README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/3000-3099/3038.Maximum%20Number%20of%20Operations%20With%20the%20Same%20Score%20I/README_EN.md	1201	Biweekly Contest 124 Q1	Array Simulation

3038. Maximum Number of Operations With the Same Score I

中文文档

Description

Given an array of integers called nums, you can perform the following operation while nums contains at least 2 elements:

• Choose the first two elements of nums and delete them.

The score of the operation is the sum of the deleted elements.

Your task is to find the maximum number of operations that can be performed, such that all operations have the same score.

Return the maximum number of operations possible that satisfy the condition mentioned above.

 

Example 1:

Input: nums = [3,2,1,4,5]
Output: 2
Explanation: We perform the following operations:
- Delete the first two elements, with score 3 + 2 = 5, nums = [1,4,5].
- Delete the first two elements, with score 1 + 4 = 5, nums = [5].
We are unable to perform any more operations as nums contain only 1 element.

Example 2:

Input: nums = [3,2,6,1,4]
Output: 1
Explanation: We perform the following operations:
- Delete the first two elements, with score 3 + 2 = 5, nums = [6,1,4].
We are unable to perform any more operations as the score of the next operation isn't the same as the previous one.

 

Constraints:

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

Solutions

Solution 1: 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 $O(1)$.

Python3

class Solution:
    def maxOperations(self, nums: List[int]) -> int:
        s = nums[0] + nums[1]
        ans, n = 0, len(nums)
        for i in range(0, n, 2):
            if i + 1 == n or nums[i] + nums[i + 1] != s:
                break
            ans += 1
        return ans

Java

class Solution {
    public int maxOperations(int[] nums) {
        int s = nums[0] + nums[1];
        int ans = 0, n = nums.length;
        for (int i = 0; i + 1 < n && nums[i] + nums[i + 1] == s; i += 2) {
            ++ans;
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxOperations(vector<int>& nums) {
        int s = nums[0] + nums[1];
        int ans = 0, n = nums.size();
        for (int i = 0; i + 1 < n && nums[i] + nums[i + 1] == s; i += 2) {
            ++ans;
        }
        return ans;
    }
};

Go

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
}

TypeScript

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;
}