Make Lexicographically Smallest Array by Swapping Elements

Make Lexicographically Smallest Array by Swapping Elements - Problem

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.

Input & Output

Example 1 — Basic Swapping

$ Input: nums = [1,5,3,9], limit = 2

› Output: [1,3,5,9]

💡 Note: We can swap 5 and 3 since |5-3|=2≤2. After swapping: [1,3,5,9] which is lexicographically smallest.

Example 2 — No Valid Swaps

$ Input: nums = [1,7,5,3], limit = 1

› Output: [1,7,5,3]

💡 Note: No adjacent values have difference ≤1, so no swaps are possible. Return original array.

Example 3 — Multiple Connected Components

$ Input: nums = [1,5,3,9,4], limit = 2

› Output: [1,3,4,9,5]

💡 Note: Elements can be grouped: 1↔3 (diff=2), 5↔3 (diff=2), 5↔4 (diff=1), 3↔4 (diff=1). This forms one connected component {1,3,4,5} at positions [0,1,2,4] and isolated element {9} at position [3]. Sorting the component gives [1,3,4,5], resulting in [1,3,4,9,5].

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹
0 ≤ limit ≤ 10⁹

Visualization

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Asked in

G Google 45 f Meta 32 a Amazon 28 M Microsoft 25

The key insight is that elements form connected components based on swapping ability. Use Union-Find to group elements that can eventually be swapped, then sort each group separately. Best approach: Union-Find with sorting - Time: O(n²α(n) + n log n), Space: O(n)

Common Approaches

✓ Brute Force - Try All Swaps

⏱️ Time: O(n³) Space: O(1)

Repeatedly scan the array and perform any valid swap that makes the array lexicographically smaller. Continue until no more beneficial swaps are possible.

Union-Find with Sorting

⏱️ Time: O(n²α(n) + n log n) Space: O(n)

Use Union-Find to identify all elements that can eventually be swapped with each other. For each connected component, sort the elements and place them in their original positions.

Brute Force - Try All Swaps — Algorithm Steps

Step 1: Scan array for valid swaps that improve lexicographic order
Step 2: Perform the swap and repeat until no improvement possible

Visualization

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Step-by-Step Walkthrough

Initial Array

Start with original array

Find Valid Swaps

Check all pairs for |nums[i] - nums[j]| <= limit

Perform Best Swap

Swap elements that improve lexicographic order

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

int* solution(int* nums, int numsSize, int limit, int* returnSize) {
    int* result = (int*)malloc(numsSize * sizeof(int));
    for (int i = 0; i < numsSize; i++) {
        result[i] = nums[i];
    }
    
    bool changed = true;
    while (changed) {
        changed = false;
        for (int i = 0; i < numsSize; i++) {
            for (int j = i + 1; j < numsSize; j++) {
                if (abs(result[i] - result[j]) <= limit && result[i] > result[j]) {
                    int temp = result[i];
                    result[i] = result[j];
                    result[j] = temp;
                    changed = true;
                }
            }
        }
    }
    
    *returnSize = numsSize;
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (token[0] >= '0' && token[0] <= '9') {
            nums[numsSize++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int limit;
    scanf("%d", &limit);
    
    int returnSize;
    int* result = solution(nums, numsSize, limit, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

In worst case, each pass checks O(n²) pairs and we may need O(n) passes

⚠ Quadratic Growth

Space Complexity

O(1)

Only uses constant extra space for variables

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Try All Swaps — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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