LeetCode #1122 — EASY

Relative Sort Array

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[2,3,1,3,2,4,6,7,9,2,19]
[2,1,4,3,9,6]

Example 2

[28,6,22,8,44,17]
[22,28,8,6]

Step 02

Core Insight

What unlocks the optimal approach

Using a hashmap, we can map the values of arr2 to their position in arr2.
After, we can use a custom sorting function.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1122: Relative Sort Array
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;
    }
}

// Accepted solution for LeetCode #1122: Relative Sort Array
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
}

# Accepted solution for LeetCode #1122: Relative Sort Array
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))

// Accepted solution for LeetCode #1122: Relative Sort Array
struct Solution;

use std::cmp::Ordering;
use std::collections::HashMap;

impl Solution {
    fn relative_sort_array(mut arr1: Vec<i32>, arr2: Vec<i32>) -> Vec<i32> {
        let mut hm: HashMap<i32, usize> = HashMap::new();
        for (i, &v) in arr2.iter().enumerate() {
            hm.insert(v, i);
        }
        arr1.sort_by(|a, b| match (hm.get(a), hm.get(b)) {
            (Some(i), Some(j)) => i.cmp(j),
            (Some(_), None) => Ordering::Less,
            (None, Some(_)) => Ordering::Greater,
            (None, None) => a.cmp(b),
        });
        arr1
    }
}

#[test]
fn test() {
    let arr1 = vec![2, 3, 1, 3, 2, 4, 6, 7, 9, 2, 19];
    let arr2 = vec![2, 1, 4, 3, 9, 6];
    let res = vec![2, 2, 2, 1, 4, 3, 3, 9, 6, 7, 19];
    assert_eq!(Solution::relative_sort_array(arr1, arr2), res);
}

// Accepted solution for LeetCode #1122: Relative Sort Array
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]);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n + m)

Space

O(n + m)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.