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
arrmust 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]) <= 1for eachiwhere1 <= 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
arrto a smaller positive integer. - Rearrange the elements of
arrto 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 rearrangingarrso it becomes[1,2,2,2,1]. The largest element inarris 2.
Example 2:
Input: arr = [100,1,1000] Output: 3 Explanation: One possible way to satisfy the conditions is by doing the following: 1. Rearrangearrso 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], whichsatisfies 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 <= 1051 <= 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)