Maximum Sum of Almost Unique Subarray - Practice Coding Problems

Maximum Sum of Almost Unique Subarray - Problem

You are given an integer array nums and two positive integers m and k.

Return the maximum sum out of all almost unique subarrays of length k of nums. If no such subarray exists, return 0.

A subarray of nums is almost unique if it contains at least m distinct elements.

A subarray is a contiguous non-empty sequence of elements within an array.

Input & Output

Example 1 — Basic Case

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

› Output: 6

💡 Note: Check subarrays of length 3: [2,2,1] has 2 distinct (sum=5), [2,1,2] has 2 distinct (sum=5), [1,2,3] has 3 distinct (sum=6). Maximum is 6.

Example 2 — Not Enough Distinct

$ Input: nums = [1,1,1,7], m = 3, k = 2

› Output: 0

💡 Note: All subarrays of length 2: [1,1] has 1 distinct, [1,1] has 1 distinct, [1,7] has 2 distinct. None have at least 3 distinct elements.

Example 3 — All Elements Distinct

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

› Output: 9

💡 Note: Subarrays: [1,2,3] has 3 distinct (sum=6), [2,3,4] has 3 distinct (sum=9). Maximum is 9.

Constraints

1 ≤ k ≤ nums.length ≤ 10⁴
1 ≤ m ≤ k
-10⁴ ≤ nums[i] ≤ 10⁴

Visualization

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Understanding the Visualization

Input

Array nums, minimum distinct count m, subarray length k

Process

Check each subarray of length k for distinct element count

Output

Maximum sum among valid subarrays

Key Takeaway

🎯 Key Insight: Use sliding window with frequency map to efficiently track distinct elements in each subarray

Asked in

a Amazon 15 G Google 12 M Microsoft 8

The key insight is to use a sliding window with a frequency map to efficiently track distinct elements. The sliding window approach maintains a window of size k and uses a hash map to count element frequencies, allowing us to determine if we have at least m distinct elements in O(1) time per window. Best approach is sliding window. Time: O(n), Space: O(k)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(n × k)	O(k)	Check all possible subarrays of length k
Sliding Window	O(n)	O(k)	Use sliding window with frequency map to efficiently track distinct elements

Brute Force — Algorithm Steps

Generate all subarrays of length k
For each subarray, count distinct elements
If distinct count >= m, calculate sum and update maximum

Visualization

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Step-by-Step Walkthrough

Generate Subarrays

Create all possible subarrays of length k

Count Distinct

For each subarray, count unique elements

Track Maximum

Keep track of maximum sum where distinct >= m

Code -

solution.c — C

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

long long solution(int* nums, int numsSize, int m, int k) {
    long long maxSum = 0;
    
    for (int i = 0; i <= numsSize - k; i++) {
        int distinct[100001] = {0};
        int distinctCount = 0;
        long long currentSum = 0;
        
        for (int j = i; j < i + k; j++) {
            if (distinct[nums[j] + 50000] == 0) {
                distinct[nums[j] + 50000] = 1;
                distinctCount++;
            }
            currentSum += nums[j];
        }
        
        if (distinctCount >= m) {
            if (currentSum > maxSum) {
                maxSum = currentSum;
            }
        }
    }
    
    return maxSum;
}

int main() {
    char line[100000];
    fgets(line, sizeof(line), stdin);
    
    int nums[10000];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        int num = atoi(token);
        nums[numsSize++] = num;
        token = strtok(NULL, "[,]");
    }
    
    int m, k;
    scanf("%d", &m);
    scanf("%d", &k);
    
    long long result = solution(nums, numsSize, m, k);
    printf("%lld\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × k)

O(n-k+1) subarrays, each taking O(k) time to process

✓ Linear Growth

Space Complexity

O(k)

Hash set to store distinct elements in current subarray

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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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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