Problem Link:
Author Roman Furko
Tester Balajiganpathi
Editorialist Utkarsh Lath
Difficulty:
Simple
Pre-requisites:
Dynamic Programming
Problem:
Given a list of N integers, each between -9 and 9, you have to put one of the operators ‘+’, ‘-’, ‘*’ between consecutive elements of the list, so the resulting number is mimimum. The result is obtained by evaluating operators from left to right.
Explanation:
Subtask 1
We have no idea which operators should be filled in each of the blanks to get minimum result, so we can try all possible combinations. This can be done via recursion:
smallest-result = 10^18;
void calc( i, res-so-far) // first i-1 operators have been chosen, chose the remaining
if i ≥ N
smallest-result = min (smallest-result, res-so-far)
else
calc(i+1, res-so-far + A[i+1]) // the ith operator can be one out of '+', '-', '*'
calc(i+1, res-so-far - A[i+1])
calc(i+1, res-so-far * A[i+1])
calc(1, A[1])
print smallest-result
NOTE: You should use 64-bit integers for this problem, because the minimum value can be upto - 910 < -231, which is out of range of 32 bit integers.
The next question is, how much time it would take ?
There are 3N-1 ways of assigning the operators. Therefore, complexity of trying out all combinations is for all test cases O(T * 3N-1). For subtasks 1 and 2, these numbers are roughly 2 * 105 and 3 * 105 respectively, and these should be easily solvable. Therefore, subtasks 1 and 2 can be done by brute force. For other subtasks, the number of operations needed in worst case would be more than 4 * 108, and would possibly need more optimizations/better algorithms.
Now lets consider the subtask 3. In this subtask, there is a restriction on numbers, that they are all positive. Can we do anything better in this case ? This brings us back to the question
Can we really place all the operators (+, -, * ) at all the positions ?
Initially, when all numbers could be positive or negative, we could construct examples where all three operators (-, +, * ) are used. But in case of positive numbers, one can quickly realize that only ’ * ’ or ‘-’ will be used. This is because intuitively, ‘+’ will always increase the value, and any ‘+’ can be replaced by a ‘-’ to construct a strictly smaller value. Therefore, we will only need to check over all combinations of ‘-’ and ’ * ', thus giving rise to O(2 N-1) possibilities. The number of steps needed for subtask # 3 would be 5 * 107, so it can be solved with our current observation.
It remains to see how to solve the subtask 4, 5 in given time limit. Let the operations applied be op1, op2 … opN-1, so the final expression becomes A1 〈op1〉 A2 〈op2〉 A3 … AN-1 〈opN-1〉 AN.
Lets develop a shorthand by denoting evali = A1 〈op1〉 A2 〈op2〉 A3 … Ai-1 〈opi-1〉 Ai, i.e. evaluation over first i numbers.
For each possibility of opN-1, lets figure out what can we say about evalN-1.
- If opN-1 = ‘+’. then it obviously makes best sense to choose the first N-2 operators so that evalN-1 is minimum.
- If opN-1 = ‘-’, then it again makes sense to choose the first N-2 operators so that evalN-1 is minimum.
- If opN-1 = ’ * ', then there are two cases:
i) AN ≥ 0 then we should choose the first N-2 operators so that evalN-1 is minimum.
ii) AN ≤ 0 then we should choose the first N-2 operators so that evalN-1 is maximum, because we will flip its sign during last operation.
Therefore, we only need to know the maximum and minimum possible values of evalN-1 to evaluate minimum possible of evalN. Similarly, if we want to maximize the value of evalN, we would again need to find only the maximum and minimum possible values of evalN-1.
To see why this really helps, consider a small test case as follows:
N = 4
A = 1 -2 4 -1
Now, we could put ‘+’ or ‘-’ or ‘*’ at the first place to obtain eval2 as ‘-1’, ‘3’ or ‘-2’. Now, to evaluate the smallest/largest possible value of eval3, only the values ‘-2’, and ‘3’ are relevant.
The smallest candidate value of eval3 can be obtained by one of the possibilities -2+4, -2-4, -2x4. The minimum being -8. Similarly, the largest possible values of eval3 can be one out of 3+4, 3-4, 3x4, best one being 12. Now in order to evaluate eval4, we only need to use eval3 = -8 and 12. Using same ideas candidates for smallest value of eval4 are only -8 + -1, -8 - -1, 12 * -1, which gives -12 as the final answer.
It is notable how a small observation has helped us eliminate so many possible assignments of operators. This approach is called Dynamic Programming.
We can summarize the ideas discussed above in the following code
best-eval[1] = A[1]
worst-eval[1] = A[1]
for i = 2 to N
if A[i] > 0
best-eval[i] = max(best-eval[i-1]+A[i], best-eval[i-1]-A[i], best-eval[i-1] * A[i])
worst-eval[i] = min(worst-eval[i-1]+A[i], worst-eval[i-1]-A[i], worst-eval[i-1] * A[i])
else
best-eval[i] = max(best-eval[i-1]+A[i], best-eval[i-1]-A[i], worst-eval[i-1] * A[i])
worst-eval[i] = min(worst-eval[i-1]+A[i], worst-eval[i-1]-A[i], best-eval[i-1] * A[i])
print worst-eval[N]
For A[i] > 0, best-eval[i]+A[i] > best-eval[i]-A[i], so we can apply some more optimizations based on these ideas, but they are not essential.
The time complexity is O(N) per test case, so it can pass all subtasks.