LeetCode #891 — HARD

Sum of Subsequence Widths

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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].

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[2,1,3]

Example 2

[2]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #891: Sum of Subsequence Widths
class Solution {
    private static final int MOD = (int) 1e9 + 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;
    }
}

// Accepted solution for LeetCode #891: Sum of Subsequence Widths
func sumSubseqWidths(nums []int) (ans int) {
	const mod int = 1e9 + 7
	sort.Ints(nums)
	p, n := 1, len(nums)
	for i, v := range nums {
		ans = (ans + (v-nums[n-i-1])*p + mod) % mod
		p = (p << 1) % mod
	}
	return
}

# Accepted solution for LeetCode #891: Sum of Subsequence Widths
class Solution:
    def sumSubseqWidths(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        nums.sort()
        ans, p = 0, 1
        for i, v in enumerate(nums):
            ans = (ans + (v - nums[-i - 1]) * p) % mod
            p = (p << 1) % mod
        return ans

// Accepted solution for LeetCode #891: Sum of Subsequence Widths
struct Solution;

const MOD: i64 = 1_000_000_007;

impl Solution {
    fn sum_subseq_widths(mut a: Vec<i32>) -> i32 {
        let n = a.len();
        a.sort_unstable();
        let mut res = 0;
        let mut c = 1;
        for i in 0..n {
            res += c * a[i] as i64;
            res %= MOD;
            c *= 2;
            c %= MOD;
        }
        c = 1;
        for i in (0..n).rev() {
            res -= c * a[i] as i64;
            res %= MOD;
            c *= 2;
            c %= MOD;
        }
        ((res + MOD) % MOD) as i32
    }
}

#[test]
fn test() {
    let a = vec![2, 1, 3];
    let res = 6;
    assert_eq!(Solution::sum_subseq_widths(a), res);
}

// Accepted solution for LeetCode #891: Sum of Subsequence Widths
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #891: Sum of Subsequence Widths
// class Solution {
//     private static final int MOD = (int) 1e9 + 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;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.