INTERVIEW C QUESTION

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Taxi application Design a Call taxi booking application -There are n number of taxi’s. For simplicity, assume 4. But it should work for any number of taxi’s. -The are 6 points(A,B,C,D,E,F) -All the points are in a straight line, and each point is 15kms away from the adjacent points. -It takes 60 mins to travel from one point to another -Each taxi charges Rs.100 minimum for the first 5 kilometers and Rs.10 for the subsequent kilometers. -For simplicity, time can be entered as absolute time. Eg: 9hrs, 15hrs etc. -All taxi’s are initially stationed at A. -When a customer books a Taxi, a free taxi at that point is allocated -If no free taxi is available at that point, a free taxi at the nearest point is allocated. -If two taxi’s are free at the same point, one with lower earning is allocated -Note that the taxi only charges the customer from the pickup point to the drop point. Not the distance it travels from an adjacent point to pickup the customer. -If no taxi is free a...

Construct a paztriangle up to the given N number. If the triangle is not formed then Print remaining elements  with X. N=23 X X  X 18  20  22 10  12  14 16 0  2  4  6  8 N=18 18 14 16 8 10 12 0 2 4 6 //Java Code import java.util.Scanner; public class Main { static void triangle( int n){ int total= 0 ; int _x= 0 ; int count= 0 ; for ( int i= 1 ;i<n;i++){ total+=i; if (( 2 *total)>n) break ; } String []a= new String[total]; if (n% 2 != 0 ) n-= 1 ; for ( int i= 0 ;i<total;i++){ if (count<=n) { a[i] = count+ "" ; count+= 2 ; } else a[i] = "X" ; } int right= 1 ; while (total!= 0 ){ _x+=right; count=a. length -_x; for ( int i= 1 ;i<=right;i++){ Sy...

There Are Two Arrays in Sorted Form .Task is to Merge the Array and Print in Sorted Order Either in Incresing or Decreasing Order NOTE:Array should be in Sorted order (donot store the array and sort) //JAVA CODE import java.util.Scanner; public class Main { static int sorted( int []a, int [] b, int value){ int min= 0 ; for ( int i= 0 ;i<a. length ;i++) { if (value == 0 && min < a[i]) min = a[i]; else if (min < a[i] && value > a[i]) min = a[i]; } for ( int i= 0 ;i<b. length ;i++) { if (value== 0 && min<b[i]) min = b[i]; else if (min < b[i] && value > b[i]) min=b[i]; } return min; } public static void main(String[] args) { Scanner sc = new Scanner(System. in ); int m=sc.nextInt(); int n=sc.nextInt(); int []a= new int [m]; int []b= ne...

Example: Input: Number of bits: 12 Bits: 1 0 1 1 0 1 1 0 1 1 1 1 Consecutive K: 2 Output: 1 0 1 1 0 0 1 1 0 0 1 1 0 1 1 0 //JAVA CODE import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc= new Scanner(System. in ); int n = sc.nextInt(); int k=sc.nextInt(); int []a = new int [n]; int temp= 0 ; //Inputs for ( int i= 0 ;i<n;i++) a[i]=sc.nextInt(); //print for ( int i= 0 ;i<n;i++){ System. out .print(a[i]+ " " ); if (a[i]== 1 ){ temp++; if (temp==k) { System. out .print( "0 " ); temp= 0 ; } } else temp= 0 ; } } }

Given a fraction. Convert it into a decimal. So simple :P eg. 10/2 = 5 3/5 = 0.6 (The Question Begins Now)  :D If the decimals are repeating recursively, then enclose them inside  (). eg. 8/3 = 2.(6) as 8/3 = 2.66666666.......  infinitly.    Input: The first line of each test case is an integer N denoting the numerator of fraction. The second line of each test case is an integer D denoting the denominator of fraction. Output: Print decimal of that fraction in separate line for each test case. Constraints: 1 ≤ N,D ≤ 2000 Example: Input 4 2 8 3 Output 2 2.(6) //JAVA CODE import java.util.Scanner; public class Main { static String simple_fraction( int a, int b){ String s=a/b+ "" ; if (a%b== 0 ) return s; a%=b; String ad= "" ; int []n= new int [b]; int count= 0 ; while (a> 0 && n[a]== 0 ){ n[a]=++count; a*= 10 ; int temp=a/b; ...

Given two expressions in the form of strings. The task is to compare them and check if they are similar. Expressions consist of lowercase alphabets, '+', '-' and  '( )'. Input: The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. Each test case contains two lines. And each line contains an expression. Output For each test case, print in a new line "YES"  if the expressions are similar else print "NO". Constraints: 1<=T<=100 3<=|Expression length|<=100 Example: Input: 2 -(a+b+c) -a-b-c a-b-(c-d) a-b-c-d Output: YES NO //JAVA CODE import java.util.ArrayList; import java.util.Scanner; public class Main { static boolean isequal(String s1,String s2){ return s1.equals(s2); } static char sign_change( int positive, int negative, int parentheses, char a){ if (negative > 0 && parentheses> 0 ){ if (a== '-' ) ...