Rod cutting problem
Posted: 24 Oct, 2020
Given a rod of length ‘N’ units. The rod can be cut into different sizes and each size has a cost associated with it. Determine the maximum cost obtained by cutting the rod and selling its pieces.
1. The sizes will range from 1 to ‘N’ and will be integers. 2. The sum of the pieces cut should be equal to ‘N’. 3. Consider 1-based indexing.
The first line of input contains an integer ‘T’ denoting the number of test cases. The next 2 * T lines represent the ‘T’ test cases. The first line of each test case contains an integer ‘N’ denoting the length of the rod. The second line of each test case contains a vector ’A’, of size ‘N’ representing the cost of different lengths, where each index of the array is the sub-length and the element at that index is the cost for that sub-length.
Since 1-based indexing is considered, the 0th index of the vector will represent sub-length 1 of the rod. Hence the (N - 1)th index would represent the cost for the length ‘N’.
For each test case, print a single line that contains a single integer which is the maximum cost obtained by selling the pieces. The output of each test case will be printed in a separate line.
You do not need to print anything; it has already been taken care of. Just implement the given function.
1 <= T <= 50 1 <= N <= 100 1 <= A[i] <= 100 Where ‘T’ is the total number of test cases, ‘N’ denotes the length of the rod, and A[i] is the cost of sub-length. Time limit: 1 sec.
- This is a recursive approach.
- Initialize a variable ‘MAXCOST’ to INT_MIN;
- In each recursive function call, run a loop where ‘I’ ranges from 0 to ‘N - 1’, and partition the rod of length ‘N’ into two parts, ‘I’ and ‘N - I’.
- Example For a rod of length 5, for 'I' = 2, the two partitions will be 2 and 3.
1. Take ‘I’ as the first cut in the rod and keep it constant. Now, a rod of ‘N - I - 1’ remains. Pass the remaining rod of size ‘N - I - 1’ to the recursion.
2. Step 1 continues until it hits the base condition, which is when the length of the rod becomes zero.
3. When the recursion hits the base condition, the cost of a particular configuration is obtained.
4. Compare it with the ‘MAXCOST’ variable and store the maximum of the cost in variable ‘MAXCOST’.
- For every ‘I’ as an initial sub-length, different configurations are obtained and compared to the ‘MAXCOST’ variable.
- Lastly, ‘MAXCOST’ contains the final answer.