Minimum Adjacent Swaps for K Consecutive Ones in Python

Suppose we have one binary array nums, and a value k. In one move, we can select two adjacent indices and swap their values. We have to find the minimum number of moves required so that nums has k consecutive 1's.

So, if the input is like nums = [1,0,0,1,0,1,0,1], k = 3, then the output will be 2 because in one swap we can make array from [1,0,0,1,0,1,0,1] to [1,0,0,0,1,1,0,1], then [1,0,0,0,1,1,1,0].

To solve this, we will follow these steps −

• j := 0

• val := 0

• ans := 999999

• loc := a new list

• for each index i, and value x in nums, do

  • if x is non-zero, then

    • insert i at the end of loc

    • m := quotient of (j + size of loc - 1) /2

    • val := val + loc[-1] - loc[m] - quotient of (size of loc -j)/2

    • if length of loc - j > k, then

      • m := quotient of (j + size of loc) /2

      • val := val - loc[m] - loc[j] - quotient of (size of loc -j)/2

      • j := j + 1

    • if size of loc -j is same as k, then

      • ans := minimum of ans and val

• return ans

Example

Let us see the following implementation to get better understanding

def solve(nums, k):
   j = val = 0
   ans = 999999
   loc = []
   for i, x in enumerate(nums):
      if x:
         loc.append(i)
         m = (j + len(loc) - 1)//2
         val += loc[-1] - loc[m] - (len(loc)-j)//2
         if len(loc) - j > k:
            m = (j + len(loc))//2
            val -= loc[m] - loc[j] - (len(loc)-j)//2
            j += 1
         if len(loc)-j == k:
            ans = min(ans, val)
   return ans

nums = [1,0,0,1,0,1,0,1]
k = 3
print(solve(nums, k))

Input

[1,0,0,1,0,1,0,1], 3

Output

2