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2659. Make Array Empty 👍

  • Time: $O(\texttt{sort})$
  • Space: $O(n)$
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class Solution {
 public:
  long long countOperationsToEmptyArray(vector<int>& nums) {
    const int n = nums.size();
    long long ans = n;
    unordered_map<int, int> numToIndex;

    for (int i = 0; i < n; ++i)
      numToIndex[nums[i]] = i;

    ranges::sort(nums);

    for (int i = 1; i < n; ++i)
      // On the i-th step we've already removed the i - 1 smallest numbers and
      // can ignore them. If an element nums[i] has smaller index in origin
      // array than nums[i - 1], we should rotate the whole left array n - i
      // times to set nums[i] element on the first position.
      if (numToIndex[nums[i]] < numToIndex[nums[i - 1]])
        ans += n - i;

    return ans;
  }
};

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class Solution {
  public long countOperationsToEmptyArray(int[] nums) {
    final int n = nums.length;
    long ans = n;
    Map<Integer, Integer> numToIndex = new HashMap<>();

    for (int i = 0; i < n; ++i)
      numToIndex.put(nums[i], i);

    Arrays.sort(nums);

    for (int i = 1; i < n; ++i)
      // On the i-th step we've already removed the i - 1 smallest numbers and
      // can ignore them. If an element nums[i] has smaller index in origin
      // array than nums[i - 1], we should rotate the whole left array n - i
      // times to set nums[i] element on the first position.
      if (numToIndex.get(nums[i]) < numToIndex.get(nums[i - 1]))
        ans += n - i;

    return ans;
  }
}

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class Solution:
  def countOperationsToEmptyArray(self, nums: List[int]) -> int:
    n = len(nums)
    ans = n
    numToIndex = {}

    for i, num in enumerate(nums):
      numToIndex[num] = i

    nums.sort()

    for i in range(1, n):
      # On the i-th step we've already removed the i - 1 smallest numbers and
      # can ignore them. If an element nums[i] has smaller index in origin
      # array than nums[i - 1], we should rotate the whole left array n - i
      # times to set nums[i] element on the first position.
      if numToIndex[nums[i]] < numToIndex[nums[i - 1]]:
        ans += n - i

    return ans