TCS Coding Question 2 | Given a maximum of four digit to the base 17 (10 – A …

July 17, 2021

Sweet Seventeen Problem Statement

Given a maximum of four digit to the base 17 (10 – A, 11 – B, 12 – C, 13 – D … 16 – G} as input, output its decimal value.

Test Cases

Case 1

  • Input – 1A
  • Expected Output – 27

Case 2

  • Input – 23GF
  • Expected Output – 10980
Given a maximum of four digit to the base 17

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18 comments on “TCS Coding Question 2 | Given a maximum of four digit to the base 17 (10 – A …”


  • Gourab

    #include
    #include
    int main(){
    char s[20];
    scanf(“%s”,s);
    int len,i,j,sum,c;
    for(len=0;s[len]!=’\0′;len++);

    for(i=len-1,j=0;i>=0;i–,j++){
    c = (int)(s[i]);
    // printf(“%d “,c);
    if(c>48 && c <58){
    c = c -48;
    sum += c*pow(17,j);
    }
    else{
    if(s[i] == 'A') {
    c= 10;
    }
    if(s[i] == 'B') {
    c= 11;
    }
    if(s[i] == 'C') {
    c= 12;
    }
    if(s[i] == 'D') {
    c= 13;
    }
    if(s[i] == 'E') {
    c= 14;
    }
    if(s[i] == 'F') {
    c= 15;
    }
    if(s[i] == 'G') {
    c= 16;
    }
    if(s[i] == 'H') {
    c= 17;
    }
    sum += c*pow(17,j);
    }
    // printf("sum: %d c: %d\n", sum,c);
    }
    printf("%d",sum);
    return 0;
    }


  • Debparna

    import java.util.*;
    import java.lang.Math;
    public class Main
    {
    public static void main(String[] args) {
    //System.out.println((int)’a’);
    Scanner sc = new Scanner(System.in);
    String str = sc.nextLine();
    int len = str.length();
    int sum=0;
    for(int i=len-1;i>=0;i–){
    if((int)str.charAt(i)>=65 && (int)str.charAt(i)=97 && (int)str.charAt(i)<=103){
    sum = sum + (int)Math.pow(17,len-i-1)*((int)str.charAt(i) – 87);
    //System.out.println(sum);
    }else{
    sum = sum + (int)Math.pow(17,len-i-1)*((int)str.charAt(i) – '0');
    //System.out.println(sum);
    }
    }

    System.out.println(sum);
    }
    }


  • Shahla

    #include
    using namespace std;

    int main()
    {
    string s;
    cin>>s;

    int mul=1,res=0;
    for(int i=s.size()-1 ; i>=0;i–)
    {
    if(s[i]>=’A’ &&s[i]=’0′ &&s[i]<='9')
    res += (s[i]-'0')*mul;

    mul*=17;

    }

    cout<<res;

    return 0;
    }


  • rohan

    #include
    #include
    #include
    int main()
    {
    int i=0,val,len; char hex[20];
    long decimal=0,place=1;
    scanf(“%s”,hex);
    len=strlen(hex)-1;
    for(i=0;hex[i]!=’\0′;i++)
    {
    if(hex[i]>=’0’&& hex[i]=’a’&& hex[i]<='z')
    val=hex[i]-97+10;
    else
    val=hex[i]-65+10;
    decimal=decimal+val*pow(17,len);
    len–;
    }
    printf("%ld",decimal);
    return 0;
    }


  • tejas

    #taking Raw Input
    userInput = raw_input()
    #saving in list so that we get character by character
    num = list(userInput)
    #creating a dictionary and mapping
    thisDict = {
    “A”:10,
    “B”:11,
    “C”:12,
    “D”:13,
    “E”:14,
    “F”:15,
    “G”:16,
    “H”:17,
    “I”:18,
    “J”:19,
    “K”:20,
    #so on…
    }

    power = len(num)-1
    final = 0

    for i in num:
    if i in thisDict:
    demo=thisDict[i] * 17**power
    final = demo + final
    power = power-1
    else:
    demo=int(i) * 17**power
    final = demo + final
    power = power-1

    print(final)
    ——————————————–
    add lowercase alphabeths to urs dictionary too.


  • Harry

    import java.io.*;
    class A{
    public static void main(String args[])throws Exception{
    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    String s=br.readLine();
    System.out.println(Integer.parseInt(s,17));
    }
    }
    This much code is enough for this question.


  • Adesh

    #include

    using namespace std;

    int main() {
    string s; cin>>s;
    int l = s.size(); int tl=l;
    //cout<<l<<endl;
    int num=0;
    for(int i=0; i<tl; i++){

    if(s[i]=='0' || s[i]=='1' || s[i]=='2' || s[i]=='3' || s[i]=='4' || s[i]=='5' || s[i]=='6' || s[i]=='7' || s[i]=='8' || s[i]=='9'){
    num = num + ((int)s[i]-48)*pow(17,l-1) ;

    }
    else{
    char ts = s[i]; int p=0;
    if(ts=='A') p=10;
    else if(ts=='B') p=11;
    else if(ts=='C') p=12;
    else if(ts=='D') p=13;
    else if(ts=='E') p=14;
    else if(ts=='F') p=15;
    else if(ts=='G') p=16;

    num=num+p*pow(17,l-1);
    }
    l–;
    }
    cout<<num;
    }


  • Himanshu

    class Question2 {
    public static void main(String[] args) {
    String number = “1G”; // Number
    System.out.println(Integer.valueOf(number, 17)); //17 is base
    }
    }


  • Amrutha

    #include
    #include
    #include
    int main()
    {
    int l,t,val,d=0,n=0,i;
    char s[32];
    scanf(“%s”,s);
    l=strlen(s)-1;
    for(i=l;i>=0;i–)
    {
    t=s[i];
    if(t>=48&&t=97&&t=65&&t<=90)
    {
    val=t-55;
    }
    d=d+val*pow(17,n);
    n++;
    }
    printf("%d",d);
    }


  • RAUSHAN

    for(int i=s.length()-1;i>=0;i–)
    {
    if((s.charAt(i)>=’A’&&s.charAt(i)=’a’ &&s.charAt(i)<='z'))
    {
    num = num + hmap.get(s.charAt(i))*(int)Math.pow(17,k++);
    }
    else
    {
    num = num+((s.charAt(i)-'0')*(int)Math.pow(17,k++));
    }
    can any one explian how is it working

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