# Program to find minimum adjacent swaps for K consecutive ones in Python

PythonServer Side ProgrammingProgramming

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
Updated on 07-Oct-2021 08:12:07