Degree of an Array - Practice Coding Problems

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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Understanding the Visualization

Input Analysis

Array [1,2,2,3,1] has degree 2 (max frequency)

Find Candidates

Elements with max frequency: 1 and 2

Calculate Spans

Find shortest span containing all occurrences

Key Takeaway

🎯 Key Insight: The shortest subarray with maximum degree spans from first to last occurrence of the most frequent element with minimum span

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

Approach	Time	Space	Notes
✓ Two-Pass Hash Map	O(n)	O(n)	First pass counts frequencies, second pass finds shortest subarray with max degree
Brute Force	O(n³)	O(n)	Check all possible subarrays to find shortest with maximum degree

Two-Pass Hash Map — Algorithm Steps

First pass: count frequency of each element
Find maximum frequency (degree)
Second pass: for each max-frequency element, find shortest span
Return minimum span length

Visualization

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Step-by-Step Walkthrough

First Pass

Count frequencies and track first/last positions

Find Max Degree

Identify elements with maximum frequency

Calculate Spans

Find shortest span among max-frequency elements

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int freq[50001] = {0};
    int first[50001];
    int last[50001];
    
    // Initialize first positions to -1
    for (int i = 0; i < 50001; i++) {
        first[i] = -1;
    }
    
    // First pass: count frequencies and track positions
    for (int i = 0; i < numsSize; i++) {
        int num = nums[i];
        if (first[num] == -1) {
            first[num] = i;
        }
        last[num] = i;
        freq[num]++;
    }
    
    // Find maximum frequency (degree)
    int degree = 0;
    for (int i = 0; i < 50001; i++) {
        if (freq[i] > degree) {
            degree = freq[i];
        }
    }
    
    // Find shortest subarray with maximum degree
    int minLength = numsSize;
    for (int i = 0; i < 50001; i++) {
        if (freq[i] == degree) {
            int length = last[i] - first[i] + 1;
            if (length < minLength) {
                minLength = length;
            }
        }
    }
    
    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(n)

Two passes through the array: O(n) + O(n) = O(n)

✓ Linear Growth

Space Complexity

O(n)

Hash map to store frequency and first/last positions of each element

⚡ Linearithmic Space

89.0K Views

Medium Frequency

~15 min Avg. Time

2.1K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Two-Pass Hash Map — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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