3318 - Find X-Sum of All K-Long Subarrays I
problem
You are given an array nums of n integers and two integers k and x.
The x-sum of an array is calculated by the following procedure:
- Count the occurrences of all elements in the array.
- Keep only the occurrences of the top
xmost frequent elements. If two elements have the same number of occurrences, the element with the bigger value is considered more frequent. - Calculate the sum of the resulting array.
Note that if an array has less than x distinct elements, its x-sum is the sum of the array.
Return an integer array answer of length n - k + 1 where answer[i] is the x-sum of the subarray nums[i..i + k - 1].
Example 1:
Input: nums = [1,1,2,2,3,4,2,3], k = 6, x = 2
Output: [6,10,12]
Explanation:
- For subarray
[1, 1, 2, 2, 3, 4], only elements 1 and 2 will be kept in the resulting array. Hence,answer[0] = 1 + 1 + 2 + 2. - For subarray
[1, 2, 2, 3, 4, 2], only elements 2 and 4 will be kept in the resulting array. Hence,answer[1] = 2 + 2 + 2 + 4. Note that 4 is kept in the array since it is bigger than 3 and 1 which occur the same number of times. - For subarray
[2, 2, 3, 4, 2, 3], only elements 2 and 3 are kept in the resulting array. Hence,answer[2] = 2 + 2 + 2 + 3 + 3.
Example 2:
Input: nums = [3,8,7,8,7,5], k = 2, x = 2
Output: [11,15,15,15,12]
Explanation:
Since k == x, answer[i] is equal to the sum of the subarray nums[i..i + k - 1].
Constraints:
1 <= n == nums.length <= 501 <= nums[i] <= 501 <= x <= k <= nums.length
submission
use std::collections::{HashMap, BinaryHeap};
// one liner just for fun
// should be straight forward to read
impl Solution {
pub fn find_x_sum(nums: Vec<i32>, k: i32, x: i32) -> Vec<i32> {
nums.windows(k as usize)
.map(|win| {
win.iter()
.fold(HashMap::<i32, i32>::new(), |mut count, &val| {
*count.entry(val).or_default() += 1;
count
})
.into_iter()
.map(|(v, c)| (c, v))
.collect::<BinaryHeap<_>>()
.into_sorted_vec()
.into_iter()
.rev()
.take(x as usize)
.map(|(c, v)| c * v)
.sum()
})
.collect()
}
}