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Minimum Swaps to Group All 1's Together

Contributed by

Deep Mavani

Last Updated: 23 Feb, 2023

Easy

0/40

Avg time to solve 15 mins

Success Rate 85 %

6 upvotes

Problem Statement

You are given an array/list ‘ARR’ of size ‘N’. ‘ARR' is binary i.e. it contains only 0s and 1s (ARR[i] = {0, 1}). Your task is to find out the minimum number of swaps required to group all 1s together.

Note: If ‘ARR’ contains only 0’s then print -1.

Example:

Let ‘ARR’ = [ 0, 1, 0, 1]. We can group all 1s together in the following ways: ‘ARR’ =[0, 0, 1, 1] or ‘ARR’ = [0, 1, 1, 0]. 

In this example, we need only 1 swap to group all 1’s together which is the minimum possible.

Detailed explanation ( Input/output format, Notes, Images )

Input Format:

The first line of input contains an integer ‘T’ which denotes the number of test cases. 

The first line of each test case contains a single integer ‘N’ representing the number of elements in the array/list ‘ARR’.

The next line of each test case contains ‘N’ single space-separated integers (0s and 1s) denoting the elements of  ‘ARR’.

Output Format :

For each test case, return the minimum number of swaps required to group all 1’s together.

Note:

You don't need to print anything, it has already been taken care of. Just implement the given function.

Constraints:

1 <= ‘T’ <= 100
2 <= ‘N’ <= 5000
0 <= ‘ARR[i]’ <= 1

Where ‘ARR[i]’ represents the elements of array/list ‘ARR’. 

Time Limit: 1 sec

Sample Input 1:

2
5
1 0 1 0 1
6
1 1 1 1 1 1

Sample Output 1:

1
0

Explanation for Sample Output 1:

In test case 1, swap ‘ARR[1]’ and ‘ARR[4]’ (0-based indexing). Then ‘ARR’ = [1, 1, 1, 0, 0]. So, the minimum swaps to group all 1s together is 1.

In test case 2, all 1s are already together in 'ARR'. So, we don’t need any swaps. Hence, the minimum swaps to group all 1s together is 0.

Sample Input 2:

2
4 
0 0 0 0
6
1 1 0 0 1 1

Sample Output 2:

-1
 2

Explanation for Sample Output 2:

In test case 1, the number of 1s in 'ARR' is 0. So we return -1.

In test case 2, first, we swap ‘ARR[2]’ and ‘ARR[4]’ (0-based indexing). Now, ‘ARR’ = [1, 1, 1, 0, 0, 1].
Then, swap ‘ARR[3]’ and ‘ARR[5]’. Now, ‘ARR’ = [1, 1, 1, 1, 0, 0]. So, the minimum swaps to group all 1s together is 2.

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