LeetCode 2202: Maximize the Topmost Element After K Moves Solution

Master LeetCode problem 2202 (Maximize the Topmost Element After K Moves), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2202. Maximize the Topmost Element After K Moves

Medium

Array Greedy

Problem Explanation

Explanation

To maximize the topmost element after k moves, we can simulate the process by keeping track of the maximum element in the current pile. We need to consider the different cases where we either remove or add elements to the pile based on the given rules.

If k < n (where n is the length of the initial pile), we can remove elements from the top of the pile starting from the largest element until we have made k moves. The maximum element left in the pile will be the answer.
If k >= n, we need to calculate the number of cycles required to reach the final state. After each cycle, the maximum element in the pile increases. We then calculate the remaining moves after the cycles and adjust the maximum element accordingly.

Time Complexity: O(n) where n is the length of the initial pile. Space Complexity: O(1)

Solution Code

class Solution {
    public int maximizeTopElementAfterKMoves(int[] nums, int k) {
        int n = nums.length;
        int maxElement = nums[0];
        
        if (k < n) {
            for (int i = 1; i < k; i++) {
                if (nums[i] > maxElement) {
                    maxElement = nums[i];
                }
            }
            return maxElement;
        } else {
            int cycles = k / n;
            maxElement = Math.max(maxElement, getMaxElement(nums));
            k %= n;
            for (int i = 0; i < k; i++) {
                if (nums[i] > maxElement) {
                    maxElement = nums[i];
                }
            }
            return maxElement + cycles;
        }
    }
    
    private int getMaxElement(int[] nums) {
        int max = nums[0];
        for (int i = 1; i < nums.length; i++) {
            max = Math.max(max, nums[i]);
        }
        return max;
    }
}

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