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Java Solution for LeetCode algorithm problems, continually updating
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2057\. Smallest Index With Equal Value
Easy
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`