LeetCode #3213 — HARD

Construct String with Minimum Cost

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

Solve on LeetCode
The Problem

Problem Statement

You are given a string target, an array of strings words, and an integer array costs, both arrays of the same length.

Imagine an empty string s.

You can perform the following operation any number of times (including zero):

  • Choose an index i in the range [0, words.length - 1].
  • Append words[i] to s.
  • The cost of operation is costs[i].

Return the minimum cost to make s equal to target. If it's not possible, return -1.

Example 1:

Input: target = "abcdef", words = ["abdef","abc","d","def","ef"], costs = [100,1,1,10,5]

Output: 7

Explanation:

The minimum cost can be achieved by performing the following operations:

  • Select index 1 and append "abc" to s at a cost of 1, resulting in s = "abc".
  • Select index 2 and append "d" to s at a cost of 1, resulting in s = "abcd".
  • Select index 4 and append "ef" to s at a cost of 5, resulting in s = "abcdef".

Example 2:

Input: target = "aaaa", words = ["z","zz","zzz"], costs = [1,10,100]

Output: -1

Explanation:

It is impossible to make s equal to target, so we return -1.

Constraints:

  • 1 <= target.length <= 5 * 104
  • 1 <= words.length == costs.length <= 5 * 104
  • 1 <= words[i].length <= target.length
  • The total sum of words[i].length is less than or equal to 5 * 104.
  • target and words[i] consist only of lowercase English letters.
  • 1 <= costs[i] <= 104
Patterns Used
📊Dynamic Programming→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given a string target, an array of strings words, and an integer array costs, both arrays of the same length. Imagine an empty string s. You can perform the following operation any number of times (including zero): Choose an index i in the range [0, words.length - 1]. Append words[i] to s. The cost of operation is costs[i]. Return the minimum cost to make s equal to target. If it's not possible, return -1.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

Example 1

"abcdef"
["abdef","abc","d","def","ef"]
[100,1,1,10,5]

Example 2

"aaaa"
["z","zz","zzz"]
[1,10,100]

Related Problems

  • Minimum Number of Valid Strings to Form Target II (minimum-number-of-valid-strings-to-form-target-ii)
  • Minimum Number of Valid Strings to Form Target I (minimum-number-of-valid-strings-to-form-target-i)
Step 02

Core Insight

What unlocks the optimal approach

  • Use Dynamic Programming along with Aho-Corasick or Hashing.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #3213: Construct String with Minimum Cost
class Hashing {
    private final long[] p;
    private final long[] h;
    private final long mod;

    public Hashing(String word, long base, int mod) {
        int n = word.length();
        p = new long[n + 1];
        h = new long[n + 1];
        p[0] = 1;
        this.mod = mod;
        for (int i = 1; i <= n; i++) {
            p[i] = p[i - 1] * base % mod;
            h[i] = (h[i - 1] * base + word.charAt(i - 1)) % mod;
        }
    }

    public long query(int l, int r) {
        return (h[r] - h[l - 1] * p[r - l + 1] % mod + mod) % mod;
    }
}

class Solution {
    public int minimumCost(String target, String[] words, int[] costs) {
        final int base = 13331;
        final int mod = 998244353;
        final int inf = Integer.MAX_VALUE / 2;

        int n = target.length();
        Hashing hashing = new Hashing(target, base, mod);

        int[] f = new int[n + 1];
        Arrays.fill(f, inf);
        f[0] = 0;

        TreeSet<Integer> ss = new TreeSet<>();
        for (String w : words) {
            ss.add(w.length());
        }

        Map<Long, Integer> d = new HashMap<>();
        for (int i = 0; i < words.length; i++) {
            long x = 0;
            for (char c : words[i].toCharArray()) {
                x = (x * base + c) % mod;
            }
            d.merge(x, costs[i], Integer::min);
        }

        for (int i = 1; i <= n; i++) {
            for (int j : ss) {
                if (j > i) {
                    break;
                }
                long x = hashing.query(i - j + 1, i);
                f[i] = Math.min(f[i], f[i - j] + d.getOrDefault(x, inf));
            }
        }

        return f[n] >= inf ? -1 : f[n];
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × sqrtL)
Space
O(n)

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.

Related Problems
#45Jump Game IImedium→#53Maximum Subarraymedium→#55Jump Gamemedium→#63Unique Paths IImedium→#64Minimum Path Summedium→