Maximum Number of Operations With the Same Score I

• If the length of the input array nums is less than 2, return 0.
• Convert the nums array into a java.util.List for easier element removal.
• Remove the first two elements from the list, calculate their sum, and store it as targetScore.
• Initialize an operations counter to 1.
• Start a while loop that continues as long as the list contains at least two elements.
• Inside the loop, calculate the sum of the first two elements of the current list.
• If the sum equals targetScore, increment the operations counter and remove the first two elements from the list.
• If the sum does not equal targetScore, break the loop.
• After the loop terminates, return the final operations count.

This method provides a straightforward translation of the problem's requirements into code. We begin by handling the edge case where the array is too small for any operations. Then, we convert the array to a List to leverage its dynamic size and removal capabilities. The score of the first operation (sum of the first two elements) sets the targetScore for all subsequent operations. We initialize our operation count to 1. The main logic resides in a while loop that continues as long as we can form a pair of elements. In each iteration, we peek at the sum of the first two elements. If it matches the targetScore, we increment our count and remove them. The critical part is that removing an element from the beginning of an ArrayList is costly because it requires shifting all subsequent elements one position to the left. This leads to a quadratic time complexity, making it unsuitable for large inputs, although it passes for the given constraints.

1import java.util.ArrayList;2import java.util.List;3import java.util.stream.Collectors;4import java.util.Arrays;5 6class Solution {7    public int maxOperations(int[] nums) {8        if (nums.length < 2) {9            return 0;10        }11 12        // Convert array to a list for easy removal of elements.13        List<Integer> numList = new ArrayList<>();14        for (int num : nums) {15            numList.add(num);16        }17        18        // Perform the first operation to set the target score.19        int first = numList.remove(0);20        int second = numList.remove(0);21        int targetScore = first + second;22        int operations = 1;23 24        // Continue performing operations as long as possible.25        while (numList.size() >= 2) {26            int currentFirst = numList.get(0);27            int currentSecond = numList.get(1);28            if (currentFirst + currentSecond == targetScore) {29                operations++;30                // These remove operations are inefficient (O(n) each).31                numList.remove(0);32                numList.remove(0);33            } else {34                break;35            }36        }37        38        return operations;39    }40}

• Handle the edge case: If the array nums has fewer than two elements, return 0.
• Calculate the targetScore by summing the first two elements, nums[0] and nums[1].
• Initialize an operationsCount to 1, for the first operation.
• Use a for loop to iterate through the rest of the array, starting from index i = 2 and incrementing by 2 in each step (i += 2).
• The loop condition must ensure that we can always access a pair of elements, i.e., i + 1 < nums.length.
• Inside the loop, calculate the sum of the current pair: nums[i] + nums[i+1].
• If this sum equals targetScore, increment operationsCount.
• If the sum does not equal targetScore, break the loop immediately.
• After the loop finishes, return the final operationsCount.

This optimal approach simulates the process without the overhead of actually removing elements from the data structure. We use an index to keep track of which pair of elements we are currently considering. First, we handle the base case where the array has fewer than two elements. We then calculate the targetScore from the first two elements (nums[0] + nums[1]) and count this as our first operation. After that, we iterate through the rest of the array with a for loop, advancing our index by two in each step to move from one pair to the next. The loop starts at index 2 and continues as long as there's a full pair of elements left to check. In each iteration, we sum the current pair and compare it to the targetScore. If they match, we increment our operation count. If at any point the sum doesn't match, we know we cannot continue, so we break the loop. This method is highly efficient as it only requires a single pass over the array.

1class Solution {2    public int maxOperations(int[] nums) {3        // If the array has fewer than 2 elements, no operations can be performed.4        if (nums.length < 2) {5            return 0;6        }7        8        // The score for all operations is determined by the first one.9        int targetScore = nums[0] + nums[1];10        int operationsCount = 1;11        12        // Iterate through the rest of the array, two elements at a time.13        for (int i = 2; i + 1 < nums.length; i += 2) {14            // Check if the current pair's sum matches the target score.15            if (nums[i] + nums[i+1] == targetScore) {16                operationsCount++;17            } else {18                // If the score doesn't match, we can't perform any more operations.19                break;20            }21        }22        23        return operationsCount;24    }25}