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 <= 1000 <= nums[i] <= 9
Solutions
Solution 1: Traversal
We directly traverse the array. For each index i, we check if it satisfies i \bmod 10 = nums[i]. If it does, we return the current index i.
If we traverse the entire array and do not find a satisfying index, we return -1.
The time complexity is O(n), where n is the length of the array. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustcj
class Solution: def smallestEqual(self, nums: List[int]) -> int: for i, x in enumerate(nums): if i % 10 == x: return i return -1(code-box)
