1122. Relative Sort Array

September 12, 2026 · View on GitHub

Description

Given two arrays arr1 and arr2, the elements of arr2 are distinct, and all elements in arr2 are also in arr1.

Sort the elements of arr1 such that the relative ordering of items in arr1 are the same as in arr2. Elements that do not appear in arr2 should be placed at the end of arr1 in ascending order.

Example 1:

Input: arr1 = [2,3,1,3,2,4,6,7,9,2,19], arr2 = [2,1,4,3,9,6]
Output: [2,2,2,1,4,3,3,9,6,7,19]

Example 2:

Input: arr1 = [28,6,22,8,44,17], arr2 = [22,28,8,6]
Output: [22,28,8,6,17,44]

Constraints:

1 <= arr1.length, arr2.length <= 1000
0 <= arr1[i], arr2[i] <= 1000
All the elements of arr2 are distinct.
Each arr2[i] is in arr1.

Solutions

Solution 1: Custom Sorting

Thinking

The sort key is the relative order in $arr2$ ; values absent from $arr2$ follow, sorted by themselves. A map stores each $arr2$ index; the comparator uses that index if present and $1000+x $otherwise, so one sort handles both groups. Values in$ arr2$ are unique, so indices are a stable order.

First, we use a hash table $pos$ to record the position of each element in array $arr2$ . Then, we map each element in array $arr1$ to a tuple $(pos.get(x, 1000 + x), x)$ , and sort these tuples. Finally, we take out the second element of all tuples and return it.

The time complexity is $O(n \times \log n + m)$ , and the space complexity is $O(n + m)$ . Here, $n$ and $m$ are the lengths of arrays $arr1$ and $arr2$ , respectively.

Python3

class Solution:
    def relativeSortArray(self, arr1: List[int], arr2: List[int]) -> List[int]:
        pos = {x: i for i, x in enumerate(arr2)}
        return sorted(arr1, key=lambda x: pos.get(x, 1000 + x))

Java

class Solution {
    public int[] relativeSortArray(int[] arr1, int[] arr2) {
        Map<Integer, Integer> pos = new HashMap<>(arr2.length);
        for (int i = 0; i < arr2.length; ++i) {
            pos.put(arr2[i], i);
        }
        int[][] arr = new int[arr1.length][0];
        for (int i = 0; i < arr.length; ++i) {
            arr[i] = new int[] {arr1[i], pos.getOrDefault(arr1[i], arr2.length + arr1[i])};
        }
        Arrays.sort(arr, (a, b) -> a[1] - b[1]);
        for (int i = 0; i < arr.length; ++i) {
            arr1[i] = arr[i][0];
        }
        return arr1;
    }
}

C++

class Solution {
public:
    vector<int> relativeSortArray(vector<int>& arr1, vector<int>& arr2) {
        unordered_map<int, int> pos;
        for (int i = 0; i < arr2.size(); ++i) {
            pos[arr2[i]] = i;
        }
        vector<pair<int, int>> arr;
        for (int i = 0; i < arr1.size(); ++i) {
            int j = pos.count(arr1[i]) ? pos[arr1[i]] : arr2.size();
            arr.emplace_back(j, arr1[i]);
        }
        sort(arr.begin(), arr.end());
        for (int i = 0; i < arr1.size(); ++i) {
            arr1[i] = arr[i].second;
        }
        return arr1;
    }
};

Go

func relativeSortArray(arr1 []int, arr2 []int) []int {
	pos := map[int]int{}
	for i, x := range arr2 {
		pos[x] = i
	}
	arr := make([][2]int, len(arr1))
	for i, x := range arr1 {
		if p, ok := pos[x]; ok {
			arr[i] = [2]int{p, x}
		} else {
			arr[i] = [2]int{len(arr2), x}
		}
	}
	sort.Slice(arr, func(i, j int) bool {
		return arr[i][0] < arr[j][0] || arr[i][0] == arr[j][0] && arr[i][1] < arr[j][1]
	})
	for i, x := range arr {
		arr1[i] = x[1]
	}
	return arr1
}

TypeScript

function relativeSortArray(arr1: number[], arr2: number[]): number[] {
    const pos: Map<number, number> = new Map();
    for (let i = 0; i < arr2.length; ++i) {
        pos.set(arr2[i], i);
    }
    const arr: number[][] = [];
    for (const x of arr1) {
        const j = pos.get(x) ?? arr2.length;
        arr.push([j, x]);
    }
    arr.sort((a, b) => a[0] - b[0] || a[1] - b[1]);
    return arr.map(a => a[1]);
}

Swift

class Solution {
    func relativeSortArray(_ arr1: [Int], _ arr2: [Int]) -> [Int] {
        var pos = [Int: Int]()
        for (i, x) in arr2.enumerated() {
            pos[x] = i
        }
        var arr = [(Int, Int)]()
        for x in arr1 {
            let j = pos[x] ?? arr2.count
            arr.append((j, x))
        }
        arr.sort { \$0.0 < \$1.0 || (\$0.0 == \$1.0 && \$0.1 < \$1.1) }
        return arr.map { \$0.1 }
    }
}

Solution 2: Counting Sort

Thinking

Method 1's comparison sort is $O(n\log n)$ . When the value range is small, count frequencies in $arr1$ , emit values in $arr2$ order, then scan the leftovers by value, which is linear.

We can use the idea of counting sort. First, count the occurrence of each element in array $arr1$ . Then, according to the order in array $arr2$ , put the elements in $arr1$ into the answer array $ans$ according to their occurrence. Finally, we traverse all elements in $arr1$ and put the elements that do not appear in $arr2$ in ascending order at the end of the answer array $ans$ .

The time complexity is $O(n + m)$ , and the space complexity is $O(n)$ . Where $n$ and $m$ are the lengths of arrays $arr1$ and $arr2$ respectively.

Python3

class Solution:
    def relativeSortArray(self, arr1: List[int], arr2: List[int]) -> List[int]:
        cnt = Counter(arr1)
        ans = []
        for x in arr2:
            ans.extend([x] * cnt[x])
            cnt.pop(x)
        mi, mx = min(arr1), max(arr1)
        for x in range(mi, mx + 1):
            ans.extend([x] * cnt[x])
        return ans

Java

class Solution {
    public int[] relativeSortArray(int[] arr1, int[] arr2) {
        int[] cnt = new int[1001];
        int mi = 1001, mx = 0;
        for (int x : arr1) {
            ++cnt[x];
            mi = Math.min(mi, x);
            mx = Math.max(mx, x);
        }
        int m = arr1.length;
        int[] ans = new int[m];
        int i = 0;
        for (int x : arr2) {
            while (cnt[x] > 0) {
                --cnt[x];
                ans[i++] = x;
            }
        }
        for (int x = mi; x <= mx; ++x) {
            while (cnt[x] > 0) {
                --cnt[x];
                ans[i++] = x;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> relativeSortArray(vector<int>& arr1, vector<int>& arr2) {
        vector<int> cnt(1001);
        for (int x : arr1) {
            ++cnt[x];
        }
        auto [mi, mx] = minmax_element(arr1.begin(), arr1.end());
        vector<int> ans;
        for (int x : arr2) {
            while (cnt[x]) {
                ans.push_back(x);
                --cnt[x];
            }
        }
        for (int x = *mi; x <= *mx; ++x) {
            while (cnt[x]) {
                ans.push_back(x);
                --cnt[x];
            }
        }
        return ans;
    }
};

Go

func relativeSortArray(arr1 []int, arr2 []int) []int {
	cnt := make([]int, 1001)
	mi, mx := 1001, 0
	for _, x := range arr1 {
		cnt[x]++
		mi = min(mi, x)
		mx = max(mx, x)
	}
	ans := make([]int, 0, len(arr1))
	for _, x := range arr2 {
		for cnt[x] > 0 {
			ans = append(ans, x)
			cnt[x]--
		}
	}
	for x := mi; x <= mx; x++ {
		for cnt[x] > 0 {
			ans = append(ans, x)
			cnt[x]--
		}
	}
	return ans
}

TypeScript

function relativeSortArray(arr1: number[], arr2: number[]): number[] {
    const cnt = Array(1001).fill(0);
    let mi = Number.POSITIVE_INFINITY;
    let mx = Number.NEGATIVE_INFINITY;

    for (const x of arr1) {
        cnt[x]++;
        mi = Math.min(mi, x);
        mx = Math.max(mx, x);
    }

    const ans: number[] = [];
    for (const x of arr2) {
        while (cnt[x]) {
            cnt[x]--;
            ans.push(x);
        }
    }

    for (let i = mi; i <= mx; i++) {
        while (cnt[i]) {
            cnt[i]--;
            ans.push(i);
        }
    }

    return ans;
}

Swift

class Solution {
    func relativeSortArray(_ arr1: [Int], _ arr2: [Int]) -> [Int] {
        var cnt = [Int](repeating: 0, count: 1001)
        for x in arr1 {
            cnt[x] += 1
        }

        guard let mi = arr1.min(), let mx = arr1.max() else {
            return []
        }

        var ans = [Int]()
        for x in arr2 {
            while cnt[x] > 0 {
                ans.append(x)
                cnt[x] -= 1
            }
        }

        for x in mi...mx {
            while cnt[x] > 0 {
                ans.append(x)
                cnt[x] -= 1
            }
        }

        return ans
    }
}