LeetCode [1151] Minimum Swaps to Group All 1's Together

 1151. Minimum Swaps to Group All 1's Together

Medium

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.

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class Solution { public int minSwaps(int[] data) { int n = data.length; int[] dp = new int[n]; int cnt1 = 0;//number of ones int minSwaps = Integer.MAX_VALUE; for(int i=0; i<n; ++i){ if(data[i]==1){ cnt1++; dp[i] = cnt1; } } if(cnt1==0) return 0; int l = 0; int left1 = 0; while(l + cnt1 - 1 < n){ int swaps = left1 + cnt1 - dp[l+cnt1-1]; minSwaps = Math.min(minSwaps, swaps); left1 +=data[l++]; } return minSwaps; } }