HackerRank Manasa and Stones problem solution

In this HackerRank Manasa and Stones problem you need to Compute all possible numbers that might occur on the last stone given a starting stone with a 0 on it, the number of additional stones found, and the possible differences between consecutive stones. Order the list ascending.

Problem solution in Python programming.

#!/usr/bin/env python

import sys


if __name__ == '__main__':
    T = int(sys.stdin.readline())
    
    for _ in range(T):
        n = int(sys.stdin.readline())
        a = int(sys.stdin.readline())
        b = int(sys.stdin.readline())
        
        print(*sorted(set(x * a + (n - 1 - x) * b for x in range(n))))

Problem solution in Java Programming.

import java.io.*;
import java.util.*;

public class Solution {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        
        int numberOfTestCases = scan.nextInt();
        for (int i = 0; i < numberOfTestCases; ++i) {
            int numberOfStones = scan.nextInt() - 1;
            int a = scan.nextInt();
            int b = scan.nextInt();
            
            if (numberOfStones <= 0) {
                System.out.println(0);
            }
            else {
                Vector<Integer> results = new Vector<Integer>();
                for (int j = 0; j <= numberOfStones; ++j) {
                    results.add((numberOfStones - j) * a + j * b);
                }

                Collections.sort(results);
                System.out.print(results.elementAt(0));
                for (int j = 1; j < results.size(); ++j) {
                    if (results.elementAt(j).equals(results.elementAt(j - 1))) { continue; }
                    else { System.out.print(" " + results.elementAt(j)); }
                }
                System.out.println();
            }
        }
    }
}

Problem solution in C++ programming.

#include <iostream>
#include <set>

using namespace std;

int main() {
  int nTests = 0; cin >> nTests;
  while (nTests--) {
    int n = 0; int a = 0; int b = 0;
    cin >> n >> a >> b;
    set<int> s;
    for (int i = 0; i <= n - 1; ++i) {
      s.insert(i * a + (n - 1 - i) * b);
    }
    for (set<int>::iterator it = s.begin(); it != s.end(); ++it) {
      if (it != s.begin()) cout << " ";
      cout << *it;
    }
    cout << "n";
  }

  return 0;
}

Problem solution in C programming.

#include <stdio.h>
#include <stdlib.h>

typedef struct path {
    long int numSteps;
    long int a;
    long int b;
} path;


int compare (const void * a, const void * b)
{
  return ( *(int*)a - *(int*)b );
}

int main()
{
    int numCases, i, j;
    path p;
    scanf("%d", &numCases);
    
    for (i = 0; i < numCases; i++) {
        long int firstStone = 0;
        
        scanf("%ld", &p.numSteps);
        scanf("%ld", &p.a);
        scanf("%ld", &p.b);
        
        int possibleLastStones[p.numSteps];
        if(p.a == p.b) {
            printf("%dn", p.a*(p.numSteps-1));
        }
        else {
            for(j = 0; j < p.numSteps; j++) {
            possibleLastStones[j] = firstStone + j*p.b + (p.numSteps - 1 - j)*p.a;
            } 
            //sort possibleLastStones array
            qsort(possibleLastStones, p.numSteps, sizeof(int), compare);
            //print out sorted array
            for(j = 0; j < p.numSteps; j++) {
               printf("%d ", possibleLastStones[j]);
            }
            printf("n");    
       }
    }
    return 0;
}

Problem solution in JavaScript programming.

'use strict';


function processData(input) {
var parse_fun = function (s) { return parseInt(s, 10); };
var answer = '';
var lines = input.split('n');
lines.splice(0,1);
while(lines.length > 2){
    var stones = parseInt(lines[0]);
    var a = Math.min(parseInt(lines[1]), parseInt(lines[2]));
    var b = Math.max(parseInt(lines[2]), parseInt(lines[1]));
    var search = {};
    var paths = {level1:[0]};
    for(var i = 2;i<=stones;i++)
    {
        paths['level'+i] = [];
        for(var j = 0; j < paths['level'+(i-1)].length;j++){
            if(paths['level'+i].indexOf(parseInt(paths['level'+(i-1)][j])+a) == -1)
                paths['level'+i].push(parseInt(paths['level'+(i-1)][j])+a);
            if(paths['level'+i].indexOf(parseInt(paths['level'+(i-1)][j])+b) == -1)
                paths['level'+i].push(parseInt(paths['level'+(i-1)][j])+b);     
        }
    }
    answer += paths['level'+stones].join(" ")+"n";
    lines.splice(0,3);
}
process.stdout.write(answer);
}

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