Maximum Unique Subarray Sum After Deletion

Maximum Unique Subarray Sum After Deletion - Problem

You are given an integer array nums. You are allowed to delete any number of elements from nums without making it empty.

After performing the deletions, select a subarray of nums such that:

All elements in the subarray are unique.
The sum of the elements in the subarray is maximized.

Return the maximum sum of such a subarray.

Input & Output

Example 1 — Basic Case

$ Input: nums = [4,2,4,5,6]

› Output: 17

💡 Note: The optimal subarray is [2,4,5,6] with sum 17. We can delete the first 4 to make all elements unique.

Example 2 — No Deletions Needed

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

› Output: 8

💡 Note: The optimal subarray is [5,2,1] with sum 8. All elements are already unique in this subarray.

Example 3 — Single Element

$ Input: nums = [1]

› Output: 1

💡 Note: The array has only one element, so the maximum sum is 1.

Constraints

1 ≤ nums.length ≤ 10⁵
-10⁴ ≤ nums[i] ≤ 10⁴

Visualization

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Asked in

G Google 25 a Amazon 18 M Microsoft 15 f Facebook 12

The key insight is to use a sliding window approach to maintain a subarray with unique elements while tracking the maximum sum. The optimal solution uses two pointers and a hash set to achieve O(n) time complexity, expanding the window when possible and shrinking it when duplicates are encountered.

Common Approaches

✓ Brute Force - Check All Subarrays

⏱️ Time: O(n²) Space: O(n)

For each starting position, extend the subarray while elements remain unique. Track the maximum sum found across all valid unique subarrays.

Sliding Window with Hash Set

⏱️ Time: O(n) Space: O(n)

Maintain a sliding window with unique elements using two pointers. Expand the right pointer to include new elements, shrink from left when duplicates are encountered.

Brute Force - Check All Subarrays — Algorithm Steps

Iterate through each starting index
For each start, extend subarray while elements are unique
Track maximum sum of all unique subarrays

Visualization

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Step-by-Step Walkthrough

Start Position

Begin from each index

Extend Subarray

Add elements while unique

Track Maximum

Record best sum found

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    int maxSum = INT_MIN;
    
    for (int i = 0; i < numsSize; i++) {
        bool seen[200001] = {false};
        int currentSum = 0;
        
        for (int j = i; j < numsSize; j++) {
            int idx = nums[j] + 100000;
            if (seen[idx]) {
                break;
            }
            seen[idx] = true;
            currentSum += nums[j];
            if (currentSum > maxSum) {
                maxSum = currentSum;
            }
        }
    }
    
    return maxSum;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[100000];
    fgets(line, sizeof(line), stdin);
    
    int nums[10000];
    int size = 0;
    char* token = strtok(line + 1, ",]");
    
    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(n²)

Nested loops: outer loop for start positions, inner loop extends subarray

⚠ Quadratic Growth

Space Complexity

O(n)

Hash set stores up to n unique elements for duplicate checking

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Subarrays — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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