Find the Minimum Possible Sum of a Beautiful Array

Find the Minimum Possible Sum of a Beautiful Array

What is this problem about?

A "beautiful" array of $n$ distinct positive integers is one where no two elements sum up to exactly $target$. The Find the Minimum Possible Sum of a Beautiful Array interview question asks you to find the smallest possible sum of such an array.

Why is this asked in interviews?

Infosys uses this to test your ability to apply a Greedy approach to set exclusion. You want to pick the smallest numbers possible (1, 2, 3...) to minimize the sum, but if you pick $x$, you are forbidden from picking $target - x$. It evaluation whether you can identify the "conflict zone" and jump over it to keep the sum optimal.

Algorithmic pattern used

This problem uses Greedy logic and Arithmetic Series formulas.

1. Start picking numbers from 1.
2. For any pair $(x, target-x)$, you can only pick one. To minimize the sum, always pick the smaller one.
3. You can pick all numbers from 1 up to $\lfloor target/2 floor$. These are all safe because their counterparts are larger.
4. Once you reach $target/2$, you must skip all numbers up to $target-1$.
5. Then, continue picking numbers starting from $target$ until you have $n$ elements in total.
6. The sum can be calculated using the formula for the sum of an arithmetic progression: $\frac{n(a+l)}{2}$.

Example explanation

$n = 3, target = 4$.

1. Pick 1. (Conflict $4-1=3$ is now forbidden).
2. Pick 2. (Conflict $4-2=2$. This is a self-conflict, so 2 is the "middle").
3. Cannot pick 3.
4. Pick 4. Array: [1, 2, 4]. Sum = 7.

Common mistakes candidates make

• Brute Force: Using a set and a loop to find $n$ numbers, which is $O(N)$ and might be too slow if $N$ is huge ($10^9$).
• Overlap: Not correctly identifying the range of numbers to skip.
• Overflow: Forgetting to use 64-bit integers for the sum of a large arithmetic series.

Interview preparation tip

Whenever you are asked for the "Minimum Sum" of distinct integers under an exclusion rule, the strategy is always to pick the smallest available numbers. Use the sum formula $S = \frac{n}{2}(2a + (n-1)d)$ to calculate the result in $O(1)$ if the ranges are contiguous.