What is the brute force approach and complexity for Travelling Salesman (bitmask DP)?

Try all (n − 1)! orders of cities 1 … n−1 and add the trip back to 0 with time complexity O(N!) and space complexity O(N).

What is the optimal approach and complexity for Travelling Salesman (bitmask DP)?

dp[mask][last] = cheapest walk from 0 that visits mask and ends at last; extend by one city with optimal time complexity O(2^N · N^2) and space complexity O(2^N · N).

How do you identify the pattern for Travelling Salesman (bitmask DP)?

Look for key signals in the prompt: "visit every city exactly once" indicates The visited set is the state, so one bit per city is enough.; "the least total cost" indicates Each state keeps the minimum cost, so the DP combines moves with min.; "n ≤ 15" indicates 2^15 · 15 states is small enough to store every mask..

Travelling Salesman (bitmask DP)

Hard·GeeksforGeeks Pro OnlyInterview-Level

GeeksforGeeks

Problem Statement

You have n cities numbered 0 to n − 1. cost[i][j] is the price of travelling directly from city i to city j. The two directions can have different prices.

Start at city 0. Visit every other city exactly once, then come back to city 0. Return the smallest total cost of such a tour.

With only one city there is no trip to make, so the answer is 0.

Input

A square matrix cost of size n × n.

Output

One integer: the least total cost of a tour that starts at city 0, visits each city once, and returns to city 0. The function is int tsp(vector<vector<int>>& cost).

Example 1

Input: cost = [[0, 111], [112, 0]]

Output: 223

There are only two cities. The tour 0 → 1 → 0 costs 111 + 112 = 223.

Example 2

Input: cost = [[0, 1000, 5000], [5000, 0, 1000], [1000, 5000, 0]]

Output: 3000

The tour 0 → 1 → 2 → 0 costs 1000 + 1000 + 1000 = 3000. The other way round, 0 → 2 → 1 → 0, costs 15000.

Constraints

• 1 ≤ n ≤ 15
• 0 ≤ cost[i][j] ≤ 10^4

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“visit every city exactly once”

→ The visited set is the state, so one bit per city is enough.

“the least total cost”

→ Each state keeps the minimum cost, so the DP combines moves with min.

“n ≤ 15”

→ 2^15 · 15 states is small enough to store every mask.

Approach Complexity Ladder

Brute Force — try every order of the other citiesO(N!)O(N)

Try all (n − 1)! orders of cities 1 … n−1 and add the trip back to 0

Optimal — Held–Karp DP over (visited set, last city)O(2^N · N^2)O(2^N · N)

dp[mask][last] = cheapest walk from 0 that visits mask and ends at last; extend by one city

Brute Force Intuition

A tour is city 0, then some order of the other cities, then back to 0. Write every order with next_permutation, add up the cost of each tour, and keep the smallest. There are (n − 1)! orders, so this only works for small n.

Recognition rule for this sub-pattern:

Represent a set of chosen items as bits of an integer, then iterate or DP over masks.

Remember this · Bit Manipulation

Trigger

"Every element appears twice except…", power of two, counting set bits, subsets as bitmasks, O(1) space tricks.

Think

What happens to this problem if I look at each bit separately?

Key question

Which bit operation (XOR, AND, OR, shift) cancels or isolates what I need?

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