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: [1,3,5,4,9]

💡 Note: Elements 5,3 can swap (diff=2). Elements 9,4 can swap (diff=5>2, but 3 can connect to both forming one group). Result: sort [5,3] to [3,5], and [9,4] to [4,9].

Constraints

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

Visualization

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Understanding the Visualization

Input

Array with elements that can swap if difference ≤ limit

Group

Find connected components of swappable elements

Sort

Sort each group to get lexicographically smallest result

Key Takeaway

🎯 Key Insight: Elements that can be swapped transitively form connected components - sort each component for optimal result

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

Approach	Time	Space	Notes
✓ Union-Find with Sorting	O(n²α(n) + n log n)	O(n)	Group elements that can be swapped using Union-Find, then sort each group
Brute Force - Try All Swaps	O(n³)	O(1)	Keep performing all possible valid swaps until no improvement can be made

Union-Find with Sorting — Algorithm Steps

Step 1: Build Union-Find structure for elements within limit distance
Step 2: Group elements by connected component
Step 3: Sort each group and place back in original positions

Visualization

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Step-by-Step Walkthrough

Build Connections

Union indices where |nums[i] - nums[j]| <= limit

Group Components

Find all connected components

Sort Each Group

Sort values within each component

Code -

solution.c — C

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

struct UnionFind {
    int* parent;
    int* rank;
    int n;
};

struct UnionFind* createUF(int n) {
    struct UnionFind* uf = (struct UnionFind*)malloc(sizeof(struct UnionFind));
    uf->parent = (int*)malloc(n * sizeof(int));
    uf->rank = (int*)calloc(n, sizeof(int));
    uf->n = n;
    for (int i = 0; i < n; i++) {
        uf->parent[i] = i;
    }
    return uf;
}

int find(struct UnionFind* uf, int x) {
    if (uf->parent[x] != x) {
        uf->parent[x] = find(uf, uf->parent[x]);
    }
    return uf->parent[x];
}

void unite(struct UnionFind* uf, int x, int y) {
    int px = find(uf, x), py = find(uf, y);
    if (px == py) return;
    if (uf->rank[px] < uf->rank[py]) {
        uf->parent[px] = py;
    } else if (uf->rank[px] > uf->rank[py]) {
        uf->parent[py] = px;
    } else {
        uf->parent[py] = px;
        uf->rank[px]++;
    }
}

int compare(const void* a, const void* b) {
    return (*(int*)a - *(int*)b);
}

int* solution(int* nums, int numsSize, int limit, int* returnSize) {
    struct UnionFind* uf = createUF(numsSize);
    
    // Union indices where elements can be swapped
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if (abs(nums[i] - nums[j]) <= limit) {
                unite(uf, i, j);
            }
        }
    }
    
    int* result = (int*)malloc(numsSize * sizeof(int));
    for (int i = 0; i < numsSize; i++) {
        result[i] = nums[i];
    }
    
    // Simple approach: for each position, find the minimum in its component
    bool* processed = (bool*)calloc(numsSize, sizeof(bool));
    
    for (int i = 0; i < numsSize; i++) {
        if (processed[i]) continue;
        
        // Find all indices in same component
        int component[1000];
        int compSize = 0;
        for (int j = 0; j < numsSize; j++) {
            if (find(uf, i) == find(uf, j)) {
                component[compSize++] = j;
                processed[j] = true;
            }
        }
        
        // Get values and sort them
        int values[1000];
        for (int j = 0; j < compSize; j++) {
            values[j] = nums[component[j]];
        }
        qsort(values, compSize, sizeof(int), compare);
        qsort(component, compSize, sizeof(int), compare);
        
        // Assign sorted values to sorted positions
        for (int j = 0; j < compSize; j++) {
            result[component[j]] = values[j];
        }
    }
    
    *returnSize = numsSize;
    free(uf->parent);
    free(uf->rank);
    free(uf);
    free(processed);
    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(n²α(n) + n log n)

O(n²α(n)) for building union-find by checking all pairs, plus O(n log n) for sorting groups

⚠ Quadratic Growth

Space Complexity

O(n)

Union-Find parent array and temporary arrays for grouping elements

⚡ Linearithmic Space

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Medium Frequency

~25 min Avg. Time

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Union-Find with Sorting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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