1331. Rank Transform of an Array 

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: Discretization

First, we copy an array $t$, then sort and deduplicate it to obtain an array of length $m$ that is strictly monotonically increasing.

Next, we traverse the original array $arr$. For each element $x$ in the array, we use binary search to find the position of $x$ in $t$. The position plus one is the rank of $x$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $arr$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Solution 2: Sorting + Hash Map

TypeScript

function arrayRankTransform(arr: number[]): number[] {
    const sorted = [...new Set(arr)].sort((a, b) => a - b);
    const map = new Map<number, number>();
    let c = 1;

    for (const x of sorted) {
        map.set(x, c++);
    }

    return arr.map(x => map.get(x)!);
}

JavaScript

function arrayRankTransform(arr) {
    const sorted = [...new Set(arr)].sort((a, b) => a - b);
    const map = new Map();
    let c = 1;

    for (const x of sorted) {
        map.set(x, c++);
    }

    return arr.map(x => map.get(x));
}