LeetCode #3149 — HARD

Find the Minimum Cost Array Permutation

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an array nums which is a permutation of [0, 1, 2, ..., n - 1]. The score of any permutation of [0, 1, 2, ..., n - 1] named perm is defined as:

score(perm) = |perm[0] - nums[perm[1]]| + |perm[1] - nums[perm[2]]| + ... + |perm[n - 1] - nums[perm[0]]|

Return the permutation perm which has the minimum possible score. If multiple permutations exist with this score, return the one that is lexicographically smallest among them.

Example 1:

Input: nums = [1,0,2]

Output: [0,1,2]

Explanation:

The lexicographically smallest permutation with minimum cost is [0,1,2]. The cost of this permutation is |0 - 0| + |1 - 2| + |2 - 1| = 2.

Example 2:

Input: nums = [0,2,1]

Output: [0,2,1]

Explanation:

The lexicographically smallest permutation with minimum cost is [0,2,1]. The cost of this permutation is |0 - 1| + |2 - 2| + |1 - 0| = 2.

Constraints:

2 <= n == nums.length <= 14
nums is a permutation of [0, 1, 2, ..., n - 1].

Step 01

Brute Force Baseline

Problem summary: You are given an array nums which is a permutation of [0, 1, 2, ..., n - 1]. The score of any permutation of [0, 1, 2, ..., n - 1] named perm is defined as: score(perm) = |perm[0] - nums[perm[1]]| + |perm[1] - nums[perm[2]]| + ... + |perm[n - 1] - nums[perm[0]]| Return the permutation perm which has the minimum possible score. If multiple permutations exist with this score, return the one that is lexicographically smallest among them.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Bit Manipulation

Example 1

[1,0,2]

Example 2

[0,2,1]

Core Insight

What unlocks the optimal approach

The score function is cyclic, so we can always set <code>perm[0] = 0</code> for the smallest lexical order.
It’s similar to the Traveling Salesman Problem. Use Dynamic Programming.
Use a bitmask to track which elements have been assigned to <code>perm</code>.

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

Largest constraint values

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 #3149: Find the Minimum Cost Array Permutation
class Solution {
    private Integer[][] f;
    private int[] nums;
    private int[] ans;
    private int n;

    public int[] findPermutation(int[] nums) {
        n = nums.length;
        ans = new int[n];
        this.nums = nums;
        f = new Integer[1 << n][n];
        g(1, 0, 0);
        return ans;
    }

    private int dfs(int mask, int pre) {
        if (mask == (1 << n) - 1) {
            return Math.abs(pre - nums[0]);
        }
        if (f[mask][pre] != null) {
            return f[mask][pre];
        }
        int res = Integer.MAX_VALUE;
        for (int cur = 1; cur < n; ++cur) {
            if ((mask >> cur & 1) == 0) {
                res = Math.min(res, Math.abs(pre - nums[cur]) + dfs(mask | 1 << cur, cur));
            }
        }
        return f[mask][pre] = res;
    }

    private void g(int mask, int pre, int k) {
        ans[k] = pre;
        if (mask == (1 << n) - 1) {
            return;
        }
        int res = dfs(mask, pre);
        for (int cur = 1; cur < n; ++cur) {
            if ((mask >> cur & 1) == 0) {
                if (Math.abs(pre - nums[cur]) + dfs(mask | 1 << cur, cur) == res) {
                    g(mask | 1 << cur, cur, k + 1);
                    break;
                }
            }
        }
    }
}

// Accepted solution for LeetCode #3149: Find the Minimum Cost Array Permutation
func findPermutation(nums []int) (ans []int) {
	n := len(nums)
	f := make([][]int, 1<<n)
	for i := range f {
		f[i] = make([]int, n)
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	var dfs func(int, int) int
	dfs = func(mask, pre int) int {
		if mask == 1<<n-1 {
			return abs(pre - nums[0])
		}
		if f[mask][pre] != -1 {
			return f[mask][pre]
		}
		res := &f[mask][pre]
		*res = math.MaxInt32
		for cur := 1; cur < n; cur++ {
			if mask>>cur&1 == 0 {
				*res = min(*res, abs(pre-nums[cur])+dfs(mask|1<<cur, cur))
			}
		}
		return *res
	}
	var g func(int, int)
	g = func(mask, pre int) {
		ans = append(ans, pre)
		if mask == 1<<n-1 {
			return
		}
		res := dfs(mask, pre)
		for cur := 1; cur < n; cur++ {
			if mask>>cur&1 == 0 {
				if abs(pre-nums[cur])+dfs(mask|1<<cur, cur) == res {
					g(mask|1<<cur, cur)
					break
				}
			}
		}
	}
	g(1, 0)
	return
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #3149: Find the Minimum Cost Array Permutation
class Solution:
    def findPermutation(self, nums: List[int]) -> List[int]:
        @cache
        def dfs(mask: int, pre: int) -> int:
            if mask == (1 << n) - 1:
                return abs(pre - nums[0])
            res = inf
            for cur in range(1, n):
                if mask >> cur & 1 ^ 1:
                    res = min(res, abs(pre - nums[cur]) + dfs(mask | 1 << cur, cur))
            return res

        def g(mask: int, pre: int):
            ans.append(pre)
            if mask == (1 << n) - 1:
                return
            res = dfs(mask, pre)
            for cur in range(1, n):
                if mask >> cur & 1 ^ 1:
                    if abs(pre - nums[cur]) + dfs(mask | 1 << cur, cur) == res:
                        g(mask | 1 << cur, cur)
                        break

        n = len(nums)
        ans = []
        g(1, 0)
        return ans

// Accepted solution for LeetCode #3149: Find the Minimum Cost Array Permutation
// 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 #3149: Find the Minimum Cost Array Permutation
// class Solution {
//     private Integer[][] f;
//     private int[] nums;
//     private int[] ans;
//     private int n;
// 
//     public int[] findPermutation(int[] nums) {
//         n = nums.length;
//         ans = new int[n];
//         this.nums = nums;
//         f = new Integer[1 << n][n];
//         g(1, 0, 0);
//         return ans;
//     }
// 
//     private int dfs(int mask, int pre) {
//         if (mask == (1 << n) - 1) {
//             return Math.abs(pre - nums[0]);
//         }
//         if (f[mask][pre] != null) {
//             return f[mask][pre];
//         }
//         int res = Integer.MAX_VALUE;
//         for (int cur = 1; cur < n; ++cur) {
//             if ((mask >> cur & 1) == 0) {
//                 res = Math.min(res, Math.abs(pre - nums[cur]) + dfs(mask | 1 << cur, cur));
//             }
//         }
//         return f[mask][pre] = res;
//     }
// 
//     private void g(int mask, int pre, int k) {
//         ans[k] = pre;
//         if (mask == (1 << n) - 1) {
//             return;
//         }
//         int res = dfs(mask, pre);
//         for (int cur = 1; cur < n; ++cur) {
//             if ((mask >> cur & 1) == 0) {
//                 if (Math.abs(pre - nums[cur]) + dfs(mask | 1 << cur, cur) == res) {
//                     g(mask | 1 << cur, cur, k + 1);
//                     break;
//                 }
//             }
//         }
//     }
// }

// Accepted solution for LeetCode #3149: Find the Minimum Cost Array Permutation
function findPermutation(nums: number[]): number[] {
    const n = nums.length;
    const ans: number[] = [];
    const f: number[][] = Array.from({ length: 1 << n }, () => Array(n).fill(-1));
    const dfs = (mask: number, pre: number): number => {
        if (mask === (1 << n) - 1) {
            return Math.abs(pre - nums[0]);
        }
        if (f[mask][pre] !== -1) {
            return f[mask][pre];
        }
        let res = Infinity;
        for (let cur = 1; cur < n; ++cur) {
            if (((mask >> cur) & 1) ^ 1) {
                res = Math.min(res, Math.abs(pre - nums[cur]) + dfs(mask | (1 << cur), cur));
            }
        }
        return (f[mask][pre] = res);
    };
    const g = (mask: number, pre: number) => {
        ans.push(pre);
        if (mask === (1 << n) - 1) {
            return;
        }
        const res = dfs(mask, pre);
        for (let cur = 1; cur < n; ++cur) {
            if (((mask >> cur) & 1) ^ 1) {
                if (Math.abs(pre - nums[cur]) + dfs(mask | (1 << cur), cur) === res) {
                    g(mask | (1 << cur), cur);
                    break;
                }
            }
        }
    };
    g(1, 0);
    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 × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.