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Formatted question description: https://leetcode.ca/all/2057.html
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].
Example 4:
Input: nums = [2,1,3,5,2] Output: 1 Explanation: 1 is the only index with i mod 10 == nums[i].
Constraints:
1 <= nums.length <= 1000 <= nums[i] <= 9
Solution 1.
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class Solution { public int smallestEqual(int[] nums) { for (int i = 0; i < nums.length; ++i) { if (i % 10 == nums[i]) { return i; } } return -1; } } -
class Solution: def smallestEqual(self, nums: List[int]) -> int: for i, v in enumerate(nums): if i % 10 == v: return i return -1 -
func smallestEqual(nums []int) int { for i, v := range nums { if i%10 == v { return i } } return -1 } -
function smallestEqual(nums: number[]): number { for (let i = 0; i < nums.length; i++) { if (i % 10 == nums[i]) return i; } return -1; }