Hello Programmers, In this post, you will learn how to solve HackerRank Manasa and Stones Solution. This problem is a part of the HackerRank Algorithms Series.
One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.
HackerRank Manasa and Stones Solution
Task
Manasa is out on a hike with friends. She finds a trail of stones with numbers on them. She starts following the trail and notices that any two consecutive stones‘ numbers differ by one of two values. Legend has it that there is a treasure trove at the end of the trail. If Manasa can guess the value of the last stone, the treasure will be hers.
Example
n = 2
a = 2
b = 3
She finds stones and their differences are a = 2 or b = 3. We know she starts with a 0 stone not included in her count. The permutations of differences for the two stones are [2, 2], [2, 3], [3, 2] or [3, 3]. Looking at each scenario, stones might have [2, 4], [2, 5], [3, 5] or [3, 6] on them. The last stone might have any of 4, 5 or 6 on its face.
Compute all possible numbers that might occur on the last stone given a starting stone with a 0 on it, a number of additional stones found, and the possible differences between consecutive stones. Order the list ascending.
Function Description
Complete the stones function in the editor below.
stones has the following parameter(s):
- int n: the number of non-zero stones
- int a: one possible integer difference
- int b: another possible integer difference
Returns
- int[]: all possible values of the last stone, sorted ascending
Input Format
The first line contains an integer T, the number of test cases.
Each test case contains 3 lines:
– The first line contains n, the number of non-zero stones found.
– The second line contains a, one possible difference
– The third line contains b, the other possible difference.
Constraints
- 1 <= T <= 10
- 1 <= n, a, b <= 103
Sample Input
STDIN Function
—– ——–
2 T = 2 (test cases)
3 n = 3 (test case 1)
1 a = 1
2 b = 2
4 n = 4 (test case 2)
10 a = 10
100 b = 100
Sample Output
2 3 4
30 120 210 300
Explanation
With differences 1 and 2, all possible series for the first test case are given below:
- 0,1,2
- 0,1,3
- 0,2,3
- 0,2,4
Hence the answer 2 3 4.
With differences 10 and 100, all possible series for the second test case are the following:
- 0, 10, 20, 30
- 0, 10, 20, 120
- 0, 10, 110, 120
- 0, 10, 110, 210
- 0, 100, 110, 120
- 0, 100, 110, 210
- 0, 100, 200, 210
- 0, 100, 200, 300
Hence the answer 30 120 210 300.
HackerRank Manasa and Stones Solution
Manasa and Stones Solution in C
#include <stdio.h>
#include <stdint.h>
int main() {
uint16_t T;
scanf("%d", &T);
for(uint16_t count=1; count<=T; count++) {
uint64_t n, a, b, x, base;
scanf("%ld", &n);
scanf("%ld", &a);
scanf("%ld", &b);
if(a>b) {
x=a;
a=b;
b=x;
}
n--;
base=n*a;
if(a!=b) {
while(n>0) {
printf("%ld ", base);
n--;
base+=b-a;
}
}
printf("%ld\n", base);
}
return 0;
}Manasa and Stones Solution in Cpp
#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;
}Manasa and Stones Solution in Java
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
for (int i = 0; i < T; i++) {
int n = in.nextInt() - 1;
int a = in.nextInt();
int b = in.nextInt();
int min = Math.min(a, b);
int max = Math.max(a, b);
int sum = n * min;
Set<Integer> used = new HashSet<Integer>();
for (int j = 0; j < n; j++) {
if (!used.contains(sum)) {
System.out.print(sum + " ");
used.add(sum);
}
sum -= min;
sum += max;
}
if (used.contains(sum))
System.out.println();
else
System.out.println(sum);
}
in.close();
}
}Manasa and Stones Solution in Python
T=int(raw_input())
for _ in range(0,T):
n=int(raw_input())
a=int(raw_input())
b=int(raw_input())
if (a<b): a,b=b,a
if (a==b): print (n-1)*a
else: print " ".join(map(str,[i*a+(n-1-i)*b for i in xrange(0,n)]))Manasa and Stones Solution using JavaScript
'use strict';
function processData(input) {
var parse_fun = function (s) { return parseInt(s, 10); };
var lines = input.split('\n');
var T = parse_fun(lines.shift());
for (var t = 0; t < T; t++) {
var n = parse_fun(lines[3 * t]);
var a = parse_fun(lines[3 * t + 1]);
var b = parse_fun(lines[3 * t + 2]);
var s = {};
for (var i = 0; i < n; i++) {
s[i * a + (n - 1 - i) * b] = 1;
}
var res = [];
for (var i in s) {
res.push(i);
}
res.sort(function(n1, n2) { return n1 - n2; });
console.log(res.join(' '));
}
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
var _input = "";
process.stdin.on("data", function (input) { _input += input; });
process.stdin.on("end", function () { processData(_input); });Manasa and Stones Solution in Scala
object Solution {
def main(arg:Array[String]) {
(1 to readInt).foreach(_ => {
def min(x:Int,y:Int) = if (x < y) x else y
def max(x:Int,y:Int) = if (x > y) x else y
val n = readInt
val a = readInt
val b = readInt
if (a == b) println((n - 1) * a)
else {
val A = min(a, b)
val B = max(a, b)
println(((n - 1) * A to (n - 1) * B by B - A).mkString(" "))
}
})
}
}Manasa and Stones Solution in Pascal
var n,a,b,t,k1,k2,i,q: integer;
begin
readln(t);
for i:=1 to t do
begin
readln(n);
readln(a);
readln(b);
if a>b then
begin
q:=a;
a:=b;
b:=q;
end;
k1:=n-1;
k2:=0;
if a=b then write(a*(n-1),' ') else
while k2<n do
begin
write(a*k1+b*k2,' ');
k1:=k1-1;
k2:=k2+1;
end;
writeln;
end;
end.Disclaimer: This problem (Manasa and Stones) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.