Chef's Ingredient Rearrangement — Problem Statement & Solution Guide

StringsMediumRandom

TimeO(n)

SpaceO(σ²) ≈ O(1) for fixed alphabet

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Strings and solve the Chef's Ingredient Rearrangement problem optimally.

TopicStrings

PatternRandom

TimeO(n)

SpaceO(σ²) ≈ O(1) for fixed alphabet

Problem Description

Given two strings of ingredients, s1 and s2, of the same length, determine the minimum number of swap operations on the characters of s1 to transform it into s2.

DSA Pattern Breakdown

"Chef's Ingredient Rearrangement"

medium

WHY DOES IT MATTER?

Understanding how to minimize swaps by exploiting reciprocal mismatches is a core combinatorial optimization pattern. It teaches candidates to look beyond greedy one‑by‑one fixes and to identify hidden symmetries that reduce work, a skill transferable to many permutation and graph‑rearrangement problems.

OPTIMIZATION CHALLENGE

The key insight is that the swap operation is symmetric; therefore, counting ordered mismatched pairs and pairing them with their reverse eliminates the need for explicit simulation of swaps, collapsing an exponential search space into a linear counting problem.

REAL-WORLD CONNECTION

In distributed databases, reconciling divergent replicas often involves swapping data blocks. Recognizing reciprocal differences lets the system perform a single network round‑trip to fix two out‑of‑sync blocks, dramatically reducing synchronization latency.

During an interview, first write the mismatch collection loop, then immediately think "which swaps fix two positions?" – that mental shortcut leads you to the reciprocal‑pair count and saves you from over‑engineering a BFS or backtracking solution.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(σ²) ≈ O(1) for fixed alphabet

Core Theory — Why This Approach?

The problem can be modeled as a transformation of one string into another using the elementary operation of swapping any two characters. A naïve view treats each mismatched position independently, leading to a linear scan and a swap per mismatch, which is sub‑optimal because certain swaps can resolve two mismatches simultaneously. The optimal insight is to recognize reciprocal mismatches: if at index i we have (a→b) and at index j we have (b→a), a single swap between i and j fixes both positions. By counting all mismatched ordered pairs and then extracting the maximum number of such reciprocal pairs, the minimal number of swaps equals the total mismatches minus the number of reciprocal pairs. This reduction transforms the problem into a simple counting exercise that runs in linear time, avoiding the combinatorial explosion of trying all possible swap sequences.

Why naive approaches fail on large inputs is evident when the strings length reaches 10^5 or more. Enumerating every possible swap or performing a breadth‑first search over string states would be exponential, quickly exhausting time and memory limits. The optimal paradigm leverages the fact that swaps are commutative and that the only interaction between mismatches is through these reciprocal relationships. By collapsing the problem to a frequency map of ordered character pairs, we achieve O(n) time and O(σ²) space (σ = alphabet size), which is effectively constant for typical English letters.

The algorithm thus follows a three‑step pipeline: (1) scan both strings once to collect mismatched ordered pairs, (2) for each distinct pair (x, y) compute the number of reciprocal pairs with (y, x) and accumulate the minimum of the two counts, and (3) compute the answer as mismatches – reciprocalPairs. This approach is both optimal and easy to implement, making it a favorite in interview settings where clarity and efficiency are prized.

Interview Questions on This Problem

Q1How would you modify the solution if each swap operation could only involve adjacent characters?

When swaps are limited to adjacent positions, the problem becomes counting the minimum number of adjacent swaps, which is equivalent to computing the inversion count needed to reorder s1 into s2. This can be solved by mapping each character in s2 to its target indices, building a list of positions from s1, and then using a Fenwick tree or BIT to count inversions in O(n log n) time.

Q2Can the algorithm be extended to handle strings of different lengths where you may also insert or delete characters?

Yes. The problem then becomes the classic edit distance with swap (or transposition) operations. A dynamic programming solution with O(n·m) time can be adapted to include a swap transition that checks if two characters are cross‑matched, but the state space grows, and specialized algorithms like the Damerau‑Levenshtein distance are used.

Q3Why does counting reciprocal pairs guarantee the minimal number of swaps, and can there be a scenario where a three‑way cycle yields a better result?

Reciprocal pairs are the only configuration where a single swap resolves two mismatches; any larger cycle (e.g., a→b, b→c, c→a) requires at least two swaps because each swap can fix at most two positions. Therefore, after exhausting all reciprocal pairs, the remaining mismatches form disjoint cycles of length ≥3, each of which needs exactly (cycle length – 1) swaps, which is captured by the formula mismatches – reciprocalPairs.

Examples

Example 1

Input

s1 = 'ab', s2 = 'ba'

Output

Explanation: Step-by-step: We need to swap 'a' and 'b' in s1 to get s2. This requires 1 operation.

Example 2

Input

s1 = 'xyz', s2 = 'zyx'

Output

Explanation: Step-by-step: We can swap 'x' and 'z' first, then 'y' and 'z' to get s2. This requires 2 operations.

Constraints

1 <= length of s1 == length of s2 <= 100
s1 and s2 contain only lowercase English letters
The input strings can be modified, and additional space can be used for bookkeeping

Optimal Approach & Strategy

Count mismatched ordered character pairs, pair each (x, y) with its reverse (y, x) to find reciprocal swaps, and compute answer as mismatches minus the number of such pairs. This runs in linear time with a constant‑size hash map.

Brute Force Approach

Try every possible pair of indices to swap, recursively explore all sequences until s1 equals s2, and keep the minimum depth. This exhaustive search explores an exponential number of states and is infeasible for large strings.

Code Solutions

Language:

JavaScript Solution

Time: O(n)

function minSwaps(s1, s2) {
    const charPositions = {};
    for (let i = 0; i < s2.length; i++) {
        if (!charPositions[s2[i]]) {
            charPositions[s2[i]] = [];
        }
        charPositions[s2[i]].push(i);
    }
    
    let swaps = 0;
    let s1Arr = s1.split('');
    
    for (let i = 0; i < s1Arr.length; i++) {
        if (s1Arr[i] !== s2[i]) {
            const targetChar = s2[i];
            const targetIndex = charPositions[targetChar].pop();
            
            // Swap s1Arr[i] and s1Arr[targetIndex]
            const temp = s1Arr[i];
            s1Arr[i] = s1Arr[targetIndex];
            s1Arr[targetIndex] = temp;
            
            // Update positions for the swapped characters
            if (s1Arr[i] !== s2[i]) {
                if (!charPositions[s1Arr[i]]) {
                    charPositions[s1Arr[i]] = [];
                }
                charPositions[s1Arr[i]].push(i);
            }
            if (s1Arr[targetIndex] !== s2[targetIndex]) {
                if (!charPositions[s1Arr[targetIndex]]) {
                    charPositions[s1Arr[targetIndex]] = [];
                }
                charPositions[s1Arr[targetIndex]].push(targetIndex);
            }
            
            swaps++;
        }
    }
    return swaps;
}

// Example usage
// const s1 = "ab";
// const s2 = "ba";
// console.log(minSwaps(s1, s2));

#include <iostream>
#include <string>
#include <unordered_map>
#include <vector>

using namespace std;

int minSwaps(string s1, string s2) {
    unordered_map<char, vector<int>> charPositions;
    for (int i = 0; i < s2.size(); ++i) {
        charPositions[s2[i]].push_back(i);
    }
    
    int swaps = 0;
    for (int i = 0; i < s1.size(); ++i) {
        if (s1[i] != s2[i]) {
            char targetChar = s2[i];
            int targetIndex = charPositions[targetChar].back();
            charPositions[targetChar].pop_back();
            
            // Swap s1[i] and s1[targetIndex]
            char temp = s1[i];
            s1[i] = s1[targetIndex];
            s1[targetIndex] = temp;
            
            // Update positions for the swapped characters
            if (s1[i] != s2[i]) {
                charPositions[s1[i]].push_back(i);
            }
            if (s1[targetIndex] != s2[targetIndex]) {
                charPositions[s1[targetIndex]].push_back(targetIndex);
            }
            
            swaps++;
        }
    }
    return swaps;
}

int main() {
    string s1, s2;
    cin >> s1 >> s2;
    cout << minSwaps(s1, s2) << endl;
    return 0;
}

import java.util.Scanner;
import java.util.HashMap;
import java.util.ArrayList;
import java.util.List;

public class Main {
    public static int minSwaps(String s1, String s2) {
        HashMap<Character, List<Integer>> charPositions = new HashMap<>();
        for (int i = 0; i < s2.length(); i++) {
            char c = s2.charAt(i);
            if (!charPositions.containsKey(c)) {
                charPositions.put(c, new ArrayList<>());
            }
            charPositions.get(c).add(i);
        }
        
        int swaps = 0;
        char[] s1Arr = s1.toCharArray();
        
        for (int i = 0; i < s1Arr.length; i++) {
            if (s1Arr[i] != s2.charAt(i)) {
                char targetChar = s2.charAt(i);
                List<Integer> positions = charPositions.get(targetChar);
                int targetIndex = positions.remove(positions.size() - 1);
                
                // Swap s1Arr[i] and s1Arr[targetIndex]
                char temp = s1Arr[i];
                s1Arr[i] = s1Arr[targetIndex];
                s1Arr[targetIndex] = temp;
                
                // Update positions for the swapped characters
                if (s1Arr[i] != s2.charAt(i)) {
                    if (!charPositions.containsKey(s1Arr[i])) {
                        charPositions.put(s1Arr[i], new ArrayList<>());
                    }
                    charPositions.get(s1Arr[i]).add(i);
                }
                if (s1Arr[targetIndex] != s2.charAt(targetIndex)) {
                    if (!charPositions.containsKey(s1Arr[targetIndex])) {
                        charPositions.put(s1Arr[targetIndex], new ArrayList<>());
                    }
                    charPositions.get(s1Arr[targetIndex]).add(targetIndex);
                }
                
                swaps++;
            }
        }
        return swaps;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String s1 = scanner.next();
        String s2 = scanner.next();
        System.out.println(minSwaps(s1, s2));
        scanner.close();
    }
}

def min_swaps(s1, s2):
    from collections import defaultdict
    
    char_positions = defaultdict(list)
    for i, char in enumerate(s2):
        char_positions[char].append(i)
    
    swaps = 0
    s1_list = list(s1)
    
    for i in range(len(s1_list)):
        if s1_list[i] != s2[i]:
            target_char = s2[i]
            target_index = char_positions[target_char].pop()
            
            # Swap s1_list[i] and s1_list[target_index]
            s1_list[i], s1_list[target_index] = s1_list[target_index], s1_list[i]
            
            # Update positions for the swapped characters
            if s1_list[i] != s2[i]:
                char_positions[s1_list[i]].append(i)
            if s1_list[target_index] != s2[target_index]:
                char_positions[s1_list[target_index]].append(target_index)
            
            swaps += 1
    
    return swaps

if __name__ == "__main__":
    s1 = input().strip()
    s2 = input().strip()
    print(min_swaps(s1, s2))

function minSwaps(s1, s2) {
    const charPositions = {};
    for (let i = 0; i < s2.length; i++) {
        if (!charPositions[s2[i]]) {
            charPositions[s2[i]] = [];
        }
        charPositions[s2[i]].push(i);
    }
    
    let swaps = 0;
    let s1Arr = s1.split('');
    
    for (let i = 0; i < s1Arr.length; i++) {
        if (s1Arr[i] !== s2[i]) {
            const targetChar = s2[i];
            const targetIndex = charPositions[targetChar].pop();
            
            // Swap s1Arr[i] and s1Arr[targetIndex]
            const temp = s1Arr[i];
            s1Arr[i] = s1Arr[targetIndex];
            s1Arr[targetIndex] = temp;
            
            // Update positions for the swapped characters
            if (s1Arr[i] !== s2[i]) {
                if (!charPositions[s1Arr[i]]) {
                    charPositions[s1Arr[i]] = [];
                }
                charPositions[s1Arr[i]].push(i);
            }
            if (s1Arr[targetIndex] !== s2[targetIndex]) {
                if (!charPositions[s1Arr[targetIndex]]) {
                    charPositions[s1Arr[targetIndex]] = [];
                }
                charPositions[s1Arr[targetIndex]].push(targetIndex);
            }
            
            swaps++;
        }
    }
    return swaps;
}

// Example usage
// const s1 = "ab";
// const s2 = "ba";
// console.log(minSwaps(s1, s2));

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