LeetCode #2948 — MEDIUM

Make Lexicographically Smallest Array by Swapping Elements

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 0-indexed array of positive integers nums and a positive integer limit.

In one operation, you can choose any two indices i and j and swap nums[i] and nums[j] if |nums[i] - nums[j]| <= limit.

Return the lexicographically smallest array that can be obtained by performing the operation any number of times.

An array a is lexicographically smaller than an array b if in the first position where a and b differ, array a has an element that is less than the corresponding element in b. For example, the array [2,10,3] is lexicographically smaller than the array [10,2,3] because they differ at index 0 and 2 < 10.

Example 1:

Input: nums = [1,5,3,9,8], limit = 2
Output: [1,3,5,8,9]
Explanation: Apply the operation 2 times:
- Swap nums[1] with nums[2]. The array becomes [1,3,5,9,8]
- Swap nums[3] with nums[4]. The array becomes [1,3,5,8,9]
We cannot obtain a lexicographically smaller array by applying any more operations.
Note that it may be possible to get the same result by doing different operations.

Example 2:

Input: nums = [1,7,6,18,2,1], limit = 3
Output: [1,6,7,18,1,2]
Explanation: Apply the operation 3 times:
- Swap nums[1] with nums[2]. The array becomes [1,6,7,18,2,1]
- Swap nums[0] with nums[4]. The array becomes [2,6,7,18,1,1]
- Swap nums[0] with nums[5]. The array becomes [1,6,7,18,1,2]
We cannot obtain a lexicographically smaller array by applying any more operations.

Example 3:

Input: nums = [1,7,28,19,10], limit = 3
Output: [1,7,28,19,10]
Explanation: [1,7,28,19,10] is the lexicographically smallest array we can obtain because we cannot apply the operation on any two indices.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹
1 <= limit <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array of positive integers nums and a positive integer limit. In one operation, you can choose any two indices i and j and swap nums[i] and nums[j] if |nums[i] - nums[j]| <= limit. Return the lexicographically smallest array that can be obtained by performing the operation any number of times. An array a is lexicographically smaller than an array b if in the first position where a and b differ, array a has an element that is less than the corresponding element in b. For example, the array [2,10,3] is lexicographically smaller than the array [10,2,3] because they differ at index 0 and 2 < 10.

Baseline thinking

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

Pattern signal: Array · Union-Find

Example 1

[1,5,3,9,8]
2

Example 2

[1,7,6,18,2,1]
3

Example 3

[1,7,28,19,10]
3

Core Insight

What unlocks the optimal approach

Construct a virtual graph where all elements in <code>nums</code> are nodes and the pairs satisfying the condition have an edge between them.
Instead of constructing all edges, we only care about the connected components.
Can we use DSU?
Sort <code>nums</code>. Now we just need to consider if the consecutive elements have an edge to check if they belong to the same connected component. Hence, all connected components become a list of position-consecutive elements after sorting.
For each index of <code>nums</code> from <code>0</code> to <code>nums.length - 1</code> we can change it to the current minimum value we have in its connected component and remove that value from the connected component.

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 #2948: Make Lexicographically Smallest Array by Swapping Elements
class Solution {
    public int[] lexicographicallySmallestArray(int[] nums, int limit) {
        int n = nums.length;
        Integer[] idx = new Integer[n];
        Arrays.setAll(idx, i -> i);
        Arrays.sort(idx, (i, j) -> nums[i] - nums[j]);
        int[] ans = new int[n];
        for (int i = 0; i < n;) {
            int j = i + 1;
            while (j < n && nums[idx[j]] - nums[idx[j - 1]] <= limit) {
                ++j;
            }
            Integer[] t = Arrays.copyOfRange(idx, i, j);
            Arrays.sort(t, (x, y) -> x - y);
            for (int k = i; k < j; ++k) {
                ans[t[k - i]] = nums[idx[k]];
            }
            i = j;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2948: Make Lexicographically Smallest Array by Swapping Elements
func lexicographicallySmallestArray(nums []int, limit int) []int {
	n := len(nums)
	idx := make([]int, n)
	for i := range idx {
		idx[i] = i
	}
	slices.SortFunc(idx, func(i, j int) int { return nums[i] - nums[j] })
	ans := make([]int, n)
	for i := 0; i < n; {
		j := i + 1
		for j < n && nums[idx[j]]-nums[idx[j-1]] <= limit {
			j++
		}
		t := slices.Clone(idx[i:j])
		slices.Sort(t)
		for k := i; k < j; k++ {
			ans[t[k-i]] = nums[idx[k]]
		}
		i = j
	}
	return ans
}

# Accepted solution for LeetCode #2948: Make Lexicographically Smallest Array by Swapping Elements
class Solution:
    def lexicographicallySmallestArray(self, nums: List[int], limit: int) -> List[int]:
        n = len(nums)
        arr = sorted(zip(nums, range(n)))
        ans = [0] * n
        i = 0
        while i < n:
            j = i + 1
            while j < n and arr[j][0] - arr[j - 1][0] <= limit:
                j += 1
            idx = sorted(k for _, k in arr[i:j])
            for k, (x, _) in zip(idx, arr[i:j]):
                ans[k] = x
            i = j
        return ans

// Accepted solution for LeetCode #2948: Make Lexicographically Smallest Array by Swapping Elements
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2948: Make Lexicographically Smallest Array by Swapping Elements
// class Solution {
//     public int[] lexicographicallySmallestArray(int[] nums, int limit) {
//         int n = nums.length;
//         Integer[] idx = new Integer[n];
//         Arrays.setAll(idx, i -> i);
//         Arrays.sort(idx, (i, j) -> nums[i] - nums[j]);
//         int[] ans = new int[n];
//         for (int i = 0; i < n;) {
//             int j = i + 1;
//             while (j < n && nums[idx[j]] - nums[idx[j - 1]] <= limit) {
//                 ++j;
//             }
//             Integer[] t = Arrays.copyOfRange(idx, i, j);
//             Arrays.sort(t, (x, y) -> x - y);
//             for (int k = i; k < j; ++k) {
//                 ans[t[k - i]] = nums[idx[k]];
//             }
//             i = j;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2948: Make Lexicographically Smallest Array by Swapping Elements
function lexicographicallySmallestArray(nums: number[], limit: number): number[] {
    const n: number = nums.length;
    const idx: number[] = Array.from({ length: n }, (_, i) => i);
    idx.sort((i, j) => nums[i] - nums[j]);
    const ans: number[] = Array(n).fill(0);
    for (let i = 0; i < n; ) {
        let j = i + 1;
        while (j < n && nums[idx[j]] - nums[idx[j - 1]] <= limit) {
            j++;
        }
        const t: number[] = idx.slice(i, j).sort((a, b) => a - b);
        for (let k: number = i; k < j; k++) {
            ans[t[k - i]] = nums[idx[k]];
        }
        i = j;
    }
    return ans;
}

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(α(n))

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND

O(α(n)) time

O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”

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.