LeetCode: Minimum Swaps to Group All 1’s Together

Posted on September 3, 2019July 26, 2020 by braindenny

Minimum Swaps to Group All 1’s Together

Similar Problems:

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

Example 3:

Input: [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].

Note:

1 <= data.length <= 10^5
0 <= data[i] <= 1

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution:

// https://code.dennyzhang.com/minimum-swaps-to-group-all-1s-together
// Basic Ideas: fixed size sliding window
//  We know the size of the final group all 1's
//  The window can start from any position
//  Use a sliding window. And the find the minimum count of 0s in the sliding window.
// Complexity: Time O(n), Space O(k)
func minSwaps(data []int) int {
    k:=0
    for _, v := range data {
        if v == 1 {  k++ }
    }
    res := 1<<31-1
    count := 0
    // sliding window. left: i-k+1, right: i
    for i:=0; i<len(data); i++ {
        // move right pointer: add a new element
        if data[i] == 0 { count++ }
        // move left pointer: remove the old element
        if i-k >= 0 {
            if data[i-k] == 0 { count-- }
        }
        // collect result
        if i-k+1 >= 0  {
            if count < res { res = count }
        }
    }
    return res
}

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