: Let's see how the file system creates folder names:
"pes" --> not assigned before, remains "pes"
"fifa" --> not assigned before, remains "fifa"
"gta" --> not assigned before, remains "gta"
"pes(2019)" --> not assigned before, remains "pes(2019)"

Example 2:

Input: names = ["gta","gta(1)","gta","avalon"]
Output: ["gta","gta(1)","gta(2)","avalon"]
Explanation: Let's see how the file system creates folder names:
"gta" --> not assigned before, remains "gta"
"gta(1)" --> not assigned before, remains "gta(1)"
"gta" --> the name is reserved, system adds (k), since "gta(1)" is also reserved, systems put k = 2. it becomes "gta(2)"
"avalon" --> not assigned before, remains "avalon"

Example 3:

Input: names = ["onepiece","onepiece(1)","onepiece(2)","onepiece(3)","onepiece"]
Output: ["onepiece","onepiece(1)","onepiece(2)","onepiece(3)","onepiece(4)"]
Explanation: When the last folder is created, the smallest positive valid k is 4, and it becomes "onepiece(4)".

Constraints:

1 <= names.length <= 5 * 10⁴
1 <= names[i].length <= 20
names[i] consists of lowercase English letters, digits, and/or round brackets.

Step 01

Brute Force Baseline

Problem summary: Given an array of strings names of size n. You will create n folders in your file system such that, at the ith minute, you will create a folder with the name names[i]. Since two files cannot have the same name, if you enter a folder name that was previously used, the system will have a suffix addition to its name in the form of (k), where, k is the smallest positive integer such that the obtained name remains unique. Return an array of strings of length n where ans[i] is the actual name the system will assign to the ith folder when you create it.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

["pes","fifa","gta","pes(2019)"]

Example 2

["gta","gta(1)","gta","avalon"]

Example 3

["onepiece","onepiece(1)","onepiece(2)","onepiece(3)","onepiece"]

Step 02

Core Insight

What unlocks the optimal approach

Keep a map of each name and the smallest valid integer that can be appended as a suffix to it.
If the name is not present in the map, you can use it without adding any suffixes.
If the name is present in the map, append the smallest proper suffix, and add the new name to the map.

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 #1487: Making File Names Unique
class Solution {
    public String[] getFolderNames(String[] names) {
        Map<String, Integer> d = new HashMap<>();
        for (int i = 0; i < names.length; ++i) {
            if (d.containsKey(names[i])) {
                int k = d.get(names[i]);
                while (d.containsKey(names[i] + "(" + k + ")")) {
                    ++k;
                }
                d.put(names[i], k);
                names[i] += "(" + k + ")";
            }
            d.put(names[i], 1);
        }
        return names;
    }
}

// Accepted solution for LeetCode #1487: Making File Names Unique
func getFolderNames(names []string) []string {
	d := map[string]int{}
	for i, name := range names {
		if k, ok := d[name]; ok {
			for {
				newName := fmt.Sprintf("%s(%d)", name, k)
				if d[newName] == 0 {
					d[name] = k + 1
					names[i] = newName
					break
				}
				k++
			}
		}
		d[names[i]] = 1
	}
	return names
}

# Accepted solution for LeetCode #1487: Making File Names Unique
class Solution:
    def getFolderNames(self, names: List[str]) -> List[str]:
        d = defaultdict(int)
        for i, name in enumerate(names):
            if name in d:
                k = d[name]
                while f'{name}({k})' in d:
                    k += 1
                d[name] = k + 1
                names[i] = f'{name}({k})'
            d[names[i]] = 1
        return names

// Accepted solution for LeetCode #1487: Making File Names Unique
struct Solution;
use std::collections::HashMap;
use std::collections::HashSet;

impl Solution {
    fn get_folder_names(names: Vec<String>) -> Vec<String> {
        let mut hs: HashSet<String> = HashSet::new();
        let mut hm: HashMap<String, usize> = HashMap::new();
        let mut res = vec![];
        'outer: for name in names {
            if !hs.insert(name.to_string()) {
                let mut i = *hm.get(&name).unwrap_or(&1);
                loop {
                    let new_name = format!("{}({})", name, i);
                    if hs.insert(new_name.to_string()) {
                        res.push(new_name);
                        hm.insert(name, i + 1);
                        continue 'outer;
                    }
                    i += 1;
                }
            } else {
                res.push(name);
            }
        }
        res
    }
}

#[test]
fn test() {
    let names = vec_string!["pes", "fifa", "gta", "pes(2019)"];
    let res = vec_string!["pes", "fifa", "gta", "pes(2019)"];
    assert_eq!(Solution::get_folder_names(names), res);
    let names = vec_string!["gta", "gta(1)", "gta", "avalon"];
    let res = vec_string!["gta", "gta(1)", "gta(2)", "avalon"];
    assert_eq!(Solution::get_folder_names(names), res);
    let names = vec_string![
        "onepiece",
        "onepiece(1)",
        "onepiece(2)",
        "onepiece(3)",
        "onepiece"
    ];
    let res = vec_string![
        "onepiece",
        "onepiece(1)",
        "onepiece(2)",
        "onepiece(3)",
        "onepiece(4)"
    ];
    assert_eq!(Solution::get_folder_names(names), res);
    let names = vec_string!["wano", "wano", "wano", "wano"];
    let res = vec_string!["wano", "wano(1)", "wano(2)", "wano(3)"];
    assert_eq!(Solution::get_folder_names(names), res);
    let names = vec_string!["kaido", "kaido(1)", "kaido", "kaido(1)"];
    let res = vec_string!["kaido", "kaido(1)", "kaido(2)", "kaido(1)(1)"];
    assert_eq!(Solution::get_folder_names(names), res);
}

// Accepted solution for LeetCode #1487: Making File Names Unique
function getFolderNames(names: string[]): string[] {
    let d: Map<string, number> = new Map();
    for (let i = 0; i < names.length; ++i) {
        if (d.has(names[i])) {
            let k: number = d.get(names[i]) || 0;
            while (d.has(names[i] + '(' + k + ')')) {
                ++k;
            }
            d.set(names[i], k);
            names[i] += '(' + k + ')';
        }
        d.set(names[i], 1);
    }
    return names;
}

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.