LeetCode-in-Java
697. Degree of an Array
Easy
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.lengthwill be between 1 and 50,000.nums[i]will be an integer between 0 and 49,999.
Solution
import java.util.HashMap;
import java.util.Map;
public class Solution {
private static class Value {
int count;
int start;
int end;
public Value(int c, int s, int e) {
count = c;
start = s;
end = e;
}
}
public int findShortestSubArray(int[] nums) {
int max = 1;
Map<Integer, Value> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int j = nums[i];
if (map.containsKey(j)) {
Value v = map.get(j);
v.count++;
max = Math.max(max, v.count);
v.end = i;
} else {
map.put(j, new Value(1, i, i));
}
}
int min = Integer.MAX_VALUE;
for (Map.Entry<Integer, Value> entry : map.entrySet()) {
Value v = entry.getValue();
if (v.count == max) {
min = Math.min(min, v.end - v.start);
}
}
return min + 1;
}
}