Leetcode #2977: Minimum Cost to Convert String II
In this guide, we solve Leetcode #2977 Minimum Cost to Convert String II in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given two 0-indexed strings source and target, both of length n and consisting of lowercase English characters. You are also given two 0-indexed string arrays original and changed, and an integer array cost, where cost[i] represents the cost of converting the string original[i] to the string changed[i].
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Graph, Trie, Array, String, Dynamic Programming, Shortest Path
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: source = "abcd", target = "acbe", original = ["a","b","c","c","e","d"], changed = ["b","c","b","e","b","e"], cost = [2,5,5,1,2,20]
Output: 28
Explanation: To convert "abcd" to "acbe", do the following operations:
- Change substring source[1..1] from "b" to "c" at a cost of 5.
- Change substring source[2..2] from "c" to "e" at a cost of 1.
- Change substring source[2..2] from "e" to "b" at a cost of 2.
- Change substring source[3..3] from "d" to "e" at a cost of 20.
The total cost incurred is 5 + 1 + 2 + 20 = 28.
It can be shown that this is the minimum possible cost.
Python Solution
class Node:
__slots__ = ["children", "v"]
def __init__(self):
self.children: List[Node | None] = [None] * 26
self.v = -1
class Solution:
def minimumCost(
self,
source: str,
target: str,
original: List[str],
changed: List[str],
cost: List[int],
) -> int:
m = len(cost)
g = [[inf] * (m << 1) for _ in range(m << 1)]
for i in range(m << 1):
g[i][i] = 0
root = Node()
idx = 0
def insert(w: str) -> int:
node = root
for c in w:
i = ord(c) - ord("a")
if node.children[i] is None:
node.children[i] = Node()
node = node.children[i]
if node.v < 0:
nonlocal idx
node.v = idx
idx += 1
return node.v
def dfs(i: int) -> int:
if i >= len(source):
return 0
res = dfs(i + 1) if source[i] == target[i] else inf
p = q = root
for j in range(i, len(source)):
p = p.children[ord(source[j]) - ord("a")]
q = q.children[ord(target[j]) - ord("a")]
if p is None or q is None:
break
if p.v < 0 or q.v < 0:
continue
res = min(res, dfs(j + 1) + g[p.v][q.v])
return res
for x, y, z in zip(original, changed, cost):
x = insert(x)
y = insert(y)
g[x][y] = min(g[x][y], z)
for k in range(idx):
for i in range(idx):
if g[i][k] >= inf:
continue
for j in range(idx):
# g[i][j] = min(g[i][j], g[i][k] + g[k][j])
if g[i][k] + g[k][j] < g[i][j]:
g[i][j] = g[i][k] + g[k][j]
ans = dfs(0)
return -1 if ans >= inf else ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.