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Maximum Number of Operations With the Same Score I - Solution & Explanation

EasyArraySimulation14 min readAsked at: Microsoft
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Problem Statement

You are given an array of integers nums. Consider the following operation:

  • Delete the first two elements nums and define the score of the operation as the sum of these two elements.

You can perform this operation until nums contains fewer than two elements. Additionally, the same score must be achieved in all operations.

Return the maximum number of operations you can perform.

 

Example 1:

Input: nums = [3,2,1,4,5]

Output: 2

Explanation:

  • We can perform the first operation with the score 3 + 2 = 5. After this operation, nums = [1,4,5].
  • We can perform the second operation as its score is 4 + 1 = 5, the same as the previous operation. After this operation, nums = [5].
  • As there are fewer than two elements, we can't perform more operations.

Example 2:

Input: nums = [1,5,3,3,4,1,3,2,2,3]

Output: 2

Explanation:

  • We can perform the first operation with the score 1 + 5 = 6. After this operation, nums = [3,3,4,1,3,2,2,3].
  • We can perform the second operation as its score is 3 + 3 = 6, the same as the previous operation. After this operation, nums = [4,1,3,2,2,3].
  • We cannot perform the next operation as its score is 4 + 1 = 5, which is different from the previous scores.

Example 3:

Input: nums = [5,3]

Output: 1

 

Constraints:

  • 2 <= nums.length <= 100
  • 1 <= nums[i] <= 1000

Approach Overview

Problem Overview: You are given an integer array nums. Each operation removes the first two elements and produces a score equal to their sum. Every operation must produce the same score. The task is to determine the maximum number of valid operations you can perform before the rule breaks.

Approach 1: Brute Force Simulation (O(n) time, O(1) space)

The direct strategy simulates the process exactly as described. Start by computing the score from the first pair: target = nums[0] + nums[1]. Then repeatedly remove the next two elements and check whether their sum equals target. If the sum differs, the sequence of operations must stop. Because each step processes exactly two elements, the algorithm scans the array once. This approach uses simple iteration and conditional checks, making it easy to implement. It fits naturally with problems categorized under Array and Simulation since you mimic the operations step by step without additional data structures.

Approach 2: Optimized Two-Pointer Scan (O(n) time, O(1) space)

A cleaner implementation uses a pointer that moves across the array in steps of two. First compute the required score using the first pair. Then maintain a pointer i starting at index 2. At each step, check whether nums[i] + nums[i+1] equals the target score. If it matches, increment the operation count and move the pointer forward by two positions. If it does not match, terminate immediately because future operations would violate the constant-score rule. This pattern resembles a simplified Two Pointers traversal where the pointer jumps across fixed-size segments instead of sliding element by element.

The key insight is that the score is fixed by the first operation. After that, the array must naturally split into consecutive pairs producing the same sum. No rearrangement or searching is allowed, so the problem reduces to verifying each pair sequentially. Because every element is visited at most once, the runtime stays linear and memory usage remains constant.

Recommended for interviews: The two-pointer style scan is the approach interviewers typically expect. It demonstrates that you recognized the greedy constraint: the first pair determines the only valid score. The brute force simulation still shows clear understanding of the problem mechanics, but the optimized pointer traversal communicates stronger pattern recognition and cleaner implementation.

Approach 1: Brute Force Approach

In this approach, we consider all possible scores that can be achieved by the sum of first two elements. Then we check for each possible score if all subsequent pairs of operations can match it.

This solution iterates over the array, using each possible score from the first pair of numbers as a potential target score. It then attempts to match that score with subsequent pairs, counting the number of successful operations. We return the maximum number of operations achieved with any target score.

Code

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Complexity

Time Complexity: O(n^2), Space Complexity: O(1), where n is the number of elements in nums.

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Approach 2: Optimized Two-Pointer Approach

Using two pointers, we explore a potential cumulative score and validate it across the array. This method reduces unnecessary re-evaluation of past cumulative scores.

This code iterates with a target starting from the first two elements, checks forward if more such operations exist, and counts them. This eliminates redundant backward checks.

Code

C

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Java

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C#

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Complexity

Time Complexity: O(n^2), but optimized and often faster in practice, Space Complexity: O(1), since only constant space is used.

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Approach 3: Traversal

First, we calculate the sum of the first two elements, denoted as s. Then we traverse the array, taking two elements at a time. If their sum is not equal to s, we stop the traversal. Finally, we return the number of operations performed.

The time complexity is O(n), where n is the length of the array nums. The space complexity is O(1).

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Complexity Comparison

ApproachComplexity
Brute Force Approach

Time Complexity: O(n^2), Space Complexity: O(1), where n is the number of elements in nums.

Optimized Two-Pointer Approach

Time Complexity: O(n^2), but optimized and often faster in practice, Space Complexity: O(1), since only constant space is used.

Traversal—

Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Brute Force SimulationO(n)O(1)When implementing the problem exactly as described using simple iteration
Two-Pointer Sequential ScanO(n)O(1)Preferred solution for interviews; cleanly checks pair sums while scanning the array once

Video Solution

Maximum Number of Operations With the Same Score I - Python - Leetcode 3038 • CheatCode Ninja • 168 views views

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Frequently Asked Questions

Is Maximum Number of Operations With the Same Score I easy or hard?
Maximum Number of Operations With the Same Score I is classified as an Easy problem on LeetCode. The challenge mainly tests careful array traversal and recognizing that the first pair fixes the only valid score for all subsequent operations.
Maximum Number of Operations With the Same Score I Python/Java solution
In Python or Java, compute the first pair sum and iterate through the array with a loop that increments by two. For each step, check nums[i] + nums[i+1] against the target sum and increment the operation counter when they match. Stop when a mismatch appears or when fewer than two elements remain.
How to solve Maximum Number of Operations With the Same Score I in O(n)?
First calculate the target score using nums[0] + nums[1]. Then iterate through the array starting from index 2, checking pairs (i, i+1). If the pair sum equals the target, count the operation and move forward by two positions; otherwise stop immediately. The single pass ensures O(n) time complexity.
What is the best approach for Maximum Number of Operations With the Same Score I?
The best approach is a linear scan using a two-pointer style traversal. Compute the target score from the first pair of elements, then move through the array in steps of two and verify that every pair produces the same sum. This runs in O(n) time and O(1) space.
Is Maximum Number of Operations With the Same Score I asked at Google/Amazon/Meta?
Problems of this type commonly appear in coding interviews at large tech companies because they test array traversal and simulation skills. While the exact question may vary, the pattern of validating pair operations and maintaining a constant constraint is common in interview rounds.
What data structure is used in Maximum Number of Operations With the Same Score I?
The problem only requires a basic array traversal. No additional data structures such as hash maps or stacks are needed because the algorithm simply compares sums of consecutive pairs while scanning the array.
What is the time complexity of Maximum Number of Operations With the Same Score I?
The optimal solution runs in O(n) time because each element in the array is checked at most once while evaluating pair sums. The algorithm uses constant extra memory, giving O(1) space complexity.

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