LeetCode #1202 — MEDIUM

Smallest String With Swaps

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

The Problem

Problem Statement

You are given a string s, and an array of pairs of indices in the string pairs where pairs[i] = [a, b] indicates 2 indices(0-indexed) of the string.

You can swap the characters at any pair of indices in the given pairs any number of times.

Return the lexicographically smallest string that s can be changed to after using the swaps.

Example 1:

Input: s = "dcab", pairs = [[0,3],[1,2]]
Output: "bacd"
Explaination: 
Swap s[0] and s[3], s = "bcad"
Swap s[1] and s[2], s = "bacd"

Example 2:

Input: s = "dcab", pairs = [[0,3],[1,2],[0,2]]
Output: "abcd"
Explaination: 
Swap s[0] and s[3], s = "bcad"
Swap s[0] and s[2], s = "acbd"
Swap s[1] and s[2], s = "abcd"

Example 3:

Input: s = "cba", pairs = [[0,1],[1,2]]
Output: "abc"
Explaination: 
Swap s[0] and s[1], s = "bca"
Swap s[1] and s[2], s = "bac"
Swap s[0] and s[1], s = "abc"

Constraints:

1 <= s.length <= 10^5
0 <= pairs.length <= 10^5
0 <= pairs[i][0], pairs[i][1] < s.length
s only contains lower case English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s, and an array of pairs of indices in the string pairs where pairs[i] = [a, b] indicates 2 indices(0-indexed) of the string. You can swap the characters at any pair of indices in the given pairs any number of times. Return the lexicographically smallest string that s can be changed to after using the swaps.

Baseline thinking

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

Pattern signal: Array · Hash Map · Union-Find

Example 1

"dcab"
[[0,3],[1,2]]

Example 2

"dcab"
[[0,3],[1,2],[0,2]]

Example 3

"cba"
[[0,1],[1,2]]

Core Insight

What unlocks the optimal approach

Think of it as a graph problem.
Consider the pairs as connected nodes in the graph, what can you do with a connected component of indices ?
We can sort each connected component alone to get the lexicographically minimum string.

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 #1202: Smallest String With Swaps
class Solution {
    private int[] p;

    public String smallestStringWithSwaps(String s, List<List<Integer>> pairs) {
        int n = s.length();
        p = new int[n];
        List<Character>[] d = new List[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            d[i] = new ArrayList<>();
        }
        for (var pair : pairs) {
            int a = pair.get(0), b = pair.get(1);
            p[find(a)] = find(b);
        }
        char[] cs = s.toCharArray();
        for (int i = 0; i < n; ++i) {
            d[find(i)].add(cs[i]);
        }
        for (var e : d) {
            e.sort((a, b) -> b - a);
        }
        for (int i = 0; i < n; ++i) {
            var e = d[find(i)];
            cs[i] = e.remove(e.size() - 1);
        }
        return String.valueOf(cs);
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

// Accepted solution for LeetCode #1202: Smallest String With Swaps
func smallestStringWithSwaps(s string, pairs [][]int) string {
	n := len(s)
	p := make([]int, n)
	d := make([][]byte, n)
	for i := range p {
		p[i] = i
	}
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	for _, pair := range pairs {
		a, b := pair[0], pair[1]
		p[find(a)] = find(b)
	}
	cs := []byte(s)
	for i, c := range cs {
		j := find(i)
		d[j] = append(d[j], c)
	}
	for i := range d {
		sort.Slice(d[i], func(a, b int) bool { return d[i][a] > d[i][b] })
	}
	for i := range cs {
		j := find(i)
		cs[i] = d[j][len(d[j])-1]
		d[j] = d[j][:len(d[j])-1]
	}
	return string(cs)
}

# Accepted solution for LeetCode #1202: Smallest String With Swaps
class Solution:
    def smallestStringWithSwaps(self, s: str, pairs: List[List[int]]) -> str:
        def find(x: int) -> int:
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        n = len(s)
        p = list(range(n))
        for a, b in pairs:
            p[find(a)] = find(b)
        d = defaultdict(list)
        for i, c in enumerate(s):
            d[find(i)].append(c)
        for i in d.keys():
            d[i].sort(reverse=True)
        return "".join(d[find(i)].pop() for i in range(n))

// Accepted solution for LeetCode #1202: Smallest String With Swaps
impl Solution {
    #[allow(dead_code)]
    pub fn smallest_string_with_swaps(s: String, pairs: Vec<Vec<i32>>) -> String {
        let n = s.as_bytes().len();
        let s = s.as_bytes();
        let mut disjoint_set: Vec<usize> = vec![0; n];
        let mut str_vec: Vec<Vec<u8>> = vec![Vec::new(); n];
        let mut ret_str = String::new();

        // Initialize the disjoint set
        for i in 0..n {
            disjoint_set[i] = i;
        }

        // Union the pairs
        for pair in pairs {
            Self::union(pair[0] as usize, pair[1] as usize, &mut disjoint_set);
        }

        // Initialize the return vector
        for (i, c) in s.iter().enumerate() {
            let p_c = Self::find(i, &mut disjoint_set);
            str_vec[p_c].push(*c);
        }

        // Sort the return vector in reverse order
        for cur_vec in &mut str_vec {
            cur_vec.sort();
            cur_vec.reverse();
        }

        // Construct the return string
        for i in 0..n {
            let index = Self::find(i, &mut disjoint_set);
            ret_str.push(str_vec[index].last().unwrap().clone() as char);
            str_vec[index].pop();
        }

        ret_str
    }

    #[allow(dead_code)]
    fn find(x: usize, d_set: &mut Vec<usize>) -> usize {
        if d_set[x] != x {
            d_set[x] = Self::find(d_set[x], d_set);
        }
        d_set[x]
    }

    #[allow(dead_code)]
    fn union(x: usize, y: usize, d_set: &mut Vec<usize>) {
        let p_x = Self::find(x, d_set);
        let p_y = Self::find(y, d_set);
        d_set[p_x] = p_y;
    }
}

// Accepted solution for LeetCode #1202: Smallest String With Swaps
function smallestStringWithSwaps(s: string, pairs: number[][]): string {
    const n = s.length;
    const p = new Array(n).fill(0).map((_, i) => i);
    const find = (x: number): number => {
        if (p[x] !== x) {
            p[x] = find(p[x]);
        }
        return p[x];
    };
    const d: string[][] = new Array(n).fill(0).map(() => []);
    for (const [a, b] of pairs) {
        p[find(a)] = find(b);
    }
    for (let i = 0; i < n; ++i) {
        d[find(i)].push(s[i]);
    }
    for (const e of d) {
        e.sort((a, b) => b.charCodeAt(0) - a.charCodeAt(0));
    }
    const ans: string[] = [];
    for (let i = 0; i < n; ++i) {
        ans.push(d[find(i)].pop()!);
    }
    return ans.join('');
}

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 × \alpha(m)

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.

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.