LC: 1151. Minimum Swaps to Group All 1's Together

https://leetcode.com/problems/minimum-swaps-to-group-all-1s-together/

1151. Minimum Swaps to Group All 1's Together

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 1:

Input: data = [1,0,1,0,1]
Output: 1
Explanation: 
There are 3 ways to group all 1's together:
[1,1,1,0,0] using 1 swap.
[0,1,1,1,0] using 2 swaps.
[0,0,1,1,1] using 1 swap.
The minimum is 1.

Example 2:

Input: data = [0,0,0,1,0]
Output: 0
Explanation: 
Since there is only one 1 in the array, no swaps needed.

Example 3:

Input: data = [1,0,1,0,1,0,0,1,1,0,1]
Output: 3
Explanation: 
One possible solution that uses 3 swaps is [0,0,0,0,0,1,1,1,1,1,1].

Example 4:

Input: data = [1,0,1,0,1,0,1,1,1,0,1,0,0,1,1,1,0,0,1,1,1,0,1,0,1,1,0,0,0,1,1,1,1,0,0,1]
Output: 8

Constraints:

1 <= data.length <= 105
data[i] is 0 or 1.

Details:

In einer ersten Schleife kann die Anzahl der 1’s gezählt werden. Die Anzahl der 0’s in jedem Intervall kann überprüft werden, indem man Präfixsummen verwendet oder in der Schleife neue und aktuelle Werte mit der aktuellen Gesamtsumme vergleicht.

Solution(s) and Further Explanation:

1151. Minimum Swaps to Group All 1's Together by ErdemT09 · Pull Request #184 · spiralgo/algorithmsGitHub

Default Code:

class Solution {
    public int minSwaps(int[] data) {
        
    }
}

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Last updated 3 years ago