Minimum Cost to Merge Stones — LeetCode 1000 Python Solution
HardArrayDynamic ProgrammingPrefix Sum
- Problem
- #1000
- Pattern
- Prefix Sum
- Reading time
- 3 min
- Source
- leetcode.com
The problem
There are n piles of stones arranged in a row. The ith pile has stones[i] stones.
Example
- Input
- stones = [3,2,4,1], k = 2
- Output
- 20
- Explanation
- We start with [3, 2, 4, 1].
Python solution
Python
class Solution:
def mergeStones(self, stones: List[int], K: int) -> int:
n = len(stones)
if (n - 1) % (K - 1):
return -1
s = list(accumulate(stones, initial=0))
f = [[[inf] * (K + 1) for _ in range(n + 1)] for _ in range(n + 1)]
for i in range(1, n + 1):
f[i][i][1] = 0
for l in range(2, n + 1):
for i in range(1, n - l + 2):
j = i + l - 1
for k in range(1, K + 1):
for h in range(i, j):
f[i][j][k] = min(f[i][j][k], f[i][h][1] + f[h + 1][j][k - 1])
f[i][j][1] = f[i][j][K] + s[j] - s[i - 1]
return f[1][n][1]Complexity
| Measure | Complexity |
|---|---|
| Time | O(n·m) (typical) |
| Space | O(n·m) or optimized auxiliary |
Pattern: Prefix Sum
Precompute running totals once so any range query becomes a single subtraction. LeetCode 1000. Minimum Cost to Merge Stones is filed here because LeetCode tags it Prefix Sum, which is the vocabulary this hub collects.
The prefix sum guide has the Python template for the pattern and the 157 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 1000. Minimum Cost to Merge Stones?
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- What topics does LeetCode 1000. Minimum Cost to Merge Stones cover?
- LeetCode 1000. Minimum Cost to Merge Stones is tagged Array, Dynamic Programming and Prefix Sum on LeetCode.