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Letter/Blog Writer Coding Problem (using Dynamic Programming)
Letter/Blog Writer Coding Problem: In this article, we are going to see how to solve a Letter/Blog Writer Coding Problem efficiently with dynamic programming? And, it can be featured in any interview rounds.
Submitted by Radib Kar, on April 19, 2020
Problem statement
Shivang is a blog writer and he is working on two websites simultaneously. He has to write two types of blogs which are:
- Technical blogs for website_1: X of this type can be written in an hour. (X will be the input)
- Fun blogs for website_2: Y of this type can be written in an hour. (Y will be input)
You are to help him to save time. Given N number of total blogs, print the minimum number of hours he needs to put for combination of both the blogs, so that no time is wasted.
Input:
N: number of total blogs
X: number of Technical blogs for website_1 per hour
Y: number of Fun blogs for website_2 per hour
Output:
Print the minimum number of hours Shivang needs to write the
combination of both the blogs (total N).
If it is NOT possible, print "−1".
Example
Input:
N: 33
X: 12
Y: 10
Output:
-1
Input:
N: 36
X: 10
Y: 2
Output:
6 (10*3+2*3)
Explanation
For the first case,
No combination is possible that's why output is -1.
For the second test case,
Possible combinations are 30 technical blogs (3 hours) + 6 fun blogs (3 hours)
20 technical blogs (2 hours) + 16 fun blogs (8 hours)
10 technical blogs (1 hours) + 26 fun blogs (13 hours)
0 technical blogs (0 hours) + 36 fun blogs (18 hours)
So, best combination is the first one which takes total 6 hours
The problem is basically solving equation,
aX + bY = N where we need to find the valid integer coefficients of X and Y. Return a+b if there exists such else return -1.
We can find a recursive function for the same too,
Say,
f(n) = minimum hours for n problems
f(n) = min(f(n-x) + f(n-y)) if f(n-x), f(n-y) is solved already
We can convert the above recursion to DP.
Solution Approach
Converting the recursion into DP:
1) Create DP[n] to store sub problem results
2) Initiate the DP with -1 except DP[0], DP[0]=0
3) Now in this step we would compute values for DP[i]
4) for i=1 to n
if i-x>=0 && we already have solution for i-x,i.e.,DP[i-x]!=-1
DP[i]=DP[i-x]+1;
end if
if i-y>=0 && we already have solution for i-y,i.e.,DP[i-y]!=-1)
if DP[i]!=-1
DP[i]=min(DP[i],DP[i-y]+1);
else
DP[i]=DP[i-y]+1;
End if
End if
End for
5) Return DP[n]
C++ Implementation
#include <bits/stdc++.h>
using namespace std;
int minimumHour(int n, int x, int y)
{
int a[n + 1];
a[0] = 0;
for (int i = 1; i <= n; i++)
a[i] = -1;
for (int i = 1; i <= n; i++) {
if (i - x >= 0 && a[i - x] != -1) {
a[i] = a[i - x] + 1;
}
if (i - y >= 0 && a[i - y] != -1) {
if (a[i] != -1)
a[i] = min(a[i], a[i - y] + 1);
else
a[i] = a[i - y] + 1;
}
}
return a[n];
}
int main()
{
int n, x, y;
cout << "Enter total number of blogs, N:\n";
cin >> n;
cout << "Enter number of techical blogs, X:\n";
cin >> x;
cout << "Enter number of fun blogs, Y:\n";
cin >> y;
cout << "Minimum hour to be dedicated: " << minimumHour(n, x, y) << endl;
return 0;
}
Output
RUN 1:
Enter total number of blogs, N:
36
Enter number of techical blogs, X:
10
Enter number of fun blogs, Y:
2
Minimum hour to be dedicated: 6
RUN 2:
Enter total number of blogs, N:
33
Enter number of techical blogs, X:
12
Enter number of fun blogs, Y:
10
Minimum hour to be dedicated: -1