Medium
A swap is defined as taking two distinct positions in an array and swapping the values in them.
A circular array is defined as an array where we consider the first element and the last element to be adjacent.
Given a binary circular array nums
, return the minimum number of swaps required to group all 1
’s present in the array together at any location.
Example 1:
Input: nums = [0,1,0,1,1,0,0]
Output: 1
Explanation: Here are a few of the ways to group all the 1’s together:
[0,0,1,1,1,0,0] using 1 swap.
[0,1,1,1,0,0,0] using 1 swap.
[1,1,0,0,0,0,1] using 2 swaps (using the circular property of the array).
There is no way to group all 1’s together with 0 swaps.
Thus, the minimum number of swaps required is 1.
Example 2:
Input: nums = [0,1,1,1,0,0,1,1,0]
Output: 2
Explanation: Here are a few of the ways to group all the 1’s together:
[1,1,1,0,0,0,0,1,1] using 2 swaps (using the circular property of the array).
[1,1,1,1,1,0,0,0,0] using 2 swaps.
There is no way to group all 1’s together with 0 or 1 swaps.
Thus, the minimum number of swaps required is 2.
Example 3:
Input: nums = [1,1,0,0,1]
Output: 0
Explanation: All the 1’s are already grouped together due to the circular property of the array.
Thus, the minimum number of swaps required is 0.
Constraints:
1 <= nums.length <= 105
nums[i]
is either 0
or 1
.class Solution {
fun minSwaps(nums: IntArray): Int {
val l = nums.size
val ones = IntArray(l)
ones[0] = if (nums[0] == 1) 1 else 0
for (i in 1 until l) {
if (nums[i] == 1) {
ones[i] = ones[i - 1] + 1
} else {
ones[i] = ones[i - 1]
}
}
if (ones[l - 1] == l || ones[l - 1] == 0) {
return 0
}
val ws = ones[l - 1]
var minSwaps = Int.MAX_VALUE
var si = 0
var ei: Int
while (si < nums.size) {
ei = (si + ws - 1) % l
var totalones: Int
totalones = if (ei >= si) {
ones[ei] - if (si == 0) 0 else ones[si - 1]
} else {
ones[ei] + (ones[l - 1] - ones[si - 1])
}
val swapsreq = ws - totalones
if (swapsreq < minSwaps) {
minSwaps = swapsreq
}
si++
}
return minSwaps
}
}