Question

2057. Smallest Index With Equal Value

中文文档

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

Solution 1

class Solution:
  def smallestEqual(self, nums: List[int]) -> int:
    for i, v in enumerate(nums):
      if i % 10 == v:
        return i
    return -1

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 {
public:
  int smallestEqual(vector<int>& nums) {
    for (int i = 0; i < nums.size(); ++i)
      if (i % 10 == nums[i])
        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;
}