How do you solve Smallest Index With Equal Value on LeetCode?

Smallest Index With Equal Value (LeetCode #2057) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The problem requires checking a simple condition for each index, making it straightforward.

What is the brute force solution for Smallest Index With Equal Value?

The brute force approach for Smallest Index With Equal Value is Brute Force, with O(n) time complexity and O(1) space complexity. We will check each index of the array to see if the condition i mod 10 == nums[i] holds true. This is straightforward but can be inefficient if the array is large. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Smallest Index With Equal Value?

Smallest Index With Equal Value uses the following concepts: Array. The recommended patterns to study are: Array, Modulo operation.

What are the interview tips for Smallest Index With Equal Value?

Always clarify the problem statement and constraints before jumping to code. Consider edge cases, such as arrays with no valid indices. Explain your thought process as you code; it helps the interviewer understand your reasoning.

What are common mistakes when solving Smallest Index With Equal Value?

Students often forget to check all indices before returning, leading to incorrect results. Some may confuse the condition and check nums[i] mod 10 instead of i mod 10.

#2057

Smallest Index With Equal Value

Easy

ArrayArrayModulo operation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The brute force approach is already efficient for this problem, but we can still optimize by breaking out of the loop early if we find the first match.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to store the smallest index found, starting with -1.
2Step 2: Loop through each index i of the array nums.
3Step 3: For each index, check if i mod 10 is equal to nums[i].
4Step 4: If the condition is true, return the current index immediately.

solution.py7 lines

1# Full working Python code
2
3def smallest_index(nums):
4    for i in range(len(nums)):
5        if i % 10 == nums[i]:
6            return i
7    return -1

ℹ

Complexity note: The time complexity remains O(n) as we traverse the array once, but we may exit early if we find a match. Space complexity is O(1) since we use a constant amount of space.

1The problem requires checking a simple condition for each index, making it straightforward.
2Finding the first valid index allows for an immediate return, optimizing performance.

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