Unique Identifier Combinations — Problem Statement & Solution Guide

BacktrackingMediumMixed

TimeO(1)

SpaceO(1)

Problem Description

In a distributed system, unique identifiers are generated by selecting a sequence of distinct characters from a provided pool. Each identifier must be exactly 7 characters long, and no character can be repeated within a single identifier. Given an array of unique characters, determine the total number of distinct 7-character identifiers that can be constructed.

The input is an array of unique characters. The output is the total count of valid identifiers. If the number of available characters is less than 7, the result is 0, as it is impossible to form a 7-character string without repetition.

This problem requires calculating the number of permutations of length 7 from a set of n distinct elements, which is mathematically defined as P(n, 7) = n! / (n-7)! for n >= 7, and 0 otherwise.

DSA Pattern Breakdown

"Unique Identifier Combinations"

medium

WHY DOES IT MATTER?

Permutations capture the essence of ordered selections, which is crucial when identifiers are sensitive to character order. Recognizing this pattern allows us to replace exponential enumeration with a closed‑form formula, drastically improving scalability.

OPTIMIZATION CHALLENGE

The key insight is that P(n,7) can be computed as a product of 7 consecutive descending integers, eliminating the need for factorials and reducing time to O(1).

REAL-WORLD CONNECTION

Generating session tokens, API keys, or distributed system IDs all require unique, ordered strings. Understanding how many such strings exist informs capacity planning and collision probability analysis.

In interviews, emphasize that you can compute the result in a single loop, discuss overflow handling, and mention that precomputing factorials is unnecessary for fixed k.

COMPLEXITY AT A GLANCE

⏱ Time:O(1)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to counting the number of ways to arrange 7 distinct characters chosen from a pool of n unique characters. This is a classic permutation problem: the number of ordered selections of size k from n distinct items is given by the falling factorial P(n,k)=n!/(n-k)!. Naïve enumeration would generate all n! permutations and then filter those of length 7, which is infeasible for large n because the number of permutations grows factorially. The optimal paradigm is to compute the product of the first 7 descending integers starting from n, i.e. n·(n-1)·…·(n-6), which yields the same result in constant time and space.

Interview Questions on This Problem

Q1At a global product company, how would you compute the number of unique 7‑character identifiers from a pool of n distinct characters, and what if n is less than 7?

I would use the permutation formula P(n,7)=n!/(n-7)! if n≥7; if n<7 the answer is 0 because we cannot pick 7 distinct characters. I would implement this as a loop multiplying the 7 descending terms to avoid computing large factorials.

Q2In a fintech platform, how would you handle potential integer overflow when n is large?

I would use a 64‑bit unsigned integer or a big integer library if the language supports it. Since the product of 7 numbers grows roughly as n^7, for n up to about 10^6 the result fits in 64 bits, but for larger n I would switch to arbitrary precision arithmetic.

Q3For a high‑growth startup that needs to generate identifiers on the fly, what strategy ensures uniqueness and performance?

I would precompute the factorial ratio once and store it, or compute the product on demand using a simple loop. To guarantee uniqueness I would maintain a hash set of generated identifiers and check before insertion, but the combinatorial count itself is independent of generation.

Examples

Example 1

Input

['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']

Output

Explanation: There are 8 unique characters. We need to form a 7-character string without repetition. The number of permutations is P(8, 7) = 8! / (8-7)! = 8! / 1! = 40320. Thus, there are 40320 unique identifiers.

Example 2

Input

['x', 'y', 'z', 'w', 'v', 'u', 't', 's', 'r', 'q']

Output

Explanation: There are 10 unique characters. The number of permutations is P(10, 7) = 10! / (10-7)! = 10! / 3! = 3628800 / 6 = 604800. Thus, there are 604800 unique identifiers.

Example 3

Input

['a', 'b', 'c', 'd', 'e', 'f']

Output

Explanation: There are only 6 unique characters. Since we need to form a 7-character string without repetition, it is impossible to do so. Therefore, the number of unique identifiers is 0.

Example 4

Input

['1', '2', '3', '4', '5', '6', '7', '8', '9', '0', 'A', 'B']

Output

39916800

Explanation: There are 12 unique characters. The number of permutations is P(12, 7) = 12! / (12-7)! = 12! / 5! = 479001600 / 120 = 3991680. Wait, 12! is 479001600. 5! is 120. 479001600 / 120 = 3991680. Let me recalculate. 12*11*10*9*8*7*6 = 3991680. Correct.

Constraints

1 <= characters.length <= 10^5
All characters in the input array are unique.
Each character is a single ASCII printable character.
The result may be large, so use 64-bit integer arithmetic.

Optimal Approach & Strategy

Compute the product of the 7 descending integers from n: result = n*(n-1)*…*(n-6). This runs in constant time and uses O(1) extra space.

Brute Force Approach

Generate all permutations of the n characters, filter those of length 7, and count them. This requires O(n!) time and is infeasible for large n.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(1)

function solution(chars) { return chars.length >= 7 ? factorial(chars.length) / factorial(chars.length - 7) : 0; function factorial(n) { let result = 1; for (let i = 2; i <= n; i++) { result *= i; } return result; } }

class Solution {
public:
    int solution(vector<char> chars) {
        if (chars.size() >= 7) {
            return factorial(chars.size()) / factorial(chars.size() - 7);
        } else {
            return 0;
        }
    }
    long long factorial(int n) {
        long long result = 1;
        for (int i = 2; i <= n; i++) {
            result *= i;
        }
        return result;
    }
};

class Solution {
    public int solution(String[] chars) {
        if (chars.length >= 7) {
            return factorial(chars.length) / factorial(chars.length - 7);
        } else {
            return 0;
        }
    }
    public long factorial(int n) {
        long result = 1;
        for (int i = 2; i <= n; i++) {
            result *= i;
        }
        return result;
    }
}

import math
def solution(chars):
    if len(chars) >= 7:
        return math.factorial(len(chars)) // math.factorial(len(chars) - 7)
    else:
        return 0

function solution(chars) { return chars.length >= 7 ? factorial(chars.length) / factorial(chars.length - 7) : 0; function factorial(n) { let result = 1; for (let i = 2; i <= n; i++) { result *= i; } return result; } }

Asked in Top Tech Interviews

uncategorizedmediumgeneric

Solve in Interative Editor

Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.

Unique Identifier Combinations Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Backtracking

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Unique Identifier Combinations — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor