Sum of Subsequence Widths

HARD

Description

The width of a sequence is the difference between the maximum and minimum elements in the sequence.

Given an array of integers nums, return the sum of the widths of all the non-empty subsequences of nums. Since the answer may be very large, return it modulo 10⁹ + 7.

A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].

Example 1:

Input: nums = [2,1,3]
Output: 6
Explanation: The subsequences are [1], [2], [3], [2,1], [2,3], [1,3], [2,1,3].
The corresponding widths are 0, 0, 0, 1, 1, 2, 2.
The sum of these widths is 6.

Example 2:

Input: nums = [2]
Output: 0

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

Approaches

Checkout 3 different approaches to solve Sum of Subsequence Widths. Click on different approaches to view the approach and algorithm in detail.

Brute Force by Generating All Subsequences

This approach involves generating every possible non-empty subsequence of the given array. For each subsequence, we find its maximum and minimum elements, calculate the width (max - min), and add it to a running total. This is the most straightforward but also the most inefficient method, as the number of subsequences grows exponentially with the size of the input array.

Algorithm

Initialize a global variable totalWidth to 0.
Create a recursive function, say findSubsequences(index, currentSubsequence), to generate all subsequences.
Base Case: When the index reaches the end of the array:
- If the currentSubsequence is not empty, find its minimum and maximum elements.
- Calculate the width (max - min).
- Add the width to totalWidth, taking modulo 10^9 + 7.
- Return.
Recursive Step: For each element nums[index], make two recursive calls:
1. One call that excludes nums[index] from the subsequence: findSubsequences(index + 1, currentSubsequence).
2. One call that includes nums[index]: add nums[index] to currentSubsequence, call findSubsequences(index + 1, currentSubsequence), and then remove it to backtrack.
Start the process by calling findSubsequences(0, new ArrayList<>()).
Return the final totalWidth.

The core of this method is a backtracking algorithm to explore all possibilities of forming a subsequence. We can define a recursive helper function that builds subsequences. At each step i in the input array nums, we decide whether to include nums[i] in the current subsequence or not. This creates a decision tree of 2^n paths, where n is the length of nums, with each path corresponding to a unique subsequence. Once a full subsequence is formed (i.e., we've made a decision for every element), we check if it's non-empty. If it is, we iterate through it to find the minimum and maximum values, compute their difference, and add this width to a global sum. All additions are performed modulo 10^9 + 7 to prevent overflow and meet the problem's requirements.

class Solution {
    long totalWidth = 0;
    int MOD = 1_000_000_007;

    public int sumSubsequenceWidths(int[] nums) {
        // Note: This approach is too slow and will time out.
        findSubsequences(nums, 0, new java.util.ArrayList<>());
        return (int) totalWidth;
    }

    private void findSubsequences(int[] nums, int index, java.util.List<Integer> currentSubsequence) {
        if (index == nums.length) {
            if (!currentSubsequence.isEmpty()) {
                int minVal = Integer.MAX_VALUE;
                int maxVal = Integer.MIN_VALUE;
                for (int num : currentSubsequence) {
                    minVal = Math.min(minVal, num);
                    maxVal = Math.max(maxVal, num);
                }
                totalWidth = (totalWidth + maxVal - minVal) % MOD;
            }
            return;
        }

        // Decision 1: Exclude nums[index]
        findSubsequences(nums, index + 1, currentSubsequence);

        // Decision 2: Include nums[index]
        currentSubsequence.add(nums[index]);
        findSubsequences(nums, index + 1, currentSubsequence);
        currentSubsequence.remove(currentSubsequence.size() - 1); // Backtrack
    }
}

Complexity Analysis

Time Complexity: O(n * 2^n). There are `2^n` subsequences in total. For each subsequence, finding the minimum and maximum can take up to O(n) time. This makes the approach infeasible for `n > 20`.Space Complexity: O(n), where n is the number of elements in `nums`. This space is used by the recursion stack and to store the current subsequence being built, both of which can go up to a depth/size of `n`.

Pros and Cons

Pros:

Simple to understand and conceptualize.

Cons:

Extremely inefficient due to its exponential time complexity.
Will result in a 'Time Limit Exceeded' (TLE) error for the given constraints.

Code Solutions

Checking out 3 solutions in different languages for Sum of Subsequence Widths. Click on different languages to view the code.

class Solution { private static final int MOD = ( int ) 1 e9 + 7 ; public int sumSubseqWidths ( int [] nums ) { Arrays . sort ( nums ); long ans = 0 , p = 1 ; int n = nums . length ; for ( int i = 0 ; i < n ; ++ i ) { ans = ( ans + ( nums [ i ] - nums [ n - i - 1 ]) * p + MOD ) % MOD ; p = ( p << 1 ) % MOD ; } return ( int ) ans ; } }

Video Solution

Watch the video walkthrough for Sum of Subsequence Widths

Algorithms:

Sorting

Patterns:

Math

Data Structures:

Array