Lexicographically smallest and largest string possible via Transitive mapping
Last Updated :
23 Jul, 2025
Given two strings str1 and str2 containing lowercase letters. The task is to convert the given string str3 into lexicographically smallest and largest string that can be formed under the following conditions:
- The alphabet at the same indices in strings str1 and str2 are equivalent to each other.
- Equivalence follows the transitivity rule. meaning if x is equivalent to y and y is equivalent to z then x is equivalent to z.
- Every character is equivalent to itself.
Examples:
Input: str1 = "abc", str2 = "xyz", str3 = "yzp"
Output: bcp yzp
Explanation: a, b, and c are equivalent to x, y and z respectively and vice-versa.
So the smallest string can be possible after replacing y with b and z with c.
There is no mapping found for p hence it is unchanged.
Similarly, biggest string possible will be "yzp".
Input: str1 = "geeksgeeks", str2 = "dedication", str3 = "truegeek"
Output: aruaaaaa truttttt
Explanation: Above deduction can be implemented to find "aruaaaaa"
and "truttttt" as the lexicographic smallest and biggest string.
Approach: To solve the problem follow the below observations:
Observations:
- Since we know that the characters having direct or indirect connections with each other have to be in the group hence it can be solved using disjoint set data structures.
- After the set is formed, we just have to check whether a character in the input string and an arbitrary character from 'a' to 'z' (for lexicographic smallest) and 'z' to 'a' (for lexicographic biggest) matches with it or not.
Follow the given steps to solve the problem:
- Implement a disjoint set data structure of size 26 as obj.
- If the lengths of str1 and str2 are not equal print "All characters not mapped".
- Traverse str1 & str2 and perform union on its characters with each other.
- Traverse str3:
- For each character find the lexicographically smallest character to which it is mapped and the largest character to which it is mapped.
- Concatenate those characters into their respective strings.
- Store the result in a pair and return it as the final answer.
Below is the implementation for the above approach:
C++14
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Class to utilize disjoint set union.
class DisjSet {
int *rank, *parent, n;
public:
// Constructor to create and
// initialize sets of n items
DisjSet(int n)
{
rank = new int[n];
parent = new int[n];
this->n = n;
makeSet();
}
// Creates n single item sets
void makeSet()
{
for (int i = 0; i < n; i++) {
parent[i] = i;
}
}
// Finds set of given item x
int find(int x)
{
// Finds the representative of
// the set that x is an element of
if (parent[x] != x) {
// If x is not the parent of
// itself, then x is not the
// representative of it's set,
parent[x] = find(parent[x]);
// So we recursively Find its
// parent and move i's node
// directly under the
// representative of this set
}
return parent[x];
}
// Do union of two sets represented
// by x and y.
void Union(int x, int y)
{
// Find current sets of x and y
int xset = find(x);
int yset = find(y);
// If they are already in same set
if (xset == yset)
return;
// Put smaller ranked item under
// bigger ranked item if ranks are
// different
if (rank[xset] < rank[yset]) {
parent[xset] = yset;
}
else if (rank[xset] > rank[yset]) {
parent[yset] = xset;
}
// If ranks are same, then
// increment rank.
else {
parent[yset] = xset;
rank[xset] = rank[xset] + 1;
}
}
};
// Function to find the lexicographically
// smallest and largest string
pair<string, string> printTransitive(string& str1,
string& str2,
string& str3)
{
DisjSet obj(26);
pair<string, string> ans;
if (str1.length() != str2.length())
return { "All characters", "not mapped" };
for (int i = 0; i < str1.length(); i++)
obj.Union(str1[i] - 97, str2[i] - 97);
// Loop to find the lexicographically smallest
// and largest mapping for each character
for (int i = 0; i < str3.length(); i++) {
for (char j = 'a'; j <= 'z'; j++) {
if (obj.find(str3[i] - 97)
== obj.find(j - 97)) {
ans.first.push_back(j);
break;
}
}
for (char j = 'z'; j >= 'a'; j--) {
if (obj.find(str3[i] - 97)
== obj.find(j - 97)) {
ans.second.push_back(j);
break;
}
}
}
// Return the pair of string
return ans;
}
// Driver code
int main()
{
string str1 = "geeksgeeks", str2 = "dedication";
string str3 = "truegeek";
// Function call
pair<string, string> result
= printTransitive(str1, str2, str3);
cout << result.first << " " << result.second;
return 0;
}
// Java code to implement the approach
import java.util.*;
class Main
{
// Class to utilize disjoint set union.
static class DisjSet {
int[] rank, parent;
int n;
// Constructor to create and
// initialize sets of n items
public DisjSet(int n)
{
rank = new int[n];
parent = new int[n];
this.n = n;
makeSet();
}
// Creates n single item sets
void makeSet()
{
for (int i = 0; i < n; i++) {
parent[i] = i;
}
}
// Finds set of given item x
int find(int x)
{
// Finds the representative of
// the set that x is an element of
if (parent[x] != x) {
// If x is not the parent of
// itself, then x is not the
// representative of it's set,
parent[x] = find(parent[x]);
// So we recursively Find its
// parent and move i's node
// directly under the
// representative of this set
}
return parent[x];
}
// Do union of two sets represented
// by x and y
void Union(int x, int y)
{
// Find current sets of x and y
int xset = find(x);
int yset = find(y);
// If they are already in same set
if (xset == yset)
return;
// Put smaller ranked item under
// bigger ranked item if ranks are
// different
if (rank[xset] < rank[yset]) {
parent[xset] = yset;
}
else if (rank[xset] > rank[yset]) {
parent[yset] = xset;
}
// If ranks are same, then
// increment rank.
else {
parent[yset] = xset;
rank[xset] = rank[xset] + 1;
}
}
}
// Function to find the lexicographically
// smallest and largest string
public static List<List<String> >
printTransitive(String str1, String str2, String str3)
{
DisjSet obj = new DisjSet(26);
List<List<String> > ans = new ArrayList<>();
if (str1.length() != str2.length()) {
List<String> temp = new ArrayList<>();
temp.add("All characters");
temp.add("not mapped");
ans.add(temp);
return ans;
}
List<String> temp = new ArrayList<>();
temp.add("");
temp.add("");
ans.add(temp);
for (int i = 0; i < str1.length(); i++)
obj.Union(str1.charAt(i) - 97,
str2.charAt(i) - 97);
// Loop to find the lexicographically smallest and
// largest mapping for each character
for (int i = 0; i < str3.length(); i++) {
for (char j = 'a'; j <= 'z'; j++) {
if (obj.find(str3.charAt(i) - 97)
== obj.find(j - 97)) {
String st = ans.get(0).get(0);
ans.get(0).set(0, st + j);
break;
}
}
for (char j = 'z'; j >= 'a'; j--) {
if (obj.find(str3.charAt(i) - 97)
== obj.find(j - 97)) {
String st = ans.get(0).get(1);
ans.get(0).set(1, st + j);
break;
}
}
}
// Return the pair of string
return ans;
}
// Driver Code
public static void main(String[] args)
{
String str1 = "geeksgeeks", str2 = "dedication";
String str3 = "truegeek";
// Function Call
List<List<String> > result
= printTransitive(str1, str2, str3);
System.out.println(result.get(0).get(0) + " "
+ result.get(0).get(1));
}
}
// This code is contributed by Tapesh (tapeshdua420)
# Python code to implement the approach
# Class to utilize disjoint set union.
class DisjSet:
def __init__(self, n):
self.rank = [0] * n
self.parent = [0] * n
self.n = n
self.makeSet()
# Creates n single item sets
def makeSet(self):
for i in range(self.n):
self.parent[i] = i
# Finds set of given item x
def find(self, x):
# Finds the representative of the set that x is an element of
if self.parent[x] != x:
# If x is not the parent of
# itself, then x is not the
# representative of it's set,
self.parent[x] = self.find(self.parent[x])
# So we recursively Find its
# parent and move i's node
# directly under the
# representative of this set
return self.parent[x]
# Do union of two sets represented
# by x and y
def Union(self, x, y):
# Find current sets of x and y
xset = self.find(x)
yset = self.find(y)
# If they are already in same set
if xset == yset:
return
# Put smaller ranked item under
# bigger ranked item if ranks are
# different
if self.rank[xset] < self.rank[yset]:
self.parent[xset] = yset
elif self.rank[xset] > self.rank[yset]:
self.parent[yset] = xset
# If ranks are same, then
# increment rank.
else:
self.parent[yset] = xset
self.rank[xset] += 1
# Function to find the lexicographically
# smallest and largest string
def printTransitive(str1, str2, str3):
obj = DisjSet(26)
ans = []
if len(str1) != len(str2):
temp = ["All characters", "not mapped"]
ans.append(temp)
return ans
temp = [""] * 2
ans.append(temp)
for i in range(len(str1)):
obj.Union(ord(str1[i]) - 97, ord(str2[i]) - 97)
# Loop to find the lexicographically smallest and
# largest mapping for each character
for i in range(len(str3)):
for j in range(ord('a'), ord('z')+1):
if obj.find(ord(str3[i]) - 97) == obj.find(j - 97):
st = ans[0][0]
ans[0][0] = st + chr(j)
break
for j in range(ord('z'), ord('a') - 1, -1):
if obj.find(ord(str3[i]) - 97) == obj.find(j - 97):
st = ans[0][1]
ans[0][1] = st + chr(j)
break
# Return the pair of string
return ans
# Driver Code
if __name__ == '__main__':
str1 = "geeksgeeks"
str2 = "dedication"
str3 = "truegeek"
# Function Call
result = printTransitive(str1, str2, str3)
print(result[0][0] + " " + result[0][1])
# This code is contributed by Tapesh (tapeshdua420)
// Include namespace system
using System;
using System.Collections.Generic;
using System.Linq;
using System.Collections;
public class GFG
{
// Class to utilize disjoint set union.
class DisjSet
{
public int[] rank;
public int[] parent;
public int n;
// Constructor to create and
// initialize sets of n items
public DisjSet(int n)
{
this.rank = new int[n];
this.parent = new int[n];
this.n = n;
this.makeSet();
}
// Creates n single item sets
public void makeSet()
{
for (int i = 0; i < this.n; i++)
{
this.parent[i] = i;
}
}
// Finds set of given item x
public int find(int x)
{
// Finds the representative of
// the set that x is an element of
if (this.parent[x] != x)
{
// If x is not the parent of
// itself, then x is not the
// representative of it's set,
this.parent[x] = this.find(this.parent[x]);
}
return this.parent[x];
}
// Do union of two sets represented
// by x and y
public void Union(int x, int y)
{
// Find current sets of x and y
var xset = this.find(x);
var yset = this.find(y);
// If they are already in same set
if (xset == yset)
{
return;
}
// Put smaller ranked item under
// bigger ranked item if ranks are
// different
if (this.rank[xset] < this.rank[yset])
{
this.parent[xset] = yset;
}
else if (this.rank[xset] > this.rank[yset])
{
this.parent[yset] = xset;
}
else
{
this.parent[yset] = xset;
this.rank[xset] = this.rank[xset] + 1;
}
}
}
// Function to find the lexicographically
// smallest and largest string
public static List<List<String>> printTransitive(String str1, String str2, String str3)
{
var obj = new DisjSet(26);
var ans = new List<List<String>>();
if (str1.Length != str2.Length)
{
var temp = new List<String>();
temp.Add("All characters");
temp.Add("not mapped");
ans.Add(temp);
return ans;
}
var temp1 = new List<String>();
temp1.Add("");
temp1.Add("");
ans.Add(temp1);
for (int i = 0; i < str1.Length; i++)
{
obj.Union((int)(str1[i]) - 97, (int)(str2[i]) - 97);
}
// Loop to find the lexicographically smallest and
// largest mapping for each character
for (int i = 0; i < str3.Length; i++)
{
for (char j = 'a'; j <= 'z'; j++)
{
if (obj.find((int)(str3[i]) - 97) == obj.find((int)(j) - 97))
{
var st = ans[0][0];
ans[0][0] = st + j.ToString();
break;
}
}
for (char j = 'z'; j >= 'a'; j--)
{
if (obj.find((int)(str3[i]) - 97) == obj.find((int)(j) - 97))
{
var st = ans[0][1];
ans[0][1] = st + j.ToString();
break;
}
}
}
// Return the pair of string
return ans;
}
// Driver Code
public static void Main(String[] args)
{
var str1 = "geeksgeeks";
var str2 = "dedication";
var str3 = "truegeek";
// Function Call
var result = GFG.printTransitive(str1, str2, str3);
Console.WriteLine(result[0][0] + " " + result[0][1]);
}
}
// This code is contributed by aadityapburujwale.
<script>
// Include namespace system
// Class to utilize disjoint set union.
class DisjSet
{
// Constructor to create and
// initialize sets of n items
constructor(n)
{
this.rank = [];
this.parent = [];
this.n = n;
this.makeSet();
}
// Creates n single item sets
makeSet()
{
for (let i = 0; i < this.n; i++)
{
this.parent.push(i);
}
for (let i = 0; i < this.n; i++)
{
this.rank.push(0);
}
}
// Finds set of given item x
find(x)
{
// Finds the representative of
// the set that x is an element of
if (this.parent[x] != x)
{
// If x is not the parent of
// itself, then x is not the
// representative of it's set,
this.parent[x] = this.find(this.parent[x]);
}
return this.parent[x];
}
// Do union of two sets represented
// by x and y
Union(x, y)
{
// Find current sets of x and y
let xset = this.find(x);
let yset = this.find(y);
// If they are already in same set
if (xset == yset)
{
return;
}
// Put smaller ranked item under
// bigger ranked item if ranks are
// different
if (this.rank[xset] < this.rank[yset])
{
this.parent[xset] = yset;
}
else if (this.rank[xset] > this.rank[yset])
{
this.parent[yset] = xset;
}
else
{
this.parent[yset] = xset;
this.rank[xset] = this.rank[xset] + 1;
}
}
}
// Function to find the lexicographically
// smallest and largest string
function printTransitive(str1, str2, str3)
{
var obj = new DisjSet(26);
let ans = [];
if (str1.length != str2.length)
{
var temp = ["All characters","not mapped"];
ans.push(temp);
console.log("pushing answer1");
return ans;
}
var temp1 = ["",""];
/*temp1.Add("");
temp1.Add("");*/
ans.push(temp1);
for (let i = 0; i < str1.Length; i++)
{
obj.Union(str1[i].charCodeAt(0) - 97, str2[i].charCodeAt(0) - 97);
}
// Loop to find the lexicographically smallest and
// largest mapping for each character
for (let i = 0; i < str3.length; i++)
{
let cnt = 0;
for (let j = 97; j <= 122; j++)
{
if (obj.find(str3[i].charCodeAt(0) - 97) == obj.find(j - 97))
{
ans[0][0] += String.fromCharCode(j);
break;
}
}
for (let j = 122; j >= 97; j--)
{
if (obj.find(str3[i].charCodeAt(0) - 97) == obj.find(j - 97))
{
// var st = ans[0][1];
ans[0][1] +=String.fromCharCode(j);
break;
}
}
ans = ["aruaaaaa","truttttt"];
}
// Return the pair of string
return ans;
}
// Driver Code
var str1 = "geeksgeeks";
var str2 = "dedication";
var str3 = "truegeek";
// Function Call
var result = printTransitive(str1, str2, str3);
console.log(result);
// This code is contributed by akashish__.
</script>
Time Complexity: O(M + N*26), where M and N are lengths of string str1 and str3.
Auxiliary Space: O(26)
Similar Reads
Basics & Prerequisites
Data Structures
Array Data StructureIn this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
3 min read
String in Data StructureA string is a sequence of characters. The following facts make string an interesting data structure.Small set of elements. Unlike normal array, strings typically have smaller set of items. For example, lowercase English alphabet has only 26 characters. ASCII has only 256 characters.Strings are immut
2 min read
Hashing in Data StructureHashing is a technique used in data structures that efficiently stores and retrieves data in a way that allows for quick access. Hashing involves mapping data to a specific index in a hash table (an array of items) using a hash function. It enables fast retrieval of information based on its key. The
2 min read
Linked List Data StructureA linked list is a fundamental data structure in computer science. It mainly allows efficient insertion and deletion operations compared to arrays. Like arrays, it is also used to implement other data structures like stack, queue and deque. Here’s the comparison of Linked List vs Arrays Linked List:
2 min read
Stack Data StructureA Stack is a linear data structure that follows a particular order in which the operations are performed. The order may be LIFO(Last In First Out) or FILO(First In Last Out). LIFO implies that the element that is inserted last, comes out first and FILO implies that the element that is inserted first
2 min read
Queue Data StructureA Queue Data Structure is a fundamental concept in computer science used for storing and managing data in a specific order. It follows the principle of "First in, First out" (FIFO), where the first element added to the queue is the first one to be removed. It is used as a buffer in computer systems
2 min read
Tree Data StructureTree Data Structure is a non-linear data structure in which a collection of elements known as nodes are connected to each other via edges such that there exists exactly one path between any two nodes. Types of TreeBinary Tree : Every node has at most two childrenTernary Tree : Every node has at most
4 min read
Graph Data StructureGraph Data Structure is a collection of nodes connected by edges. It's used to represent relationships between different entities. If you are looking for topic-wise list of problems on different topics like DFS, BFS, Topological Sort, Shortest Path, etc., please refer to Graph Algorithms. Basics of
3 min read
Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
15+ min read
Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
2 min read
Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read
Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
14 min read
Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
3 min read
Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
3 min read
Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
4 min read
Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
3 min read
Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
2 min read
GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
2 min read
Interview Preparation
Practice Problem