#3618

Medium

LeetCode Problem Workspace

Split Array by Prime Indices

Split Array by Prime Indices challenges you to separate elements at prime positions and minimize the absolute sum difference efficiently.

array math number theory

Problem Statement

You are given an integer array nums. Split the array into two groups based on their indices: one containing elements at prime indices, the other containing all remaining elements.

Return the absolute difference between the sum of the elements in the prime index group and the sum of the elements in the remaining group. The array may contain negative numbers and large values, so consider efficient handling of prime indices and sums.

Examples

Example 1

Input: nums = [2,3,4]

Output: 1

Example 2

Input: nums = [-1,5,7,0]

Output: 3

Constraints

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

Solution Approach

Generate Prime Indices

Use the Sieve of Eratosthenes to identify all prime numbers up to nums.length. This creates a boolean mask to quickly check if an index is prime, which is crucial for splitting the array efficiently.

Separate Elements into Two Arrays

Iterate through nums and assign elements at prime indices to array A and all others to array B. Keep running sums to avoid extra space usage and to prepare for the final difference calculation.

Compute Absolute Difference

After summing both arrays, return the absolute value of the difference |sum(A) - sum(B)|. This final step directly solves the problem while adhering to the array plus math pattern.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n log log n) for the sieve plus O(n) for iterating the array, totaling O(n log log n). Space complexity is O(n) for the prime index mask, but the algorithm can reduce auxiliary storage by summing on the fly.

What Interviewers Usually Probe

Check if you can optimize by avoiding repeated prime checks per index.
Be ready to handle negative and large values while calculating sums.
Consider edge cases where nums.length is very small or consists entirely of primes.

Common Pitfalls or Variants

Common pitfalls

Confusing array indices with element values when separating by prime indices.
Failing to account for the first index being 0 or 1, which are not prime.
Using inefficient prime checks instead of a sieve, leading to timeouts.

Follow-up variants

Split array using odd or even indices instead of prime indices for a simpler variant.
Compute maximum absolute difference rather than minimum, testing sum strategy changes.
Use a sliding window of prime indices to handle dynamic array updates efficiently.

FAQ

What is the main challenge in Split Array by Prime Indices?

The key challenge is correctly identifying prime indices and efficiently computing the absolute difference of sums without iterating redundantly.

Can negative numbers in nums affect the result?

Yes, negative values impact the sums directly, so the solution must account for both positive and negative numbers accurately.

Is using a Sieve of Eratosthenes necessary?

Using a sieve is the most efficient way to identify all prime indices up to nums.length, avoiding repeated primality checks.

Does the algorithm work for arrays of length 1 or 2?

Yes, edge cases with very small arrays are handled by the prime index check, where only valid primes contribute to one array.

How does this problem relate to Array plus Math pattern?

It combines array iteration with number theory by using prime index logic and arithmetic operations to compute the absolute difference.

terminal

Solution

Solution 1: Sieve of Eratosthenes + Simulation

We can use the Sieve of Eratosthenes to preprocess all prime numbers in the range $[0, 10^5]$. Then we iterate through the array $\textit{nums}$. For $\textit{nums}[i]$, if $i$ is a prime number, we add $\textit{nums}[i]$ to the answer; otherwise, we add $-\textit{nums}[i]$ to the answer. Finally, we return the absolute value of the answer.

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m = 10**5 + 10
primes = [True] * m
primes[0] = primes[1] = False
for i in range(2, m):
    if primes[i]:
        for j in range(i + i, m, i):
            primes[j] = False


class Solution:
    def splitArray(self, nums: List[int]) -> int:
        return abs(sum(x if primes[i] else -x for i, x in enumerate(nums)))

Continue Practicing

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