2023-11-15 Daily Challenge
Today I have done leetcode's November LeetCoding Challenge with cpp
.
November LeetCoding Challenge 15
Description
Maximum Element After Decreasing and Rearranging
You are given an array of positive integers arr
. Perform some operations (possibly none) on arr
so that it satisfies these conditions:
- The value of the first element in
arr
must be1
. - The absolute difference between any 2 adjacent elements must be less than or equal to
1
. In other words,abs(arr[i] - arr[i - 1]) <= 1
for eachi
where1 <= i < arr.length
(0-indexed).abs(x)
is the absolute value ofx
.
There are 2 types of operations that you can perform any number of times:
- Decrease the value of any element of
arr
to a smaller positive integer. - Rearrange the elements of
arr
to be in any order.
Return the maximum possible value of an element in arr
after performing the operations to satisfy the conditions.
Example 1:
Input: arr = [2,2,1,2,1] Output: 2 Explanation: We can satisfy the conditions by rearrangingarr
so it becomes[1,2,2,2,1]
. The largest element inarr
is 2.
Example 2:
Input: arr = [100,1,1000] Output: 3 Explanation: One possible way to satisfy the conditions is by doing the following: 1. Rearrangearr
so it becomes[1,100,1000]
. 2. Decrease the value of the second element to 2. 3. Decrease the value of the third element to 3. Nowarr = [1,2,3], which
satisfies the conditions. The largest element inarr is 3.
Example 3:
Input: arr = [1,2,3,4,5] Output: 5 Explanation: The array already satisfies the conditions, and the largest element is 5.
Constraints:
1 <= arr.length <= 105
1 <= arr[i] <= 109
Solution
class Solution {
public:
int maximumElementAfterDecrementingAndRearranging(vector<int>& arr) {
sort(arr.begin(), arr.end());
arr[0] = 1;
for(int i = 1; i < arr.size(); ++i) {
if(arr[i] - arr[i - 1] > 1) {
arr[i] = arr[i - 1] + 1;
}
}
return arr.back();
}
};
// Accepted
// 49/49 cases passed (86 ms)
// Your runtime beats 21.64 % of cpp submissions
// Your memory usage beats 56.72 % of cpp submissions (51.6 MB)