LeetCode in Kotlin

3936. Minimum Swaps to Move Zeros to End

Easy

You are given an integer array nums.

In one operation, you can choose any two distinct indices i and j and swap nums[i] and nums[j].

Return an integer denoting the minimum number of operations required to move all 0s to the end of the array.

Example 1:

Input: nums = [0,1,0,3,12]

Output: 2

Explanation:

We perform the following swap operations:

• Swap nums[0] and nums[3], giving nums = [3, 1, 0, 0, 12].
• Swap nums[2] and nums[4], giving nums = [3, 1, 12, 0, 0].

Thus, the answer is 2.

Example 2:

Input: nums = [0,1,0,2]

Output: 1

Explanation:

We perform the following swap operations:

• Swap nums[0] and nums[3], giving nums = [2, 1, 0, 0].

Thus, the answer is 1.

Example 3:

Input: nums = [1,2,0]

Output: 0

Explanation:

The array already satisfies the condition. Therefore, no swap operations are needed.

Constraints:

• 1 <= nums.length <= 100
• 0 <= nums[i] <= 100

Solution

class Solution {
    private fun swap(nums: IntArray, i: Int, j: Int) {
        val temp = nums[i]
        nums[i] = nums[j]
        nums[j] = temp
    }

    fun minimumSwaps(nums: IntArray): Int {
        val n = nums.size
        var count = 0
        var l = 0
        var r = n - 1
        while (l < r) {
            if (nums[l] == 0 && nums[r] != 0) {
                swap(nums, l, r)
                count++
                l++
                r--
            } else if (nums[l] != 0) {
                l++
            } else if (nums[r] == 0) {
                r--
            }
        }
        return count
    }
}

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