You are given an integer array arr and an integer requests.
Repeat the following operation exactly requests times:
- Find the current maximum value and the current minimum value in
arr. - Add their sum to the answer.
- Choose one occurrence of the maximum value and decrease it by 1.
Return the final accumulated answer.
Complete the function sumMaxPlusMinAfterOperations in the editor below.
sumMaxPlusMinAfterOperations has the following parameters:
int[] arr: the initial valuesint requests: the number of operations
Returns
long: the accumulated sum
Examples
01 · Example 1
arr = [1, 2] requests = 2 return = 5
First add 1 + 2 = 3 and decrement the 2 to 1. Then add 1 + 1 = 2. The total is 5.
02 · Example 2
arr = [3, 3, 3] requests = 1 return = 6
The current maximum and minimum are both 3, so the answer increases by 6.
Constraints
1 <= arr.length <= 1051 <= arr[i] <= 1090 <= requests <= 109- Each operation decreases one occurrence of the current maximum value by exactly 1.
- Use a wide enough integer type (e.g.,
long) for the accumulated total.
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public long sumMaxPlusMinAfterOperations(int[] arr, int requests) {
// write your code here
}
arr[1, 2]
requests2
expected5
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