Degree of an Array

Degree of an Array - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,2,3,1]

› Output: 2

💡 Note: The degree is 2 (both 1 and 2 appear twice). The shortest subarray with degree 2 is [2,2] with length 2.

Example 2 — Multiple Same Degree

$ Input: nums = [1,2,2,3,1]

› Output: 2

💡 Note: Elements 1 and 2 both have degree 2. Element 1 spans from index 0 to 4 (length 5), element 2 spans from index 1 to 2 (length 2). Shortest is 2.

Example 3 — All Different Elements

$ Input: nums = [1,2,3,4,5]

› Output: 1

💡 Note: All elements appear once, so degree is 1. Any single element subarray has degree 1, so minimum length is 1.

Constraints

1 ≤ nums.length ≤ 50,000
0 ≤ nums[i] ≤ 49,999

Visualization

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Asked in

G Google 15 F Facebook 12 A Amazon 8

The key insight is that the shortest subarray with maximum degree must span from the first to last occurrence of some most frequent element. The optimal approach uses a hash map to track frequencies and positions in one pass. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force

⏱️ Time: O(n³) Space: O(n)

Generate all possible contiguous subarrays, calculate the degree of each, and track the shortest subarray that has the same degree as the original array.

Two-Pass Hash Map

⏱️ Time: O(n) Space: O(n)

Use two passes: first to find the degree and elements with maximum frequency, second to find the shortest subarray containing all occurrences of any max-frequency element.

Brute Force — Algorithm Steps

Calculate degree of original array
Generate all subarrays
For each subarray, calculate its degree
Track shortest subarray with maximum degree

Visualization

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Step-by-Step Walkthrough

Calculate Original Degree

Count frequency of each element in full array

Generate All Subarrays

Create every possible contiguous subarray

Find Shortest

Track shortest subarray with same degree

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int numsSize) {
    int freq[50001] = {0};
    for (int i = 0; i < numsSize; i++) {
        freq[nums[i]]++;
    }
    int originalDegree = 0;
    for (int i = 0; i < 50001; i++) {
        if (freq[i] > originalDegree) {
            originalDegree = freq[i];
        }
    }
    int minLength = numsSize;
    
    for (int i = 0; i < numsSize; i++) {
        for (int j = i; j < numsSize; j++) {
            int subFreq[50001] = {0};
            for (int k = i; k <= j; k++) {
                subFreq[nums[k]]++;
            }
            int degree = 0;
            for (int k = 0; k < 50001; k++) {
                if (subFreq[k] > degree) {
                    degree = subFreq[k];
                }
            }
            if (degree == originalDegree) {
                int len = j - i + 1;
                if (len < minLength) {
                    minLength = len;
                }
            }
        }
    }
    
    return minLength;
}

int main() {
    char line[100000];
    fgets(line, sizeof(line), stdin);
    
    int nums[50000];
    int size = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        nums[size++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, size);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops: O(n²) subarrays × O(n) to calculate degree of each

⚠ Quadratic Growth

Space Complexity

O(n)

Hash map to store frequency count for each subarray

⚡ Linearithmic Space

89.0K Views

Medium Frequency

~15 min Avg. Time

2.1K Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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