Problem Link:

Practice
Contest

AuthorRoman Furko
TesterBalajiganpathi
EditorialistUtkarsh 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 3^N-1 ways of assigning the operators. Therefore, complexity of trying out all combinations is for all test cases O(T * 3^N-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 ... op_N-1, so the final expression becomes A₁⟨op₁⟩ A₂⟨op₂⟩ A₃ ... A_N-1⟨op_N-1⟩ A_N.

Lets develop a shorthand by denoting eval_i = A₁⟨op₁⟩ A₂⟨op₂⟩ A₃ ... A_i-1⟨op_i-1⟩ A_i, i.e. evaluation over first i numbers.

For each possibility of op_N-1, lets figure out what can we say about eval_N-1.
* If op_N-1 = '+'. then it obviously makes best sense to choose the first N-2 operators so that eval_N-1 is minimum.
* If op_N-1 = '-', then it again makes sense to choose the first N-2 operators so that eval_N-1 is minimum.
* If op_N-1 = ' * ', then there are two cases:
i) A_N≥ 0 then we should choose the first N-2 operators so that eval_N-1 is minimum.
ii) A_N≤ 0 then we should choose the first N-2 operators so that eval_N-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 eval_N-1 to evaluate minimum possible of eval_N. Similarly, if we want to maximize the value of eval_N, we would again need to find only the maximum and minimum possible values of eval_N-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.

Solutions:

Setter's Solution
Tester's Solution
Editorialist's Solution