Given 7 empty boxes b1, b2, b3, b4, b5, b6, b7, and an integer N, the task is to find the total amount of money that can be placed in the boxes after N days based on the following conditions:
- Each day, the money can be put only in one box in circular fashion b1, b2, b3, b4, b5, b6, b7, b1, b2, ….. and so on.
- In box b1, put 1 more than the money already present in box b1.
- In each box except b1, put 1 more than the money present in the previous box.
Examples:
Input: N = 4
Output: 15
Explanation:
Putting money in the box b1 on day 1 = 1
Putting money in the box b2 on day 2 = 2
Putting money in the box b3 on day 3 = 3
Putting money in the box b4 on day 4 = 4
Putting money in the box b5 on day 5 = 5
After the 5th day, total amount = 1 + 2 + 3 + 4 + 5 = 15Input: N = 15
Output: 66
Explanation: After the 15th day, the total amount = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 3 = 66
Approach: Follow the steps below to solve the problem
- The money spent on ith day is ((i – 1)/ 7) + ((i – 1) % 7 + 1), where i lies in the range [1, N]
- Simulate the same for days [1, N]
- Print the total cost.
Below is the implementation of the above approach:
// C++ program for // the above approach #include <iostream> using namespace std;
// Function to find the total money // placed in boxes after N days int totalMoney( int N)
{ // Stores the total money
int ans = 0;
// Iterate for N days
for ( int i = 0; i < N; i++)
{
// Adding the Week number
ans += i / 7;
// Adding previous amount + 1
ans += (i % 7 + 1);
}
// Return the total amount
return ans;
} // Driver code int main()
{ // Input
int N = 15;
// Function call to find
// total money placed
cout << totalMoney(N);
} // This code is contributed khushboogoyal499 |
// Java program for // the above approach import java.io.*;
class GFG {
// Function to find the total money
// placed in boxes after N days
public static int totalMoney( int N)
{
// Stores the total money
int ans = 0 ;
// Iterate for N days
for ( int i = 0 ; i < N; i++) {
// Adding the Week number
ans += i / 7 ;
// Adding previous amount + 1
ans += (i % 7 + 1 );
}
// Return the total amount
return ans;
}
// Driver Code
public static void main(String[] args)
{
// Input
int N = 15 ;
// Function call to find
// total money placed
System.out.println(
totalMoney(N));
}
} |
# Python program for # the above approach # Function to find the total money # placed in boxes after N days def totalMoney(N):
# Stores the total money
ans = 0
# Iterate for N days
for i in range ( 0 , N):
# Adding the Week number
ans + = i / 7
# Adding previous amount + 1
ans + = (i % 7 + 1 )
# Return the total amount
return ans
# Driver code # Input N = 15
# Function call to find # total money placed print (totalMoney(N))
# This code is contributed by shivanisinghss2110 |
// C# program for // the above approach using System;
class GFG{
// Function to find the total money // placed in boxes after N days public static int totalMoney( int N)
{ // Stores the total money
int ans = 0;
// Iterate for N days
for ( int i = 0; i < N; i++)
{
// Adding the Week number
ans += i / 7;
// Adding previous amount + 1
ans += (i % 7 + 1);
}
// Return the total amount
return ans;
} // Driver code static public void Main()
{ // Input
int N = 15;
// Function call to find
// total money placed
Console.WriteLine(totalMoney(N));
} } // This code is contributed by offbeat |
<script> // JavaScript Program to implement // the above approach // Function to find the total money
// placed in boxes after N days
function totalMoney(N)
{
// Stores the total money
let ans = 0;
// Iterate for N days
for (let i = 0; i < N; i++) {
// Adding the Week number
ans += Math.floor(i / 7);
// Adding previous amount + 1
ans += (i % 7 + 1);
}
// Return the total amount
return ans;
}
// Driver Code // Input
let N = 15;
// Function call to find
// total money placed
document.write(
totalMoney(N));
</script> |
66
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by finding out the number of completed weeks and the number of days remaining in the last week.
Follow the steps to solve the problem:
- Initialize variables X and Y, to store the amount of money that can be placed in the complete weeks and partial weeks respectively.
- The money in each week can be calculated as:
- 1st Week: 1 2 3 4 5 6 7 = 28 + (7 x 0)
- 2nd Week: 2 3 4 5 6 7 8 = 28 + (7 x 1)
- 3rd Week: 3 4 5 6 7 8 9 = 28 + (7 x 2)
- 4th Week: 4 5 6 7 8 9 10 = 28 + (7 x 3) and so on.
- Therefore, update:
X = 28 + 7 x (Number of completed weeks – 1)
Y = Sum of remaining days + ( Number of complete weeks * Number of days remaining in the last week) - Therefore, total amount is equal to X + Y. Print the total amount placed.
Below is the implementation of the above approach:
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to find total // money placed in the box int totalMoney( int N)
{ // Number of complete weeks
int CompWeeks = N / 7;
// Remaining days in
// the last week
int RemDays = N % 7;
int X = 28 * CompWeeks
+ 7 * (CompWeeks
* (CompWeeks - 1) / 2);
int Y = RemDays
* (RemDays + 1) / 2
+ CompWeeks * RemDays;
int cost = X + Y;
cout << cost << '\n' ;
} // Driver Code int main()
{ // Input
int N = 15;
// Function call to find
// the total money placed
totalMoney(N);
return 0;
} |
// Java program for above approach import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Function to find total // money placed in the box static void totalMoney( int N)
{ // Number of complete weeks
int CompWeeks = N / 7 ;
// Remaining days in
// the last week
int RemDays = N % 7 ;
int X = 28 * CompWeeks
+ 7 * (CompWeeks
* (CompWeeks - 1 ) / 2 );
int Y = RemDays
* (RemDays + 1 ) / 2
+ CompWeeks * RemDays;
int cost = X + Y;
System.out.print(cost);
} // Driver Code
public static void main(String[] args)
{
// Input
int N = 15 ;
// Function call to find
// the total money placed
totalMoney(N);
}
} // This code is contributed by souravghosh0416. |
# Python3 code to implement the approach # Function to find total money # placed in the box def total_money(N):
# Number of complete weeks
CompWeeks = N / / 7
# Remaining days in the last week
RemDays = N % 7
X = 28 * CompWeeks + 7 * (CompWeeks * (CompWeeks - 1 ) / / 2 )
Y = RemDays * (RemDays + 1 ) / / 2 + CompWeeks * RemDays
cost = X + Y
print (cost)
# Driver Code # Input N = 15
# Function call to find the total money placed total_money(N) # This code is contributed by phasing17 |
// C# program for the above approach using System;
class GFG{
// Function to find total // money placed in the box static void totalMoney( int N)
{ // Number of complete weeks
int CompWeeks = N / 7;
// Remaining days in
// the last week
int RemDays = N % 7;
int X = 28 * CompWeeks + 7 *
(CompWeeks * (CompWeeks - 1) / 2);
int Y = RemDays * (RemDays + 1) / 2 +
CompWeeks * RemDays;
int cost = X + Y;
Console.WriteLine(cost);
} // Driver Code public static void Main()
{ // Input
int N = 15;
// Function call to find
// the total money placed
totalMoney(N);
} } // This code is contriobuted by sanjoy_62 |
<script> // JavaScript Program to implement // the above approach // Function to find total // money placed in the box function totalMoney( N)
{ // Number of complete weeks
let CompWeeks = Math.floor(N / 7);
// Remaining days in
// the last week
let RemDays = N % 7;
let X = 28 * CompWeeks
+ 7 * Math.floor((CompWeeks
* (CompWeeks - 1) / 2));
let Y = RemDays
* Math.floor((RemDays + 1) / 2)
+ CompWeeks * RemDays;
let cost = X + Y;
document.write(cost , '<br>' );
} // Driver Code // Input
let N = 15;
// Function call to find
// the total money placed
totalMoney(N);
</script> |
66
Time Complexity: O(1)
Auxiliary Space: O(1)