Minimum Swaps to Group All 1's Together — LeetCode 1151 Python Solution
- Problem
- #1151
- Pattern
- Sliding Window
- Reading time
- 2 min
- Source
- leetcode.com
The problem
Given a binary array data, return the minimum number of swaps required to group all 1’s present in the array together in any place in the array.
Example
- Input
- data = [1,0,1,0,1]
- Output
- 1
- Explanation
- There are 3 ways to group all 1's together:
Python solution
class Solution:
def minSwaps(self, data: List[int]) -> int:
k = data.count(1)
mx = t = sum(data[:k])
for i in range(k, len(data)):
t += data[i]
t -= data[i - k]
mx = max(mx, t)
return k - mxComplexity
| Measure | Complexity |
|---|---|
| Time | O(n) |
| Space | O(1) auxiliary |
Pattern: Sliding Window
Collapse a nested loop over every subarray into a single pass with two indices. LeetCode 1151. Minimum Swaps to Group All 1's Together is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Sliding Window.
The sliding window guide has the Python template for the pattern and the 116 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 1151. Minimum Swaps to Group All 1's Together?
- LeetCode 1151. Minimum Swaps to Group All 1's Together is rated Medium on LeetCode.
- What is the time complexity of LeetCode 1151. Minimum Swaps to Group All 1's Together?
- The Python solution on this page runs in O(n).
- What is the space complexity of LeetCode 1151. Minimum Swaps to Group All 1's Together?
- The Python solution on this page uses O(1) auxiliary space.
- What topics does LeetCode 1151. Minimum Swaps to Group All 1's Together cover?
- LeetCode 1151. Minimum Swaps to Group All 1's Together is tagged Array and Sliding Window on LeetCode.
- Is LeetCode 1151. Minimum Swaps to Group All 1's Together a premium problem?
- Yes. LeetCode 1151. Minimum Swaps to Group All 1's Together is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.