https://leetcode.com/problems/degree-of-an-array/description/
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: [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: [1,2,2,3,1,4,2] Output: 6
Note:
nums.length will be between 1 and 50,000.nums[i] will be an integer between 0 and 49,999.public int findShortestSubArray(int[] nums) { Map<Integer, Integer> left = new HashMap(), right = new HashMap(), count = new HashMap(); for (int i = 0; i < nums.length; i++) { int x = nums[i]; if (left.get(x) == null) left.put(x, i); right.put(x, i); count.put(x, count.getOrDefault(x, 0) + 1); } int ans = nums.length; int degree = Collections.max(count.values()); for (int x: count.keySet()) { if (count.get(x) == degree) { ans = Math.min(ans, right.get(x) - left.get(x) + 1); } } return ans; }
public int findShortestSubArray(int[] nums) {
Map<Integer, Element> map = new HashMap<>();
int max = scan(nums, map);
int result = Integer.MAX_VALUE;
for (Element el : map.values()) {
if (el.count == max) {
// 1,2
result = Math.min(result, el.lastPos - el.firstPos + 1);
}
}
return result;
}
private int scan(int[] nums, Map<Integer, Element> map) {
int max = 0;
for (int i = 0; i < nums.length; i++) {
int value = nums[i];
Element el = map.get(value);
if (el == null) {
el = new Element(i, i, 1);
map.put(value, el);
} else {
el.lastPos = i;
el.count++;
}
max = Math.max(max, el.count);
}
return max;
}
private static class Element {
public int firstPos, lastPos, count;
public Element(int firstPos, int lastPos, int count) {
super();
this.firstPos = firstPos;
this.lastPos = lastPos;
this.count = count;
}
}
