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2294. Partition Array Such That Maximum Difference Is K 👍

  • Time: $O(n\log n)$
  • Space: $O(n)$
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class Solution {
 public:
  int partitionArray(vector<int>& nums, int k) {
    sort(begin(nums), end(nums));

    int ans = 1;
    int min = nums[0];

    for (int i = 1; i < nums.size(); ++i)
      if (min + k < nums[i]) {
        ++ans;
        min = nums[i];
      }

    return ans;
  }
};

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class Solution {
  public int partitionArray(int[] nums, int k) {
    Arrays.sort(nums);

    int ans = 1;
    int min = nums[0];

    for (int i = 1; i < nums.length; ++i)
      if (min + k < nums[i]) {
        ++ans;
        min = nums[i];
      }

    return ans;
  }
}

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class Solution:
  def partitionArray(self, nums: List[int], k: int) -> int:
    nums.sort()

    ans = 1
    min = nums[0]

    for i in range(1, len(nums)):
      if min + k < nums[i]:
        ans += 1
        min = nums[i]

    return ans