Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.
Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.
Example 1:
Input: nums = [1,2,2,3,1] Output: 2 Explanation: The input array has a degree of 2 because both elements 1 and 2 appear twice. Of the subarrays that have the same degree: [1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2] The shortest length is 2. So return 2.
Example 2:
Input: nums = [1,2,2,3,1,4,2] Output: 6 Explanation: The degree is 3 because the element 2 is repeated 3 times. So [2,2,3,1,4,2] is the shortest subarray, therefore returning 6.
Constraints:
nums.length will be between 1 and 50,000.nums[i] will be an integer between 0 and 49,999.struct Solution;
struct Degree {
left: usize,
right: usize,
count: usize,
}
use std::collections::HashMap;
use std::usize;
impl Solution {
fn find_shortest_sub_array(nums: Vec<i32>) -> i32 {
let mut hm: HashMap<i32, Degree> = HashMap::new();
let n = nums.len();
let mut max_degree: usize = 0;
for i in 0..n {
let x = nums[i];
let e = hm.entry(x).or_insert(Degree {
left: i,
right: i,
count: 0,
});
e.left = usize::min(e.left, i);
e.right = usize::max(e.right, i);
e.count += 1;
max_degree = usize::max(e.count, max_degree);
}
let mut min_width: usize = n;
for d in hm.values() {
if d.count == max_degree {
min_width = usize::min(d.right - d.left + 1, min_width);
}
}
min_width as i32
}
}
#[test]
fn test() {
let nums = vec![1, 2, 2, 3, 1];
assert_eq!(Solution::find_shortest_sub_array(nums), 2);
let nums = vec![1, 2, 2, 3, 1, 4, 2];
assert_eq!(Solution::find_shortest_sub_array(nums), 6);
}