Solution

Method 1 – Prefix Sum Simulation

Intuition

The score for each operation is the sum of the first two elements. To maximize the number of operations, try all possible initial scores (from the first two elements), and simulate the process for each, counting how many times you can keep removing pairs with the same score.

Approach

For each possible initial score (sum of the first two elements), simulate the process:
- Remove the first two elements and check if their sum equals the score.
- Continue as long as the sum matches and at least two elements remain.
Track the maximum number of operations found for any possible initial score.
Return the maximum number of operations.

Code

[{"language":"C++","code":"class Solution {\npublic:\n    int maxOperations(vector<int>& nums) {\n        int n = nums.size(), ans = 0;\n        for (int i = 1; i < n; ++i) {\n            int score = nums[0] + nums[i];\n            int cnt = 0, l = 0, r = i;\n            while (r + 1 < n) {\n                if (nums[l] + nums[r] == score) {\n                    cnt++;\n                    l = r + 1;\n                    r = l + 1;\n                } else {\n                    break;\n                }\n            }\n            ans = max(ans, cnt + 1);\n        }\n        return ans;\n    }\n};","lang":"cpp"},{"language":"Go","code":"func maxOperations(nums []int) int {\n    n, ans := len(nums), 0\n    for i := 1; i < n; i++ {\n        score := nums[0] + nums[i]\n        cnt, l, r := 0, 0, i\n        for r+1 < n {\n            if nums[l]+nums[r] == score {\n                cnt++\n                l = r + 1\n                r = l + 1\n            } else {\n                break\n            }\n        }\n        if cnt+1 > ans {\n            ans = cnt + 1\n        }\n    }\n    return ans\n}","lang":"go"},{"language":"Java","code":"class Solution {\n    public int maxOperations(int[] nums) {\n        int n = nums.length, ans = 0;\n        for (int i = 1; i < n; i++) {\n            int score = nums[0] + nums[i];\n            int cnt = 0, l = 0, r = i;\n            while (r + 1 < n) {\n                if (nums[l] + nums[r] == score) {\n                    cnt++;\n                    l = r + 1;\n                    r = l + 1;\n                } else {\n                    break;\n                }\n            }\n            ans = Math.max(ans, cnt + 1);\n        }\n        return ans;\n    }\n}","lang":"java"},{"language":"Kotlin","code":"class Solution {\n    fun maxOperations(nums: IntArray): Int {\n        val n = nums.size\n        var ans = 0\n        for (i in 1 until n) {\n            val score = nums[0] + nums[i]\n            var cnt = 0\n            var l = 0\n            var r = i\n            while (r + 1 < n) {\n                if (nums[l] + nums[r] == score) {\n                    cnt++\n                    l = r + 1\n                    r = l + 1\n                } else {\n                    break\n                }\n            }\n            ans = maxOf(ans, cnt + 1)\n        }\n        return ans\n    }\n}","lang":"kotlin"},{"language":"Python","code":"class Solution:\n    def maxOperations(self, nums: list[int]) -> int:\n        n = len(nums)\n        ans = 0\n        for i in range(1, n):\n            score = nums[0] + nums[i]\n            cnt, l, r = 0, 0, i\n            while r + 1 < n:\n                if nums[l] + nums[r] == score:\n                    cnt += 1\n                    l = r + 1\n                    r = l + 1\n                else:\n                    break\n            ans = max(ans, cnt + 1)\n        return ans","lang":"python"},{"language":"Rust","code":"impl Solution {\n    pub fn max_operations(nums: Vec<i32>) -> i32 {\n        let n = nums.len();\n

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