LeetCode-in-Java

3819. Rotate Non Negative Elements

Medium

You are given an integer array nums and an integer k.

Rotate only the non-negative elements of the array to the left by k positions, in a cyclic manner.

All negative elements must stay in their original positions and must not move.

After rotation, place the non-negative elements back into the array in the new order, filling only the positions that originally contained non-negative values and skipping all negative positions.

Return the resulting array.

Example 1:

Input: nums = [1,-2,3,-4], k = 3

Output: [3,-2,1,-4]

Explanation:

• The non-negative elements, in order, are [1, 3].
• Left rotation with k = 3 results in:
  • [1, 3] -> [3, 1] -> [1, 3] -> [3, 1]
• Placing them back into the non-negative indices results in [3, -2, 1, -4].

Example 2:

Input: nums = [-3,-2,7], k = 1

Output: [-3,-2,7]

Explanation:

• The non-negative elements, in order, are [7].
• Left rotation with k = 1 results in [7].
• Placing them back into the non-negative indices results in [-3, -2, 7].

Example 3:

Input: nums = [5,4,-9,6], k = 2

Output: [6,5,-9,4]

Explanation:

• The non-negative elements, in order, are [5, 4, 6].
• Left rotation with k = 2 results in [6, 5, 4].
• Placing them back into the non-negative indices results in [6, 5, -9, 4].

Constraints:

• 1 <= nums.length <= 105
• -105 <= nums[i] <= 105
• 0 <= k <= 105

Solution

public class Solution {
    public int[] rotateElements(int[] nums, int k) {
        int n = nums.length;
        int m = 0;
        int[] a = new int[n];
        for (int x : nums) {
            if (x >= 0) {
                a[m++] = x;
            }
        }
        if (m == 0) {
            return nums;
        }
        k %= m;
        int j = 0;
        for (int i = 0; i < n; i++) {
            if (nums[i] >= 0) {
                int index = (j + k) % m;
                nums[i] = a[index];
                j++;
            }
        }
        return nums;
    }
}

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