Record each value's count and first occurrence. On every occurrence, update the best span when its count reaches or exceeds the current degree.
Complexity
O(n) time and O(n) space.
09 Sept 2026•Easy•Go / Python / TypeScript
Degree of an Array
Find the shortest subarray with the same frequency degree as the complete array.
Solution files
Go•degree-of-an-array/solution.go
package main
func findShortestSubArray(nums []int) int {
count, first := map[int]int{}, map[int]int{}
degree, shortest := 0, len(nums)
for index, value := range nums {
if _, ok := first[value]; !ok {
first[value] = index
}
count[value]++
span := index - first[value] + 1
if count[value] > degree {
degree, shortest = count[value], span
} else if count[value] == degree && span < shortest {
shortest = span
}
}
return shortest
}
Python•degree-of-an-array/solution.py
class Solution:
def findShortestSubArray(self, nums: list[int]) -> int:
count: dict[int, int] = {}
first: dict[int, int] = {}
degree, shortest = 0, len(nums)
for index, value in enumerate(nums):
first.setdefault(value, index)
count[value] = count.get(value, 0) + 1
span = index - first[value] + 1
if count[value] > degree: degree, shortest = count[value], span
elif count[value] == degree: shortest = min(shortest, span)
return shortest
TypeScript•degree-of-an-array/solution.ts
function findShortestSubArray(nums: number[]): number {
const count = new Map<number, number>();
const first = new Map<number, number>();
let degree = 0;
let shortest = nums.length;
nums.forEach((value, index) => {
if (!first.has(value)) first.set(value, index);
const nextCount = (count.get(value) ?? 0) + 1;
count.set(value, nextCount);
const span = index - first.get(value)! + 1;
if (nextCount > degree) { degree = nextCount; shortest = span; }
else if (nextCount === degree) shortest = Math.min(shortest, span);
});
return shortest;
}