You are given a 0-indexed integer array nums
and an integer k
. Your task is to perform the following operation exactly k
times in order to maximize your score:
- Select an element
m
from nums
. - Remove the selected element
m
from the array. - Add a new element with a value of
m + 1
to the array. - Increase your score by
m
.
Return the maximum score you can achieve after performing the operation exactly k
times.
Example 1:
Input: nums = [1,2,3,4,5], k = 3
Output: 18
Explanation: We need to choose exactly 3 elements from nums to maximize the sum.
For the first iteration, we choose 5. Then sum is 5 and nums = [1,2,3,4,6]
For the second iteration, we choose 6. Then sum is 5 + 6 and nums = [1,2,3,4,7]
For the third iteration, we choose 7. Then sum is 5 + 6 + 7 = 18 and nums = [1,2,3,4,8]
So, we will return 18.
It can be proven, that 18 is the maximum answer that we can achieve.
Example 2:
Input: nums = [5,5,5], k = 2
Output: 11
Explanation: We need to choose exactly 2 elements from nums to maximize the sum.
For the first iteration, we choose 5. Then sum is 5 and nums = [5,5,6]
For the second iteration, we choose 6. Then sum is 5 + 6 = 11 and nums = [5,5,7]
So, we will return 11.
It can be proven, that 11 is the maximum answer that we can achieve.
Constraints:
1 <= nums.length <= 100
1 <= nums[i] <= 100
1 <= k <= 100
Solution: Greedy
Always to chose the largest element from the array.
We can find the largest element of the array m, then the total score will be
m + (m + 1) + (m + 2) + … + (m + k – 1),
We can use summation formula of arithmetic sequence to compute that in O(1)
ans = (m + (m + k – 1)) * k / 2
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int maximizeSum(vector<int>& nums, int k) { int m = *max_element(begin(nums), end(nums)); return (m + m + k - 1) * k / 2; } }; |