Question

1331. Rank Transform of an Array

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Description

Given an array of integers arr, replace each element with its rank.

The rank represents how large the element is. The rank has the following rules:

• Rank is an integer starting from 1.
• The larger the element, the larger the rank. If two elements are equal, their rank must be the same.
• Rank should be as small as possible.

 

Example 1:

Input: arr = [40,10,20,30]
Output: [4,1,2,3]
Explanation: 40 is the largest element. 10 is the smallest. 20 is the second smallest. 30 is the third smallest.

Example 2:

Input: arr = [100,100,100]
Output: [1,1,1]
Explanation: Same elements share the same rank.

Example 3:

Input: arr = [37,12,28,9,100,56,80,5,12]
Output: [5,3,4,2,8,6,7,1,3]

 

Constraints:

• 0 <= arr.length <= 105
• -109 <= arr[i] <= 109

Solutions

Solution 1

class Solution:
  def arrayRankTransform(self, arr: List[int]) -> List[int]:
    t = sorted(set(arr))
    return [bisect_right(t, x) for x in arr]

class Solution {
  public int[] arrayRankTransform(int[] arr) {
    int n = arr.length;
    int[] t = arr.clone();
    Arrays.sort(t);
    int m = 0;
    for (int i = 0; i < n; ++i) {
      if (i == 0 || t[i] != t[i - 1]) {
        t[m++] = t[i];
      }
    }
    int[] ans = new int[n];
    for (int i = 0; i < n; ++i) {
      ans[i] = Arrays.binarySearch(t, 0, m, arr[i]) + 1;
    }
    return ans;
  }
}

class Solution {
public:
  vector<int> arrayRankTransform(vector<int>& arr) {
    vector<int> t = arr;
    sort(t.begin(), t.end());
    t.erase(unique(t.begin(), t.end()), t.end());
    vector<int> ans;
    for (int x : arr) {
      ans.push_back(upper_bound(t.begin(), t.end(), x) - t.begin());
    }
    return ans;
  }
};

func arrayRankTransform(arr []int) (ans []int) {
  t := make([]int, len(arr))
  copy(t, arr)
  sort.Ints(t)
  m := 0
  for i, x := range t {
    if i == 0 || x != t[i-1] {
      t[m] = x
      m++
    }
  }
  t = t[:m]
  for _, x := range arr {
    ans = append(ans, sort.SearchInts(t, x)+1)
  }
  return
}

function arrayRankTransform(arr: number[]): number[] {
  const t = [...arr].sort((a, b) => a - b);
  let m = 0;
  for (let i = 0; i < t.length; ++i) {
    if (i === 0 || t[i] !== t[i - 1]) {
      t[m++] = t[i];
    }
  }
  const search = (t: number[], right: number, x: number) => {
    let left = 0;
    while (left < right) {
      const mid = (left + right) >> 1;
      if (t[mid] > x) {
        right = mid;
      } else {
        left = mid + 1;
      }
    }
    return left;
  };
  const ans: number[] = [];
  for (const x of arr) {
    ans.push(search(t, m, x));
  }
  return ans;
}