2202. Maximize the Topmost Element After K Moves

2202. Maximize the Topmost Element After K Moves (Medium)

You are given a 0-indexed integer array nums representing the contents of a pile, where nums[0] is the topmost element of the pile.

In one move, you can perform either of the following:

• If the pile is not empty, remove the topmost element of the pile.
• If there are one or more removed elements, add any one of them back onto the pile. This element becomes the new topmost element.

You are also given an integer k, which denotes the total number of moves to be made.

Return the maximum value of the topmost element of the pile possible after exactly k moves. In case it is not possible to obtain a non-empty pile after k moves, return -1.

 

Example 1:

Input: nums = [5,2,2,4,0,6], k = 4
Output: 5
Explanation:
One of the ways we can end with 5 at the top of the pile after 4 moves is as follows:
- Step 1: Remove the topmost element = 5. The pile becomes [2,2,4,0,6].
- Step 2: Remove the topmost element = 2. The pile becomes [2,4,0,6].
- Step 3: Remove the topmost element = 2. The pile becomes [4,0,6].
- Step 4: Add 5 back onto the pile. The pile becomes [5,4,0,6].
Note that this is not the only way to end with 5 at the top of the pile. It can be shown that 5 is the largest answer possible after 4 moves.

Example 2:

Input: nums = [2], k = 1
Output: -1
Explanation: 
In the first move, our only option is to pop the topmost element of the pile.
Since it is not possible to obtain a non-empty pile after one move, we return -1.

 

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i], k <= 109

Companies:
American Express

Related Topics:
Array

Similar Questions:

• Gas Station (Medium)

Solution 1.

See comments in the code.

A note on the removing min(k - 1, N) elements case:

What if k > N + 1 -- there are still steps left after removing N elements and putting back the greatest one? We can always waste these steps by putting another element in and out. Since N >= 2 in this case, it's guaranteed to have another element to waste steps.

// OJ: https://leetcode.com/problems/maximize-the-topmost-element-after-k-moves/
// Author: github.com/lzl124631x
// Time: O(min(N, K))
// Space: O(1)
class Solution {
public:
    int maximumTop(vector<int>& A, int k) {
        int N = A.size();
        if (k == 0) return N >= 1 ? A[0] : -1; // if no moves allowed, return the topmost element if any
        if (k == 1) return N == 1 ? -1 : A[1]; // if only one move is allowed, we can only remove the topmost element
        if (N == 1) return k % 2 == 0 ? A[0] : -1; // if `N == 1`, we can return the topmost element if `k` is a even number (keep removing the topmost element and adding it back).
        int mx = *max_element(begin(A), begin(A) + min(k - 1, N)); // we can take `min(k-1, N)` elements and put back the largest one on the top
        if (k < N) mx = max(mx, A[k]); // If `k < N`, we can take all the topmost `k` elements and return the one left at the top
        return mx;
    }
};

Discuss

https://leetcode.com/problems/maximize-the-topmost-element-after-k-moves/discuss/1844102/