HardNo. 0465LeetCode Premium
Optimal Account Balancing — LeetCode 465 Python Solution
- Pattern
- Backtracking
- Topics
- Bit Manipulation, Array, Dynamic Programming, Backtracking, Bitmask
- Reading time
- 4 min
- Source
- leetcode.com
The problem
You are given an array of transactions transactions where transactions[i] = [fromi, toi, amounti] indicates that the person with ID = fromi gave amounti $ to the person with ID = toi. Return the minimum number of transactions required to settle the debt.
Example
- Input
- transactions = [[0,1,10],[2,0,5]]
- Output
- 2
- Explanation
- Person #0 gave person #1 $10.
Python solution
class Solution:
def minTransfers(self, transactions: List[List[int]]) -> int:
g = defaultdict(int)
for f, t, x in transactions:
g[f] -= x
g[t] += x
nums = [x for x in g.values() if x]
m = len(nums)
f = [inf] * (1 << m)
f[0] = 0
for i in range(1, 1 << m):
s = 0
for j, x in enumerate(nums):
if i >> j & 1:
s += x
if s == 0:
f[i] = i.bit_count() - 1
j = (i - 1) & i
while j > 0:
f[i] = min(f[i], f[j] + f[i ^ j])
j = (j - 1) & i
return f[-1]Complexity
- Time
- O(n·m) (typical)
- Space · auxiliary
- O(n·m)or optimized
Pattern: Backtracking
Build candidates one choice at a time and abandon a branch the moment it cannot work. LeetCode 465. Optimal Account Balancing is filed here because LeetCode tags it Backtracking, which is the vocabulary this hub collects.
The backtracking guide has the Python template for the pattern and the 105 LeetCode problems that use it.
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Frequently asked questions
- How hard is LeetCode 465. Optimal Account Balancing?
LeetCode 465. Optimal Account Balancing is rated Hard on LeetCode.
- What topics does LeetCode 465. Optimal Account Balancing cover?
LeetCode 465. Optimal Account Balancing is tagged Bit Manipulation, Array, Dynamic Programming, Backtracking and Bitmask on LeetCode.
- Is LeetCode 465. Optimal Account Balancing a premium problem?
Yes. LeetCode 465. Optimal Account Balancing is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.