LeetCode #1785 — MEDIUM

Minimum Elements to Add to Form a Given Sum

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an integer array nums and two integers limit and goal. The array nums has an interesting property that abs(nums[i]) <= limit.

Return the minimum number of elements you need to add to make the sum of the array equal to goal. The array must maintain its property that abs(nums[i]) <= limit.

Note that abs(x) equals x if x >= 0, and -x otherwise.

Example 1:

Input: nums = [1,-1,1], limit = 3, goal = -4
Output: 2
Explanation: You can add -2 and -3, then the sum of the array will be 1 - 1 + 1 - 2 - 3 = -4.

Example 2:

Input: nums = [1,-10,9,1], limit = 100, goal = 0
Output: 1

Constraints:

1 <= nums.length <= 10⁵
1 <= limit <= 10⁶
-limit <= nums[i] <= limit
-10⁹ <= goal <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and two integers limit and goal. The array nums has an interesting property that abs(nums[i]) <= limit. Return the minimum number of elements you need to add to make the sum of the array equal to goal. The array must maintain its property that abs(nums[i]) <= limit. Note that abs(x) equals x if x >= 0, and -x otherwise.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[1,-1,1]
3
-4

Example 2

[1,-10,9,1]
100
0

Step 02

Core Insight

What unlocks the optimal approach

Try thinking about the problem as if the array is empty. Then you only need to form goal using elements whose absolute value is <= limit.
You can greedily set all of the elements except one to limit or -limit, so the number of elements you need is ceil(abs(goal)/ limit).
You can "normalize" goal by offsetting it by the sum of the array. For example, if the goal is 5 and the sum is -3, then it's exactly the same as if the goal is 8 and the array is empty.
The answer is ceil(abs(goal-sum)/limit) = (abs(goal-sum)+limit-1) / limit.

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 #1785: Minimum Elements to Add to Form a Given Sum
class Solution {
    public int minElements(int[] nums, int limit, int goal) {
        // long s = Arrays.stream(nums).asLongStream().sum();
        long s = 0;
        for (int v : nums) {
            s += v;
        }
        long d = Math.abs(s - goal);
        return (int) ((d + limit - 1) / limit);
    }
}

// Accepted solution for LeetCode #1785: Minimum Elements to Add to Form a Given Sum
func minElements(nums []int, limit int, goal int) int {
	s := 0
	for _, v := range nums {
		s += v
	}
	d := abs(s - goal)
	return (d + limit - 1) / limit
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #1785: Minimum Elements to Add to Form a Given Sum
class Solution:
    def minElements(self, nums: List[int], limit: int, goal: int) -> int:
        d = abs(sum(nums) - goal)
        return (d + limit - 1) // limit

// Accepted solution for LeetCode #1785: Minimum Elements to Add to Form a Given Sum
impl Solution {
    pub fn min_elements(nums: Vec<i32>, limit: i32, goal: i32) -> i32 {
        let limit = limit as i64;
        let goal = goal as i64;
        let mut sum = 0;
        for &num in nums.iter() {
            sum += num as i64;
        }
        let diff = (goal - sum).abs();
        ((diff + limit - 1) / limit) as i32
    }
}

// Accepted solution for LeetCode #1785: Minimum Elements to Add to Form a Given Sum
function minElements(nums: number[], limit: number, goal: number): number {
    const sum = nums.reduce((r, v) => r + v, 0);
    const diff = Math.abs(goal - sum);
    return Math.floor((diff + limit - 1) / limit);
}

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

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.