Smallest Index With Equal Value
LeetCode 2181 | Difficulty: Easyβ
EasyProblem Descriptionβ
Given a 0-indexed integer array nums, return the smallest index i of nums such that i mod 10 == nums[i], or -1 if such index does not exist.
x mod y denotes the remainder when x is divided by y.
Example 1:
Input: nums = [0,1,2]
Output: 0
Explanation:
i=0: 0 mod 10 = 0 == nums[0].
i=1: 1 mod 10 = 1 == nums[1].
i=2: 2 mod 10 = 2 == nums[2].
All indices have i mod 10 == nums[i], so we return the smallest index 0.
Example 2:
Input: nums = [4,3,2,1]
Output: 2
Explanation:
i=0: 0 mod 10 = 0 != nums[0].
i=1: 1 mod 10 = 1 != nums[1].
i=2: 2 mod 10 = 2 == nums[2].
i=3: 3 mod 10 = 3 != nums[3].
2 is the only index which has i mod 10 == nums[i].
Example 3:
Input: nums = [1,2,3,4,5,6,7,8,9,0]
Output: -1
Explanation: No index satisfies i mod 10 == nums[i].
Constraints:
- `1 <= nums.length <= 100`
- `0 <= nums[i] <= 9`
Topics: Array
Approachβ
Direct Approachβ
This problem can typically be solved with straightforward iteration or simple data structure usage. Focus on correctness first, then optimize.
When to use
Basic problems that test fundamental programming skills.
Solutionsβ
Solution 1: C# (Best: 84 ms)β
| Metric | Value |
|---|---|
| Runtime | 84 ms |
| Memory | 41.2 MB |
| Date | 2021-11-25 |
Solution
public class Solution {
public int SmallestEqual(int[] nums) {
for(int i=0;i<nums.Length;i++)
{
if(i%10 == nums[i])
{
return i;
}
}
return -1;
}
}
Complexity Analysisβ
| Approach | Time | Space |
|---|---|---|
| Solution | $O(n)$ | $O(1) to O(n)$ |
Interview Tipsβ
Key Points
- Start by clarifying edge cases: empty input, single element, all duplicates.
- LeetCode provides 1 hint(s) for this problem β try solving without them first.
π‘ Hints
Hint 1: Starting with i=0, check the condition for each index. The first one you find to be true is the smallest index.