How do you solve Find the Smallest Balanced Index on LeetCode?

Find the Smallest Balanced Index (LeetCode #3862) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Prefix sums help in calculating left sums efficiently.

What is the time complexity of Find the Smallest Balanced Index?

The optimal solution for Find the Smallest Balanced Index runs in O(n) time and uses O(n) space. We traverse the array twice: once for prefix sums and once for suffix products, leading to linear complexity.

What is the brute force solution for Find the Smallest Balanced Index?

The brute force approach for Find the Smallest Balanced Index is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each index by calculating the left sum and right product separately. This method is straightforward but inefficient as it requires nested loops. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Smallest Balanced Index?

Find the Smallest Balanced Index uses the following concepts: Array, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the Smallest Balanced Index?

Explain your thought process clearly. Discuss edge cases upfront. Practice similar problems to build intuition.

What are common mistakes when solving Find the Smallest Balanced Index?

Forgetting to handle edge cases with empty or single-element arrays. Incorrectly calculating the right product when it involves zero.

#3862

Find the Smallest Balanced Index

Medium

ArrayPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use prefix sums for left sums and suffix products for right products to efficiently determine balanced indices in a single pass.

⚙️

Algorithm

3 steps

1Step 1: Calculate prefix sums for left sums in a single pass.
2Step 2: Calculate suffix products for right products in reverse.
3Step 3: Compare left sums and right products to find the smallest balanced index.

solution.py11 lines

1def smallestBalancedIndex(nums):
2    n = len(nums)
3    left_sum = [0] * n
4    for i in range(1, n):
5        left_sum[i] = left_sum[i - 1] + nums[i - 1]
6    right_product = 1
7    for i in range(n - 1, -1, -1):
8        if left_sum[i] == right_product:
9            return i
10        right_product *= nums[i]
11    return -1

ℹ

Complexity note: We traverse the array twice: once for prefix sums and once for suffix products, leading to linear complexity.

1Prefix sums help in calculating left sums efficiently.
2Suffix products allow for quick right product calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.