LeetCode 3038 - Maximum Number of Operations With the Same Score I
The problem asks us to repeatedly perform an operation on an integer array nums. In each operation, we remove the first two elements of the array and compute a score, which is simply the sum of those two removed values.
Difficulty: 🟢 Easy
Topics: Array, Simulation
Solution
Problem Understanding
The problem asks us to repeatedly perform an operation on an integer array nums. In each operation, we remove the first two elements of the array and compute a score, which is simply the sum of those two removed values.
The key restriction is that every operation must have the exact same score. We are free to continue performing operations as long as there are at least two elements remaining, but the moment a pair produces a different sum, we must stop.
The goal is to return the maximum number of operations that can be performed while maintaining a consistent score.
Restating the problem in simpler terms, we are trying to determine how many consecutive pairs from the front of the array can be removed such that every pair has the same sum as the very first pair.
For example, if the array starts as:
[3,2,1,4,5]
The first operation removes 3 and 2, producing a score of 5. Since this score becomes fixed for all future operations, every subsequent pair must also sum to 5.
After removing [3,2], the array becomes:
[1,4,5]
The next pair is [1,4], which also sums to 5, so the operation is valid. At that point, only [5] remains, meaning we cannot continue.
The input consists of:
nums, an integer array.- Each operation always removes the first two elements.
- The score must remain identical across all operations.
The output is a single integer representing the maximum number of valid operations.
Understanding the Constraints
The constraints are:
2 <= nums.length <= 1001 <= nums[i] <= 1000
These constraints are very small. Since the array length is at most 100, even less efficient approaches would work comfortably within limits.
However, the simplicity of the problem suggests there is a very direct and efficient simulation-based solution.
Important Edge Cases
One important edge case occurs when the array has exactly two elements. Since one operation can always be performed, the answer must be 1.
Another case happens when the second pair immediately produces a different score. In that scenario, only the first operation is valid.
We should also consider arrays where every pair has the same sum, allowing operations to continue until fewer than two elements remain.
The problem guarantees that the array always contains at least two elements, so we never need to handle an empty input.
Approaches
Brute Force Approach
A brute force approach would simulate the process exactly as described.
We first compute the score using the first two elements. Then we repeatedly remove the first two elements and check whether their sum matches the target score.
One naive implementation could literally remove elements from the front of the array using operations like pop(0) in Python or repeated slicing. Since removing from the front of an array is expensive, this introduces unnecessary overhead.
Although this approach still works for n <= 100, it is inefficient because repeatedly shifting elements costs extra time.
The approach is correct because it exactly follows the rules of the problem. Every operation is validated in order, and the process stops as soon as a mismatch occurs.
Key Insight for an Optimal Solution
The important observation is that we never actually need to modify the array.
Since operations always happen in pairs from the front, we can simply iterate through the array two elements at a time.
The first pair determines the required score:
target_score = nums[0] + nums[1]
Then, for every subsequent pair:
nums[i] + nums[i+1]
we compare the sum with target_score.
If the sums match, we increment the operation count. If they differ, we stop immediately.
Because each element is visited at most once, this gives us a simple linear-time solution.
| Approach | Time Complexity | Space Complexity | Notes |
|---|---|---|---|
| Brute Force | O(n²) | O(1) | Repeatedly removing from the front causes shifting |
| Optimal | O(n) | O(1) | Scan the array in pairs without modifying it |
Algorithm Walkthrough
- Compute the target score using the first two elements.
The first operation is mandatory because the array length is guaranteed to be at least 2. The sum of the first pair becomes the score that every future operation must match.
2. Initialize the operation counter.
Since the first operation is always valid, start the count at 1.
3. Iterate through the remaining array in steps of two.
Start from index 2, because the first pair has already been used. Move through the array using a step size of 2 so that each iteration represents one operation.
4. Compute the current pair's sum.
For every index i, calculate:
nums[i] + nums[i+1]
- Compare with the target score.
If the sum matches the target score, increment the operation counter.
If it does not match, stop immediately because no further operations are allowed. 6. Return the total operation count.
Once all valid pairs have been checked or a mismatch occurs, return the count.
Why it works
The algorithm works because the score is fixed after the very first operation. Since operations must happen strictly from left to right using the first two remaining elements, there is no choice or branching involved.
At every step, we verify whether the next required pair satisfies the invariant:
Every operation must produce the same score as the first operation.
The moment this invariant fails, no further operations are valid. Since we process pairs in order and count every valid one exactly once, the result is guaranteed to be correct.
Python Solution
from typing import List
class Solution:
def maxOperations(self, nums: List[int]) -> int:
target_score = nums[0] + nums[1]
operations = 1
for index in range(2, len(nums) - 1, 2):
current_score = nums[index] + nums[index + 1]
if current_score != target_score:
break
operations += 1
return operations
The implementation starts by computing the target score from the first pair. Since the problem guarantees at least two elements, this access is always safe.
We initialize operations to 1 because the first operation is guaranteed to happen.
The loop begins at index 2, which represents the next available pair after removing the first two elements conceptually. The step size of 2 ensures we only inspect valid operation boundaries.
For each pair, we calculate the current score and compare it with the target score. If the values differ, the process stops immediately because future operations are invalid. Otherwise, we increment the count.
Finally, we return the total number of valid operations.
Go Solution
func maxOperations(nums []int) int {
targetScore := nums[0] + nums[1]
operations := 1
for i := 2; i < len(nums)-1; i += 2 {
currentScore := nums[i] + nums[i+1]
if currentScore != targetScore {
break
}
operations++
}
return operations
}
The Go implementation follows the same logic as the Python version.
Since Go slices provide constant-time indexing, we iterate through the array using index arithmetic instead of modifying the slice.
Integer overflow is not a concern here because nums[i] <= 1000, meaning the largest possible pair sum is only 2000, which comfortably fits within Go's int type.
No special handling for nil or empty slices is necessary because the constraints guarantee at least two elements.
Worked Examples
Example 1
Input:
nums = [3,2,1,4,5]
The target score comes from the first pair:
3 + 2 = 5
| Step | Pair | Pair Sum | Target Score | Valid? | Operations |
|---|---|---|---|---|---|
| Initial | [3,2] | 5 | 5 | Yes | 1 |
| 2 | [1,4] | 5 | 5 | Yes | 2 |
At this point, only [5] remains, which contains fewer than two elements.
Final Answer: 2
Example 2
Input:
nums = [1,5,3,3,4,1,3,2,2,3]
The target score is:
1 + 5 = 6
| Step | Pair | Pair Sum | Target Score | Valid? | Operations |
|---|---|---|---|---|---|
| Initial | [1,5] | 6 | 6 | Yes | 1 |
| 2 | [3,3] | 6 | 6 | Yes | 2 |
| 3 | [4,1] | 5 | 6 | No | Stop |
Final Answer: 2
Example 3
Input:
nums = [5,3]
The target score is:
5 + 3 = 8
| Step | Pair | Pair Sum | Target Score | Valid? | Operations |
|---|---|---|---|---|---|
| Initial | [5,3] | 8 | 8 | Yes | 1 |
No more elements remain.
Final Answer: 1
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(n) | Each element is visited at most once |
| Space | O(1) | Only a few variables are used |
The algorithm scans the array once in increments of two, checking each pair exactly one time. Since there are n / 2 possible operations, the running time grows linearly with the size of the input.
The space complexity is constant because we only store a target score, a counter, and temporary variables, regardless of input size.
Test Cases
solution = Solution()
assert solution.maxOperations([3, 2, 1, 4, 5]) == 2 # Example 1
assert solution.maxOperations([1, 5, 3, 3, 4, 1, 3, 2, 2, 3]) == 2 # Example 2
assert solution.maxOperations([5, 3]) == 1 # Minimum valid input size
assert solution.maxOperations([2, 2, 2, 2]) == 2 # All pairs match
assert solution.maxOperations([1, 2, 4, 5]) == 1 # Second pair mismatches immediately
assert solution.maxOperations([1, 1, 2, 0, 3, -1]) == 3 # Multiple matching sums
assert solution.maxOperations([10, 5, 8, 7, 4, 11]) == 3 # Every pair matches
assert solution.maxOperations([1, 2, 3, 4, 5, 6]) == 1 # First mismatch after initial pair
assert solution.maxOperations([1000, 1000, 500, 1500]) == 2 # Large values
assert solution.maxOperations([7, 3, 6]) == 1 # Odd length array
| Test | Why |
|---|---|
[3,2,1,4,5] |
Verifies Example 1 |
[1,5,3,3,4,1,3,2,2,3] |
Verifies Example 2 |
[5,3] |
Minimum allowed input |
[2,2,2,2] |
Every pair matches |
[1,2,4,5] |
Immediate mismatch |
[1,1,2,0,3,-1] |
Multiple matching operations |
[10,5,8,7,4,11] |
Full traversal succeeds |
[1,2,3,4,5,6] |
Early stopping behavior |
[1000,1000,500,1500] |
Large values within constraints |
[7,3,6] |
Odd-length array |
Edge Cases
Minimum Input Size
When nums contains exactly two elements, there is only one possible operation. A buggy implementation might accidentally return 0 if it initializes the counter incorrectly or skips processing.
The implementation handles this correctly by immediately treating the first pair as one valid operation and initializing operations = 1.
Immediate Score Mismatch
An array such as:
[1,2,4,5]
creates a target score of 3, but the second pair sums to 9.
A naive implementation might continue checking later pairs even though operations must stop immediately after the first mismatch. The algorithm avoids this by breaking out of the loop as soon as a mismatch occurs.
Odd-Length Arrays
Arrays with an odd number of elements can leave one leftover value after valid operations.
For example:
[3,2,1,4,5]
After removing valid pairs, one element remains.
The implementation safely handles this by iterating only while i < len(nums) - 1, ensuring we never attempt to access a non-existent second element in a pair.
Constraint Note About Test Cases
The official constraints guarantee:
1 <= nums[i] <= 1000
so negative-number test cases are outside the problem limits. They were included only to stress the pairing logic. Under official inputs, all values are positive integers.