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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 interview prepration kit

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