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3550. Smallest Index With Digit Sum Equal to Index

Description

You are given an integer array nums.

Return the smallest index i such that the sum of the digits of nums[i] is equal to i.

If no such index exists, return -1.

 

Example 1:

Input: nums = [1,3,2]

Output: 2

Explanation:

  • For nums[2] = 2, the sum of digits is 2, which is equal to index i = 2. Thus, the output is 2.

Example 2:

Input: nums = [1,10,11]

Output: 1

Explanation:

  • For nums[1] = 10, the sum of digits is 1 + 0 = 1, which is equal to index i = 1.
  • For nums[2] = 11, the sum of digits is 1 + 1 = 2, which is equal to index i = 2.
  • Since index 1 is the smallest, the output is 1.

Example 3:

Input: nums = [1,2,3]

Output: -1

Explanation:

  • Since no index satisfies the condition, the output is -1.

 

Constraints:

  • 1 <= nums.length <= 100
  • 0 <= nums[i] <= 1000

Solutions

Solution 1: Enumeration + Digit Sum

We can start from index \(i = 0\) and iterate through each element \(x\) in the array, calculating the digit sum \(s\) of \(x\). If \(s = i\), return the index \(i\). If no such index is found after traversing all elements, return -1.

The time complexity is \(O(n)\), where \(n\) is the length of the array. The space complexity is \(O(1)\), as only constant extra space is used.

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class Solution:
    def smallestIndex(self, nums: List[int]) -> int:
        for i, x in enumerate(nums):
            s = 0
            while x:
                s += x % 10
                x //= 10
            if s == i:
                return i
        return -1

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class Solution {
    public int smallestIndex(int[] nums) {
        for (int i = 0; i < nums.length; ++i) {
            int s = 0;
            while (nums[i] != 0) {
                s += nums[i] % 10;
                nums[i] /= 10;
            }
            if (s == i) {
                return i;
            }
        }
        return -1;
    }
}

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class Solution {
public:
    int smallestIndex(vector<int>& nums) {
        for (int i = 0; i < nums.size(); ++i) {
            int s = 0;
            while (nums[i]) {
                s += nums[i] % 10;
                nums[i] /= 10;
            }
            if (s == i) {
                return i;
            }
        }
        return -1;
    }
};

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func smallestIndex(nums []int) int {
    for i, x := range nums {
        s := 0
        for ; x > 0; x /= 10 {
            s += x % 10
        }
        if s == i {
            return i
        }
    }
    return -1
}

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function smallestIndex(nums: number[]): number {
    for (let i = 0; i < nums.length; ++i) {
        let s = 0;
        for (; nums[i] > 0; nums[i] = Math.floor(nums[i] / 10)) {
            s += nums[i] % 10;
        }
        if (s === i) {
            return i;
        }
    }
    return -1;
}

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