Skip to content
Programmingoneonone
Programmingoneonone

Learn everything about programming

  • Home
  • CS Subjects
    • Internet of Things (IoT)
    • Digital Communication
    • Human Values
    • Cybersecurity
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
  • HackerRank Solutions
    • HackerRank Algorithms Solutions
    • HackerRank C problems solutions
    • HackerRank C++ problems solutions
    • HackerRank Java problems solutions
    • HackerRank Python problems solutions
Programmingoneonone
Programmingoneonone

Learn everything about programming

HackerRank Abbreviation problem solution

YASH PAL, 31 July 202410 September 2024

In this HackerRank Abbreviation Interview preparation kit problem you need to complete the function abbreviation.

HackerRank Abbreviation Interview preparation kit solution

Problem solution in Python programming.

import sys

q = int(input().strip())
for i in range(q):
    a = input().strip()
    b = input().strip()
    #store all possibilities
    bpos = {}
    for i in range(len(b)):
        bpos[b[i]] = (bpos[b[i]] | set([i])) if b[i] in bpos else set([i])
    possibilities = set([0])
    for i in range(len(a)):
        if a[i].upper() in bpos:
            intersection = bpos[a[i].upper()] & possibilities
            advancement = set([i + 1 for i in intersection])
        else:
            advancement = set([])
        if a[i].upper() == a[i]:#capitals must follow the intersection
            possibilities = advancement
        else:
            possibilities = possibilities | advancement
    print("YES" if (len(b)) in possibilities else "NO")

Problem solution in Java Programming.

import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;

public class Solution {

    // Complete the abbreviation function below.
    static String abbreviation(String a, String b) {
        boolean[][] isValid = new boolean[a.length()+1][b.length()+1];
        isValid[0][0] = true;

        for (int i= 1; i <= a.length(); i++) {
            if (Character.isUpperCase(a.charAt(i - 1))) {
                isValid[i][0] = false;
            }
            else isValid[i][0] = true;
        }
        // tabulation from start of string
        for (int i = 1; i <= a.length(); i++) {
            for (int j = 1; j <= b.length(); j++) {
                if (a.charAt(i-1) == b.charAt(j-1)) {
                    isValid[i][j] = isValid[i-1][j-1];
                }else if (Character.toUpperCase(a.charAt(i-1))  ==  b.charAt(j-1)) {
                    isValid[i][j] = isValid[i-1][j-1] || isValid[i-1][j];
                }else if (Character.isUpperCase(a.charAt(i-1))) {
                    isValid[i][j] = false;
                }else {
                    isValid[i][j] = isValid[i-1][j];
                }
            }
        }
        return isValid[a.length()][b.length()]? "YES" : "NO";


    }

    private static final Scanner scanner = new Scanner(System.in);

    public static void main(String[] args) throws IOException {
        BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));

        int q = scanner.nextInt();
        scanner.skip("(rn|[nru2028u2029u0085])?");

        for (int qItr = 0; qItr < q; qItr++) {
            String a = scanner.nextLine();

            String b = scanner.nextLine();

            String result = abbreviation(a, b);

            bufferedWriter.write(result);
            bufferedWriter.newLine();
        }

        bufferedWriter.close();

        scanner.close();
    }
}

Problem solution in C++ programming.

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <ctime>
#include <deque>
#include <unordered_set>
using namespace std;

char A[2000], B[2000];
int Q, n, m;
bool ok[1100][1100];

int main() {
	scanf("%d", &Q);
	while (Q--) {
		scanf("%s%s", A + 1, B + 1);
		n = strlen(A + 1);
		m = strlen(B + 1);
		memset(ok, false, sizeof ok);
		ok[0][0] = true;
		for (int i = 0; i <= n; i++)
			for (int j = 0; j <= m; j++)
	if (ok[i][j]) {
		if ('a' <= A[i + 1] && A[i + 1] <= 'z')
			ok[i + 1][j] = true;
			if (A[i + 1] == B[j + 1])
			ok[i + 1][j + 1] = true;
	if ('a' <= A[i + 1] && A[i + 1] <= 'z' && A[i + 1] - 'a' + 'A' == B[j + 1])
			ok[i + 1][j + 1] = true;
				}
		if (ok[n][m])
			printf("YESn");
		else
			printf("NOn");
	}
}

Problem solution in C programming.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <ctype.h>

int main() {
    char a[1000];
    char b[1000];
    int q,n;
    scanf("%d",&q);
    for (int i = 0; i < q; i++) {
        scanf("%s",a);
        scanf("%s",b);

        int curr = 0;
        n = strlen(a);
        for (int j = 0; j < n; j++) {
            //if (b[curr] == toupper(a[j]) || b[curr] == a[j]) {
            if (b[curr] == a[j]) {  
                curr++;
            }
            else {
                if (isupper(a[j])) {
                    curr = 0;
                    break;
                }
            }
        }
        if (curr == strlen(b)) {
            printf("YESn");
        }
        else {
            for (int j = 0; j < n; j++) {
                if (b[curr] == toupper(a[j]) || b[curr] == a[j]) {
                //if (b[curr] == a[j]) {  
                    curr++;
                }
                else {
                    if (isupper(a[j])) {
                        curr = 0;
                        break;
                    }
                }
            }
            if (curr == strlen(b)) {
                printf("YESn");
            }
            else 
                printf("NOn");
        }
    }
    
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    return 0;
}

Problem solution in JavaScript programming.

function processData(input) {
    var lines = input.split("n");

    function readLine(){
        var n = 0;
        this.nxLn = function(){
            return n++;
        }
    }

    var rd = new readLine();

    var q = parseInt(lines[rd.nxLn()]);

    var isLowerCase = function(value){
        if(value.charCodeAt(0)>96){
            return true;
        } else {
            return false;	
        }
    }

    var containsUpperCase = function(value){
        for(var i=0;i<value.length;i++){
            if(!isLowerCase(value.substr(i,1))){
                return true;
            }
        }
        return false;
    }

    while (q!==0) {
        var a = lines[rd.nxLn()];
        var b = lines[rd.nxLn()];

        var firstCondition = true;

        b.split("").forEach(val=>{
            var ind = a.indexOf(val);
            if(ind==-1){
                ind = a.indexOf(val.toLowerCase());
            }
            if(ind==-1){
                firstCondition = false;
            }
            if(containsUpperCase(a.substr(0,ind))){
                firstCondition = false;
            }
            a = a.substr(ind+1);
        });

        if(containsUpperCase(a)){
            firstCondition = false;
        }

        if(firstCondition){
            console.log("YES");
        } else {
            console.log("NO");
        }

        q--;
    }
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});

coding problems solutions interview prepration kit

Post navigation

Previous post
Next post

Are you a student and stuck with your career or worried about real-time things, and don't know how to manage your learning phase? Which profession to choose? and how to learn new things according to your goal, and land a dream job. Then this might help to you.

Hi My name is YASH PAL, founder of this Blog and a Senior Software engineer with 5+ years of Industry experience. I personally helped 40+ students to make a clear goal in their professional lives. Just book a one-on-one personal call with me for 30 minutes for 300 Rupees. Ask all your doubts and questions related to your career to set a clear roadmap for your professional life.

Book session - https://wa.me/qr/JQ2LAS7AASE2M1

Pages

  • About US
  • Contact US
  • Privacy Policy

Follow US

  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2026 Programmingoneonone | WordPress Theme by SuperbThemes