How do you solve Minimum Cost to Convert String I on LeetCode?

Minimum Cost to Convert String I (LeetCode #2976) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Constructing a graph helps visualize transformations and their costs.

What is the brute force solution for Minimum Cost to Convert String I?

The brute force approach for Minimum Cost to Convert String I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach tries to convert each character in the source string to the corresponding character in the target string by checking all possible transformations. This method is straightforward but inefficient as it explores all transformations without optimizing the cost. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Cost to Convert String I?

Minimum Cost to Convert String I uses the following concepts: Array, String, Graph Theory, Shortest Path. The recommended patterns to study are: Graph Theory, Dijkstra's Algorithm.

What are the interview tips for Minimum Cost to Convert String I?

Clearly explain your thought process and the graph construction. Discuss edge cases, such as characters that cannot be transformed. Practice Dijkstra's algorithm beforehand, as it is a common interview topic.

What are common mistakes when solving Minimum Cost to Convert String I?

Not handling cases where transformations are impossible. Overlooking the need to minimize costs when multiple transformations exist.

#2976

Minimum Cost to Convert String I

Medium

ArrayStringGraph TheoryShortest PathGraph TheoryDijkstra's Algorithm

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal approach constructs a graph where each character is a node and transformations are edges with weights. By using Dijkstra's algorithm, we can efficiently find the minimum cost to convert each character in the source string to the corresponding character in the target string.

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Algorithm

3 steps

1Step 1: Construct a graph where each character is a node and edges represent possible transformations with their associated costs.
2Step 2: For each character pair (source[i], target[i]), use Dijkstra's algorithm to find the minimum cost path from source[i] to target[i].
3Step 3: Sum the costs for all character transformations. If any transformation is impossible, return -1.

solution.py38 lines

1# Full working Python code
2import heapq
3from collections import defaultdict
4
5def minCost(source, target, original, changed, cost):
6    graph = defaultdict(dict)
7    for i in range(len(original)):
8        if changed[i] not in graph[original[i]]:
9            graph[original[i]][changed[i]] = cost[i]
10        else:
11            graph[original[i]][changed[i]] = min(graph[original[i]][changed[i]], cost[i])
12
13    total_cost = 0
14    for s, t in zip(source, target):
15        if s == t:
16            continue
17        min_cost = dijkstra(graph, s, t)
18        if min_cost == float('inf'):
19            return -1
20        total_cost += min_cost
21    return total_cost
22
23
24def dijkstra(graph, start, end):
25    pq = [(0, start)]
26    min_cost = {start: 0}
27    while pq:
28        cost, node = heapq.heappop(pq)
29        if node == end:
30            return cost
31        for neighbor, weight in graph[node].items():
32            new_cost = cost + weight
33            if new_cost < min_cost.get(neighbor, float('inf')):
34                min_cost[neighbor] = new_cost
35                heapq.heappush(pq, (new_cost, neighbor))
36    return float('inf')
37
38print(minCost('abcd', 'acbe', ['a','b','c','c','e','d'], ['b','c','b','e','b','e'], [2,5,5,1,2,20]))

ℹ

Complexity note: The time complexity is O(n log n) due to the priority queue operations in Dijkstra's algorithm, which is more efficient than the brute-force approach. The space complexity is O(n) for storing the graph and the minimum costs.

1Constructing a graph helps visualize transformations and their costs.
2Using Dijkstra's algorithm optimizes the search for minimum costs.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.