LeetCode #2834 — MEDIUM

Find the Minimum Possible Sum of a Beautiful Array

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

Solve on LeetCode

The Problem

Problem Statement

You are given positive integers n and target.

An array nums is beautiful if it meets the following conditions:

nums.length == n.
nums consists of pairwise distinct positive integers.
There doesn't exist two distinct indices, i and j, in the range [0, n - 1], such that nums[i] + nums[j] == target.

Return the minimum possible sum that a beautiful array could have modulo 10⁹ + 7.

Example 1:

Input: n = 2, target = 3
Output: 4
Explanation: We can see that nums = [1,3] is beautiful.
- The array nums has length n = 2.
- The array nums consists of pairwise distinct positive integers.
- There doesn't exist two distinct indices, i and j, with nums[i] + nums[j] == 3.
It can be proven that 4 is the minimum possible sum that a beautiful array could have.

Example 2:

Input: n = 3, target = 3
Output: 8
Explanation: We can see that nums = [1,3,4] is beautiful.
- The array nums has length n = 3.
- The array nums consists of pairwise distinct positive integers.
- There doesn't exist two distinct indices, i and j, with nums[i] + nums[j] == 3.
It can be proven that 8 is the minimum possible sum that a beautiful array could have.

Example 3:

Input: n = 1, target = 1
Output: 1
Explanation: We can see, that nums = [1] is beautiful.

Constraints:

1 <= n <= 10⁹
1 <= target <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given positive integers n and target. An array nums is beautiful if it meets the following conditions: nums.length == n. nums consists of pairwise distinct positive integers. There doesn't exist two distinct indices, i and j, in the range [0, n - 1], such that nums[i] + nums[j] == target. Return the minimum possible sum that a beautiful array could have modulo 109 + 7.

Baseline thinking

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

Pattern signal: Math · Greedy

Example 1

2
3

Example 2

3
3

Example 3

1
1

Step 02

Core Insight

What unlocks the optimal approach

<div class="_1l1MA">Greedily try to add the smallest possible number in the array <code>nums</code>, such that <code>nums</code> contains distinct positive integers, and there are no two indices <code>i</code> and <code>j</code> with <code>nums[i] + nums[j] == target</code>.</div>

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 #2834: Find the Minimum Possible Sum of a Beautiful Array
class Solution {
    public int minimumPossibleSum(int n, int target) {
        final int mod = (int) 1e9 + 7;
        int m = target / 2;
        if (n <= m) {
            return (int) ((1L + n) * n / 2 % mod);
        }
        long a = (1L + m) * m / 2 % mod;
        long b = ((1L * target + target + n - m - 1) * (n - m) / 2) % mod;
        return (int) ((a + b) % mod);
    }
}

// Accepted solution for LeetCode #2834: Find the Minimum Possible Sum of a Beautiful Array
func minimumPossibleSum(n int, target int) int {
	const mod int = 1e9 + 7
	m := target / 2
	if n <= m {
		return (n + 1) * n / 2 % mod
	}
	a := (m + 1) * m / 2 % mod
	b := (target + target + n - m - 1) * (n - m) / 2 % mod
	return (a + b) % mod
}

# Accepted solution for LeetCode #2834: Find the Minimum Possible Sum of a Beautiful Array
class Solution:
    def minimumPossibleSum(self, n: int, target: int) -> int:
        mod = 10**9 + 7
        m = target // 2
        if n <= m:
            return ((1 + n) * n // 2) % mod
        return ((1 + m) * m // 2 + (target + target + n - m - 1) * (n - m) // 2) % mod

// Accepted solution for LeetCode #2834: Find the Minimum Possible Sum of a Beautiful Array
/**
 * [2834] Find the Minimum Possible Sum of a Beautiful Array
 */
pub struct Solution {}

// submission codes start here
impl Solution {
    pub fn minimum_possible_sum(n: i32, target: i32) -> i32 {
        let n = n as i64;
        let target = target as i64;
        let m = 1e9 as i64 + 7;

        let length = target / 2;

        if length >= n {
            ((1 + n) * n / 2 % m) as i32
        } else {
            (((1 + length) * length / 2 + (target + target + n - length - 1) * (n - length) / 2)
                % m) as i32
        }
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2834() {
        assert_eq!(4, Solution::minimum_possible_sum(2, 3));
        assert_eq!(8, Solution::minimum_possible_sum(3, 3));
        assert_eq!(1, Solution::minimum_possible_sum(1, 1));
    }
}

// Accepted solution for LeetCode #2834: Find the Minimum Possible Sum of a Beautiful Array
function minimumPossibleSum(n: number, target: number): number {
    const mod = 10 ** 9 + 7;
    const m = target >> 1;
    if (n <= m) {
        return (((1 + n) * n) / 2) % mod;
    }
    return (((1 + m) * m) / 2 + ((target + target + n - m - 1) * (n - m)) / 2) % mod;
}

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(1)

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.

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.

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.