How do you solve Sum of Subsequence Widths on LeetCode?

Sum of Subsequence Widths (LeetCode #891) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting the array allows us to efficiently calculate contributions as max and min.

What is the brute force solution for Sum of Subsequence Widths?

The brute force approach for Sum of Subsequence Widths is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible non-empty subsequences of the array and calculating their widths. This method is straightforward but inefficient for larger arrays due to the exponential growth of subsequences. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Sum of Subsequence Widths?

Sum of Subsequence Widths uses the following concepts: Array, Math, Sorting. The recommended patterns to study are: Sorting, Combinatorial Counting.

What are the interview tips for Sum of Subsequence Widths?

Always consider edge cases, such as arrays of length 1. Explain your thought process clearly while coding, especially when transitioning from brute force to optimal solutions. Practice combinatorial problems to become comfortable with counting techniques.

What are common mistakes when solving Sum of Subsequence Widths?

Failing to consider the modulo operation when summing widths. Not recognizing that subsequences grow exponentially, leading to inefficient solutions.

#891

Sum of Subsequence Widths

Hard

ArrayMathSortingSortingCombinatorial Counting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

💡

Intuition

Time O(n log n)Space O(1)

The optimal solution leverages sorting and combinatorial counting. By sorting the array, we can determine how many times each element contributes as a maximum and minimum in various subsequences, allowing us to compute the total width efficiently.

⚙️

Algorithm

4 steps

1Step 1: Sort the array in ascending order.
2Step 2: For each element nums[i], calculate its contribution as a maximum and minimum in subsequences based on its position.
3Step 3: Use combinatorial counting to determine how many subsequences include nums[i] as the maximum and how many include it as the minimum.
4Step 4: Sum these contributions and return the result modulo 10^9 + 7.

solution.py11 lines

1# Full working Python code
2def sum_of_subsequence_widths(nums):
3    MOD = 10**9 + 7
4    nums.sort()
5    n = len(nums)
6    total_width = 0
7    for i in range(n):
8        max_contrib = nums[i] * (1 << i) % MOD
9        min_contrib = nums[i] * (1 << (n - 1 - i)) % MOD
10        total_width = (total_width + max_contrib - min_contrib) % MOD
11    return total_width

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, and the space complexity is O(1) since we are using a constant amount of extra space.

1Sorting the array allows us to efficiently calculate contributions as max and min.
2Each element's contribution can be determined using combinatorial counting.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.