## 697. Degree of an Array

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.

## Rust Solution

``````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);
}
``````

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