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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

Thinking

Replace each value by its \(1\)-based rank, sharing a rank on ties. Pairwise comparison is quadratic for \(n \le 10^5\). Rank depends only on the position in the sorted unique list \(t\); \(\mathrm{bisect\_right}\) on \(t\) yields that rank for every \(x\).

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 \(arr\).

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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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impl Solution {
    pub fn array_rank_transform(arr: Vec<i32>) -> Vec<i32> {
        let mut q1 = arr.clone();
        q1.sort_unstable();
        q1.dedup();
        arr.iter()
            .map(|q2| q1.binary_search(q2).unwrap() as i32 + 1)
            .collect()
    }
}

Solution 2: Sorting + Hash Map

Thinking

A binary search per element repeats the same work. After sorting the unique values, a hash map from value to rank answers every lookup in expected constant time.

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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)!);
}

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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));
}

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