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2161. Partition Array According to Given Pivot 👍

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  vector<int> pivotArray(vector<int>& nums, int pivot) {
    vector<int> ans;

    for (const int num : nums)
      if (num < pivot)
        ans.push_back(num);

    for (const int num : nums)
      if (num == pivot)
        ans.push_back(num);

    for (const int num : nums)
      if (num > pivot)
        ans.push_back(num);

    return ans;
  }
};
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class Solution {
  public int[] pivotArray(int[] nums, int pivot) {
    int[] ans = new int[nums.length];
    int i = 0; // ans's index

    for (final int num : nums)
      if (num < pivot)
        ans[i++] = num;

    for (final int num : nums)
      if (num == pivot)
        ans[i++] = num;

    for (final int num : nums)
      if (num > pivot)
        ans[i++] = num;

    return ans;
  }
}
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class Solution:
  def pivotArray(self, nums: List[int], pivot: int) -> List[int]:
    return [num for num in nums if num < pivot] + \
        [num for num in nums if num == pivot] + \
        [num for num in nums if num > pivot]