Minimum operations (Medium Complexity) - Problem From HackerEarth

Index of "Algorithms: Design and Analysis"

Can you explain the solution to this problem:

Problem
You are given an array A of size N. You can perform the following operation on array A:

Select two indices i and j such that 1 ≤  i, j ≤  N. (note that i and j can be equal)
Assign Ai = Ai + 2
Then assign Aj = Aj - 1
You need to make all the elements of the array equal to zero.

Task
Determine the minimum number of operations required to make all the elements of A equal to zero. Else, print -1 if it is not possible to do so.

Function description

Complete the function solve() provided in the editor. This function takes the following parameters and returns the required answer:

N: Represents the size of the array A
A: Represents the elements of array A.

Input format

Note: This is the input format that you must use to provide custom input (available above the Compile and Test button).

The first line contains T, denoting the number of test cases. T also specifies the number of times you have to run the solve() function on a different set of inputs.
For each test case:
The first line contains N, denoting the size of array A.
The second line contains space-separated values, denoting the elements of array A.
Output format For each test case, print the output on a new line. Either the minimum number of operations required to make all the elements of A equal to zero or print -1 if it is impossible to do so.


Sample Input
2
1
-2
2
1 -1
Sample Output
2
-1
Time Limit: 2
Memory Limit: 256
Source Limit:
Explanation
The first line contains the number of test cases, T = 2.

The first test case

Given

N = 1
A = [-2]
Approach

We can perform the following operations on A:

Select i = 1, j = 1, A1 = -2 + 2 = 0 and A2 = 0 - 1 = -1, A becomes [-1].
Select i = 1, j = 1, A1 = -1 + 2 = 1 and A2 = 1 - 1 = 0, A becomes [0].
Hence, all the elements are equal to zero after 2 operations. Thus, the answer is 2.

The second test case

Given

N = 2
A = [1, -1]
Approach

It can be shown that it is not possible to reduce all the elements of A equal to zero simultaneously. Thus, the answer is -1.

Download Test Cases
Solve on HackerEarth

This is a great problem because it looks simple, but it’s really about invariants and feasibility, not simulation.

Let’s break it down cleanly.

---

🔁 The operation (very important)

In one operation:

• Pick indices i and j (they can be the same)

• Do:

  • A[i] += 2

  • A[j] -= 1

So each operation changes the array in a fixed way.

---

🔑 Step 1: Understand what never changes (invariants)

🔹 Invariant 1: Total sum increases by +1

Each operation does:

diff
+2  and  -1  → net change = +1

So after k operations:

sql
sum(final) = sum(initial) + k

But your goal is:

sql
final array = [0, 0, ..., 0]
⇒ sum(final) = 0

Therefore:

bash
0 = sum(A) + k
⇒ k = -sum(A)

📌 Conclusion 1

• If sum(A) > 0, then k < 0 → impossible

• So sum(A) must be ≤ 0

---

🔹 Invariant 2: What happens to individual elements?

An element:

• Can increase by 2

• Or decrease by 1

• Or both (if i == j, net +1)

So to reduce a positive value, you must apply -1 operations to it.
To increase a negative value, you use +2.

This creates a cost structure.

---

🔢 Step 2: What must the number of operations be?

From invariant 1:

ini
k = -sum(A)

So the number of operations is fixed, not something we can choose.

If a solution exists, the answer must be:

ini
answer = -sum(A)

No more, no less.

---

🧠 Step 3: Feasibility condition (the real trick)

Each operation gives:

• one -1

• one +2

Across k operations:

• Total negative capacity = k

• Total positive capacity = 2k

Now think element-wise:

For positive elements A[i] > 0

They must be reduced to zero using -1 operations:

bash
Total required decrements = sum of positive A[i]

But we only have k decrements available.

So we need:

nginx
sum_positive ≤ k

Substitute k = -sum(A):

scss
sum_positive ≤ -sum(A)

Since:

bash
sum(A) = sum_positive + sum_negative

This becomes:

scss
sum_positive ≤ -(sum_positive + sum_negative)
⇒ 2 * sum_positive ≤ -sum_negative

📌 This is the key feasibility condition

---

✅ Final Conditions for a Solution

A solution exists if and only if:

1. sum(A) ≤ 0

2. 2 * sum_positive ≤ -sum_negative

If both are satisfied:

bash
minimum operations = -sum(A)

Otherwise:

ini
answer = -1

---

🧪 Apply to the sample cases

✅ Case 1

makefile
A = [-2]
sum = -2 → k = 2
sum_positive = 0
sum_negative = -2

Check:
2 * 0 ≤ 2 ✔
Answer = 2

❌ Case 2

makefile
A = [1, -1]
sum = 0 → k = 0
sum_positive = 1
sum_negative = -1

Check:
2 * 1 ≤ 1 ❌
Answer = -1

---

🧠 What this problem teaches you

This is very important conceptually:

🧩 1. Don’t simulate when operations are large

Look for invariants instead.

🧩 2. Operations define resource constraints

Here:

• Each operation provides limited + and − power

🧩 3. Sometimes the answer is forced

The number of operations wasn’t something to minimize — it was fixed by math.

---

🧾 Final Pseudocode

python
sumA = sum(A)
if sumA > 0:
    return -1

sum_pos = sum(x for x in A if x > 0)
sum_neg = sum(x for x in A if x < 0)  # negative number

if 2 * sum_pos > -sum_neg:
    return -1

return -sumA

Failing Test Case

1
10
858 268 -73 -83 -613 -88 -240 -463 -605 -90

Tags: Algorithms,