HackerEarth Lexicographically minimal string problem solution

YASH PAL, 31 July 202414 February 2026

In this HackerEarth Lexicographically minimal string problem solution You are given three strings named as A, B, and C. Here, the length of the strings A and B is equal. All strings contain lowercase English letters. The string A and B contain the following properties:

The characters located at the same indexes in the string A and B are equivalent to each other.
If character a is equivalent to character b, then the character b is also equivalent to the character .
If character a is equivalent to character b and character b is equivalent to character c, then character a is also equivalent to character c.
Every character is equivalent to itself.

HackerEarh Lexicographically minimal string problem solution

HackerEarth Lexicographically minimal string problem solution.

import java.io.IOException;
import java.util.InputMismatchException;
import java.util.Scanner;

public class hacker
{
    static int parent[];
    static int size[];

    public static int find(int i)
    {
        if (parent[i] != i)
        {
            return find(parent[i]);
        }

        return i;
    }

    public static void union(int a, int b)
    {
        int x = find(a);
        int y = find(b);

        if (x != y)
        {
            if (size[x] > size[y])
            {
                size[x] += size[y];
                parent[y] = x;
            } else
            {
                size[y] += size[x];
                parent[x] = y;
            }
        }
    }

    public String solution(String a, String c, String b)
    {
        parent = new int[26];
        size = new int[26];

        for (int i = 0; i < 26; i++)
        {
            parent[i] = i;
        }

        for (int i = 0; i < a.length(); i++)
        {
            int x = (int) (a.charAt(i) - 'a');
            int y = (int) (b.charAt(i) - 'a');
            union(x, y);
        }

        String ans = "";

        for (int i = 0; i < c.length(); i++)
        {
            int f = find((int) (c.charAt(i) - 'a'));

            for (int j = 0; j < 26; j++)
            {
                if (find(j) == f)
                {
                    ans += (char) (j + 'a');
                    break;
                }
            }

        }

        return ans;

    }

    void solve()
    {
        Scanner in = new Scanner(System.in);

        String a = in.nextLine();
        String b = in.nextLine();
        String c = in.nextLine();

        System.out.println(solution(a, c, b));
    }

    void run()
    {

        solve();

    }

    public static void main(String[] args) throws NumberFormatException, InputMismatchException, IOException
    {

        new hacker().run();
    }

}

Second solution

#include<bits/stdc++.h>
using namespace std;
char par[150];
char find(char x)
{
    if(par[x]==x)return par[x];
    return par[x]=find(par[x]);
}
void unionset(char x,char y)
{
    char parx=find(x),pary=find(y);
    if(parx==pary)return;
    if(parx<pary)
        par[pary]=parx;
    else par[parx]=pary;
}
int main()
{
    string a,b,c;
    cin>>a>>b>>c;
    assert(a.size()==b.size());
    for(char i='a';i<='z';i++)
    {
        par[i]=i;
    }
    for(int i=0;i<a.size();i++)
    {
        unionset(a[i],b[i]);
    }
    for(int i=0;i<c.size();i++)
    {
        cout<<find(c[i]);
    }
    cout<<"n";
    return 0;
}

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