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2057. Smallest Index With Equal Value
Description
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
Solutions
-
class Solution { public int smallestEqual(int[] nums) { for (int i = 0; i < nums.length; ++i) { if (i % 10 == nums[i]) { return i; } } return -1; } }
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class Solution { public: int smallestEqual(vector<int>& nums) { for (int i = 0; i < nums.size(); ++i) if (i % 10 == nums[i]) return i; return -1; } };
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class Solution: def smallestEqual(self, nums: List[int]) -> int: for i, v in enumerate(nums): if i % 10 == v: return i return -1
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func smallestEqual(nums []int) int { for i, v := range nums { if i%10 == v { return i } } return -1 }
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function smallestEqual(nums: number[]): number { for (let i = 0; i < nums.length; i++) { if (i % 10 == nums[i]) return i; } return -1; }
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impl Solution { pub fn smallest_equal(nums: Vec<i32>) -> i32 { for (i, &x) in nums.iter().enumerate() { if i % 10 == x as usize { return i as i32; } } -1 } }