Algorithm
Problem Name: Mathematics -
https://www.hackerrank.com/challenges/filling-jars/problem?isFullScreen=true
In this HackerRank in Mathematics -
Animesh has n empty candy jars, numbered from 1 to n, with infinite capacity. He performs m operations. Each operation is described by 3 integers, a,b and k. Here, a and b are indices of the jars, and k is the number of candies to be added inside each jar whose index lies between a and b (both inclusive). Can you tell the average number of candies after m operations?
Example
n = 5
operations = [[1,2,10],[5,2,10]]
The array has 5 elements that all start at 0 . In the first operation, add 10 to the first 2 elements. Now the array is [10,10,0,0,0] . In the second operation, add 10 to the last 3 elements (3 - 5). Now the array is [10,10,10,10,10] and the average is 10. Sincd 10 is already an integer value, it does not need to be rounded.
Function Description
Complete the solve function in the editor below.
solve has the following parameters:
- int n: the number of candy jars
- int operations[m][3]: a 2-dimensional array of operations
Returns
- int: the floor of the average number of canidies in all jars
Input Format
The first line contains two integers, n and m , separated by a single space. m lines follow. Each of them contains three integers, a,b and k, separated by spaces.
Constraints
3 <= n <= 107
1 <= m <= 105
1 <= a <= b <= N
0 <= k <= 106
Sample Input
STDIN Function ----- -------- 5 3 n = 5, operations[] size = 3 1 2 100 operations = [[1, 2, 100], [2, 5, 100], [3, 4, 100]] 2 5 100 3 4 100
Sample Output
160
Explanation
Initially each of the jars contains 0 candies
0 0 0 0 0
First operation:
100 100 0 0 0
Second operation:
100 200 100 100 100
Third operation:
100 200 200 200 100
Total = 800, Average = 800/5 = 160
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
int main ()
{
long unsigned int a, b, i, k, n, m, sum = 0;
scanf("%lu %lu", &n, &m);
for (i = 1; i < = m; i++)
{
scanf("%lu %lu %lu", &a, &b, &k);
sum = sum+(b-a+1)*k;
}
printf("%lu", sum/n);
}
#2 Code Example with C++ Programming
Code -
C++ Programming
#include<bits/stdc++.h>
using namespace std;
int main()
{
uint64_t n,m,a,b,k,i;
cin >> n >> m;
uint64_t sum=0;
while(m--)
{
cin >> a >>b >> k;
sum+=(b-a+1)*k;
}
uint64_t z;
z=floor(sum/n);
cout << z;
}
#3 Code Example with Java Programming
Code -
Java Programming
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in );
long n,t,sum=0;
n=sc.nextLong();
t=sc.nextLong();
while(t-->0)
{
long a,b,k;
a=sc.nextLong();
b=sc.nextLong();
k=sc.nextLong();
sum+=(b-a+1)*k;
}
System.out.println(sum/n);
}
}
#4 Code Example with Python Programming
Code -
Python Programming
#!/bin/python3
import math
import os
import random
import re
import sys
#
# Complete the 'solve' function below.
#
# The function is expected to return an INTEGER.
# The function accepts following parameters:
# 1. INTEGER n
# 2. 2D_INTEGER_ARRAY operations
#
def solve(n, operations):
output = 0
for o in operations:
output += o[2] * (o[1]-o[0]+1)
return output//n
if __name__ == '__main__':
fptr = open(os.environ['OUTPUT_PATH'], 'w')
first_multiple_input = input().rstrip().split()
n = int(first_multiple_input[0])
m = int(first_multiple_input[1])
operations = []
for _ in range(m):
operations.append(list(map(int, input().rstrip().split())))
result = solve(n, operations)
fptr.write(str(result) + '\n')
fptr.close()