Given a binary array, task is to sort this binary array using minimum swaps. We are allowed to swap only adjacent elements
Input : [0, 0, 1, 0, 1, 0, 1, 1] Output : 3 1st swap : [0, 0, 1, 0, 0, 1, 1, 1] 2nd swap : [0, 0, 0, 1, 0, 1, 1, 1] 3rd swap : [0, 0, 0, 0, 1, 1, 1, 1] Input : Array = [0, 1, 0, 1, 0] Output : 3
This can be done by finding number of zeroes to the right side of every 1 and add them. In order to sort the array every one always has to perform a swap operation with every zero on its right side. So the total number of swap operations for a particular 1 in array is the number of zeroes on its right hand side. Find the number of zeroes on right side for every one i.e. the number of swaps and add them all to obtain the total number of swaps.
Time Complexity : O(n)
Auxiliary Space : O(n)
Space Optimized Solution :
An auxiliary space is not needed. We just need to start reading the list from the back and keep track of number of zeros we encounter. If we encounter a 1 the number of zeros is the number of swaps needed to put the 1 in correct place.
- Minimum number of swaps required to sort an array
- Minimum number of swaps required to sort an array | Set 2
- Minimum number of swaps required for arranging pairs adjacent to each other
- Minimum number of swaps required to sort an array of first N number
- Minimum swaps required to convert one binary string to another
- Minimum swaps required to make a binary string alternating
- Minimum swaps required to make a binary string divisible by 2^k
- Minimum operations required to modify the array such that parity of adjacent elements is different
- Number of swaps to sort when only adjacent swapping allowed
- Minimum swaps to reach permuted array with at most 2 positions left swaps allowed
- Minimum Swaps required to group all 1's together
- Minimum number of adjacent swaps for arranging similar elements together
- Minimum swaps required to bring all elements less than or equal to k together
- Minimum inversions required so that no two adjacent elements are same
- Minimum swaps so that binary search can be applied
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