Maximum Number of Operations With the Same Score I solution and explanation
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The problem
You are given an array of integers nums. Consider the following operation:
- Delete the first two elements
numsand 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 <= 1001 <= 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
def max_operations(nums):
max_ops = 0
for i in range(len(nums) - 1):
target = nums[i] + nums[i + 1]
count = 0
for j in range(0, len(nums) - 1, 2):
if j + 1 < len(nums) and nums[j] + nums[j + 1] == target:
count += 1
else:
break
max_ops = max(max_ops, count)
return max_ops
if __name__ == "__main__":
nums = [3, 2, 1, 4, 5]
print(max_operations(nums))public class MaxOperations {
public static int maxOperations(int[] nums) {
int maxOps = 0;
for (int i = 0; i < nums.length - 1; i++) {
int target = nums[i] + nums[i + 1];
int count = 0;
for (int j = 0; j < nums.length - 1; j += 2) {
if (j + 1 < nums.length && nums[j] + nums[j + 1] == target) {
count++;
} else {
break;
}
}
maxOps = Math.max(maxOps, count);
}
return maxOps;
}
public static void main(String[] args) {
int[] nums = {3, 2, 1, 4, 5};
System.out.println(maxOperations(nums));
}
}#include <iostream>
#include <vector>
using namespace std;
int maxOperations(vector<int>& nums) {
int max_ops = 0;
for (int i = 0; i < nums.size() - 1; ++i) {
int target = nums[i] + nums[i + 1];
int count = 0;
for (int j = 0; j < nums.size() - 1; j += 2) {
if (j + 1 < nums.size() && nums[j] + nums[j + 1] == target) {
count++;
} else {
break;
}
}
max_ops = max(max_ops, count);
}
return max_ops;
}
int main() {
vector<int> nums = {3, 2, 1, 4, 5};
cout << maxOperations(nums) << endl;
return 0;
}function maxOperations(nums) {
let maxOps = 0;
for (let i = 0; i < nums.length - 1; i++) {
const target = nums[i] + nums[i + 1];
let count = 0;
for (let j = 0; j < nums.length - 1; j += 2) {
if (j + 1 < nums.length && nums[j] + nums[j + 1] === target) {
count++;
} else {
break;
}
}
maxOps = Math.max(maxOps, count);
}
return maxOps;
}
console.log(maxOperations([3, 2, 1, 4, 5]));#include <stdio.h>
int maxOperations(int* nums, int numsSize) {
int max_ops = 0;
for (int i = 0; i < numsSize - 1; i++) {
int target = nums[i] + nums[i + 1];
int count = 0;
for (int j = 0; j < numsSize - 1; j += 2) {
if (j + 1 < numsSize && nums[j] + nums[j + 1] == target) {
count++;
} else {
break;
}
}
if (count > max_ops) {
max_ops = count;
}
}
return max_ops;
}
int main() {
int nums[] = {3, 2, 1, 4, 5};
int size = sizeof(nums) / sizeof(nums[0]);
printf("%d\n", maxOperations(nums, size));
return 0;
}using System;
public class Program
{
public static int MaxOperations(int[] nums)
{
int maxOps = 0;
for (int i = 0; i < nums.Length - 1; i++)
{
int target = nums[i] + nums[i + 1];
int count = 0;
for (int j = 0; j < nums.Length - 1; j += 2)
{
if (j + 1 < nums.Length && nums[j] + nums[j + 1] == target)
{
count++;
}
else
{
break;
}
}
maxOps = Math.Max(maxOps, count);
}
return maxOps;
}
public static void Main()
{
int[] nums = { 3, 2, 1, 4, 5 };
Console.WriteLine(MaxOperations(nums));
}
}Python code iterating through nums to set targets and calculate operation count considering each score achievable by the beginning pair for maximum operations.
Time Complexity: O(n^2), Space Complexity: O(1), for n elements in nums.
Approach 2: Optimized Two-Pointer Approach
def max_operations(nums):
max_ops = 0
for i in range(len(nums) - 1):
target = nums[i] + nums[i + 1]
count = 0
for j in range(i + 2, len(nums) - 1, 2):
if j + 1 < len(nums) and nums[j] + nums[j + 1] == target:
count += 1
else:
break
max_ops = max(max_ops, count + 1)
return max_ops
if __name__ == "__main__":
nums = [3, 2, 1, 4, 5]
print(max_operations(nums))public class MaxOperations {
public static int maxOperations(int[] nums) {
int maxOps = 0;
for (int i = 0; i < nums.length - 1; i++) {
int target = nums[i] + nums[i + 1];
int count = 0;
for (int j = i + 2; j < nums.length - 1; j += 2) {
if (j + 1 < nums.length && nums[j] + nums[j + 1] == target) {
count++;
} else {
break;
}
}
maxOps = Math.max(maxOps, count + 1);
}
return maxOps;
}
public static void main(String[] args) {
int[] nums = {3, 2, 1, 4, 5};
System.out.println(maxOperations(nums));
}
}#include <iostream>
#include <vector>
using namespace std;
int maxOperations(vector<int>& nums) {
int max_ops = 0;
for (int i = 0; i < nums.size() - 1; ++i) {
int target = nums[i] + nums[i + 1];
int count = 0;
for (int j = i + 2; j < nums.size() - 1; j += 2) {
if (j + 1 < nums.size() && nums[j] + nums[j + 1] == target) {
count++;
} else {
break;
}
}
max_ops = max(max_ops, count + 1);
}
return max_ops;
}
int main() {
vector<int> nums = {3, 2, 1, 4, 5};
cout << maxOperations(nums) << endl;
return 0;
}function maxOperations(nums) {
let maxOps = 0;
for (let i = 0; i < nums.length - 1; i++) {
const target = nums[i] + nums[i + 1];
let count = 0;
for (let j = i + 2; j < nums.length - 1; j += 2) {
if (j + 1 < nums.length && nums[j] + nums[j + 1] === target) {
count++;
} else {
break;
}
}
maxOps = Math.max(maxOps, count + 1);
}
return maxOps;
}
console.log(maxOperations([3, 2, 1, 4, 5]));#include <stdio.h>
int maxOperations(int* nums, int numsSize) {
int max_ops = 0;
for (int i = 0; i < numsSize - 1; i++) {
int target = nums[i] + nums[i + 1];
int count = 0;
for (int j = i + 2; j < numsSize - 1; j += 2) {
if (j + 1 < numsSize && nums[j] + nums[j + 1] == target) {
count++;
} else {
break;
}
}
if (count + 1 > max_ops) {
max_ops = count + 1;
}
}
return max_ops;
}
int main() {
int nums[] = {3, 2, 1, 4, 5};
int size = sizeof(nums) / sizeof(nums[0]);
printf("%d\n", maxOperations(nums, size));
return 0;
}using System;
public class Program
{
public static int MaxOperations(int[] nums)
{
int maxOps = 0;
for (int i = 0; i < nums.Length - 1; i++)
{
int target = nums[i] + nums[i + 1];
int count = 0;
for (int j = i + 2; j < nums.Length - 1; j += 2)
{
if (j + 1 < nums.Length && nums[j] + nums[j + 1] == target)
{
count++;
}
else
{
break;
}
}
maxOps = Math.Max(maxOps, count + 1);
}
return maxOps;
}
public static void Main()
{
int[] nums = { 3, 2, 1, 4, 5 };
Console.WriteLine(MaxOperations(nums));
}
}This Python solution makes a first pass for each potential score starting point, and checks forwards for subsequent score consistency without tracking unnecessary previous pairs.
Time Complexity: O(n^2) with aforementioned optimizations, Space Complexity: O(1).
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).
function maxOperations(nums: number[]): number {
const s = nums[0] + nums[1];
const n = nums.length;
let ans = 0;
for (let i = 0; i + 1 < n && nums[i] + nums[i + 1] === s; i += 2) {
++ans;
}
return ans;
}func maxOperations(nums []int) (ans int) {
s, n := nums[0]+nums[1], len(nums)
for i := 0; i+1 < n && nums[i]+nums[i+1] == s; i += 2 {
ans++
}
return
}Complexity comparison
| Approach | Complexity |
|---|---|
| Brute Force Approach | Time Complexity: O(n^2), Space Complexity: O(1), for n elements in |
| Optimized Two-Pointer Approach | Time Complexity: O(n^2) with aforementioned optimizations, Space Complexity: O(1). |
| Traversal | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Simulation | O(n) | O(1) | When implementing the problem exactly as described using simple iteration |
| Two-Pointer Sequential Scan | O(n) | O(1) | Preferred solution for interviews; cleanly checks pair sums while scanning the array once |
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.