findShortestSubArray.java

/*
Intuition and Algorithm

An array that has degree d, must have some element x occur d times. If some 
subarray has the same degree, then some element x (that occured d times), 
still occurs d times. The shortest such subarray would be from the first 
occurrence of x until the last occurrence.

For each element in the given array, let's know left, the index of its 
first occurrence; and right, the index of its last occurrence. 
For example, with nums = [1,2,3,2,5] we have left[2] = 1 and right[2] = 3.

Then, for each element x that occurs the maximum number of times, 
right[x] - left[x] + 1 will be our candidate answer, and we'll take the minimum of those candidates.
*/
import java.util.*;

public class findShortestSubArray{
	public static void main(String[] args){
		int array[]={1,2,2,3,1,4,2};
		int degree=find_degree(array);
		System.out.println(degree);

	}

	public static int find_degree(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;
    }
}