LeetCode #2535 — EASY

Difference Between Element Sum and Digit Sum of an Array

Build confidence with an intuition-first walkthrough focused on array fundamentals.

Solve on LeetCode

The Problem

Problem Statement

You are given a positive integer array nums.

The element sum is the sum of all the elements in nums.
The digit sum is the sum of all the digits (not necessarily distinct) that appear in nums.

Return the absolute difference between the element sum and digit sum of nums.

Note that the absolute difference between two integers x and y is defined as |x - y|.

Example 1:

Input: nums = [1,15,6,3]
Output: 9
Explanation: 
The element sum of nums is 1 + 15 + 6 + 3 = 25.
The digit sum of nums is 1 + 1 + 5 + 6 + 3 = 16.
The absolute difference between the element sum and digit sum is |25 - 16| = 9.

Example 2:

Input: nums = [1,2,3,4]
Output: 0
Explanation:
The element sum of nums is 1 + 2 + 3 + 4 = 10.
The digit sum of nums is 1 + 2 + 3 + 4 = 10.
The absolute difference between the element sum and digit sum is |10 - 10| = 0.

Constraints:

1 <= nums.length <= 2000
1 <= nums[i] <= 2000

Step 01

Brute Force Baseline

Problem summary: You are given a positive integer array nums. The element sum is the sum of all the elements in nums. The digit sum is the sum of all the digits (not necessarily distinct) that appear in nums. Return the absolute difference between the element sum and digit sum of nums. Note that the absolute difference between two integers x and y is defined as |x - y|.

Baseline thinking

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

Pattern signal: Array · Math

Example 1

[1,15,6,3]

Example 2

[1,2,3,4]

Core Insight

What unlocks the optimal approach

Use a simple for loop to iterate each number.
How you can get the digit for each number?

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

Upper-end input sizes

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 #2535: Difference Between Element Sum and Digit Sum of an Array
class Solution {
    public int differenceOfSum(int[] nums) {
        int x = 0, y = 0;
        for (int v : nums) {
            x += v;
            for (; v > 0; v /= 10) {
                y += v % 10;
            }
        }
        return x - y;
    }
}

// Accepted solution for LeetCode #2535: Difference Between Element Sum and Digit Sum of an Array
func differenceOfSum(nums []int) int {
	var x, y int
	for _, v := range nums {
		x += v
		for ; v > 0; v /= 10 {
			y += v % 10
		}
	}
	return x - y
}

# Accepted solution for LeetCode #2535: Difference Between Element Sum and Digit Sum of an Array
class Solution:
    def differenceOfSum(self, nums: List[int]) -> int:
        x = y = 0
        for v in nums:
            x += v
            while v:
                y += v % 10
                v //= 10
        return x - y

// Accepted solution for LeetCode #2535: Difference Between Element Sum and Digit Sum of an Array
impl Solution {
    pub fn difference_of_sum(nums: Vec<i32>) -> i32 {
        let mut x = 0;
        let mut y = 0;

        for &v in &nums {
            x += v;
            let mut num = v;
            while num > 0 {
                y += num % 10;
                num /= 10;
            }
        }

        x - y
    }
}

// Accepted solution for LeetCode #2535: Difference Between Element Sum and Digit Sum of an Array
function differenceOfSum(nums: number[]): number {
    let [x, y] = [0, 0];
    for (let v of nums) {
        x += v;
        for (; v; v = Math.floor(v / 10)) {
            y += v % 10;
        }
    }
    return x - y;
}

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.