Smallest Index With Digit Sum Equal to Index

Smallest Index With Digit Sum Equal to Index - Problem

Array Math Easy

You are given an integer array nums. Your task is to find the smallest index i such that the sum of the digits of nums[i] equals the index i itself.

For example, if nums = [18, 7, 23, 4]:

At index 0: nums[0] = 18, digit sum = 1 + 8 = 9, but 9 ≠ 0
At index 1: nums[1] = 7, digit sum = 7, but 7 ≠ 1
At index 2: nums[2] = 23, digit sum = 2 + 3 = 5, but 5 ≠ 2
At index 3: nums[3] = 4, digit sum = 4, but 4 ≠ 3

If no such index exists, return -1.

Note: The digit sum is calculated by adding all individual digits of a number. For instance, the digit sum of 1234 is 1 + 2 + 3 + 4 = 10.

Input & Output

example_1.py — Basic Example

$ Input: nums = [18, 7, 23, 4]

› Output: -1

💡 Note: Index 0: digit sum of 18 = 1+8 = 9 ≠ 0. Index 1: digit sum of 7 = 7 ≠ 1. Index 2: digit sum of 23 = 2+3 = 5 ≠ 2. Index 3: digit sum of 4 = 4 ≠ 3. No valid index found.

example_2.py — Match Found

$ Input: nums = [10, 1, 23, 6]

› Output: 1

💡 Note: Index 0: digit sum of 10 = 1+0 = 1 ≠ 0. Index 1: digit sum of 1 = 1 = 1 ✓. Found match at index 1, which is the smallest valid index.

example_3.py — Multiple Matches

$ Input: nums = [0, 1, 5, 12]

› Output: 0

💡 Note: Index 0: digit sum of 0 = 0 = 0 ✓. Found match at index 0, which is the smallest possible index, so we return 0 immediately.

Constraints

1 ≤ nums.length ≤ 10⁴
0 ≤ nums[i] ≤ 10⁹
nums[i] consists of positive integers only

Visualization

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Understanding the Visualization

Initialize scanner

Start checking from index 0, moving left to right

Calculate digit sum

For each number, add up all its individual digits

Compare with index

Check if the digit sum equals the current index position

Return or continue

Return index if match found, otherwise move to next position

Key Takeaway

🎯 Key Insight: Since we need the smallest index with the matching condition, we must scan the array sequentially from left to right and return immediately when we find the first valid match.

Asked in

G Google 23 a Amazon 18 ⊞ Microsoft 15 f Meta 12

The optimal approach uses a single linear scan through the array, calculating the digit sum of each element and comparing it with the current index. Since we need the smallest index, we return immediately upon finding the first match. Time complexity is O(n × d) where d is the average number of digits, and space complexity is O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Linear Scan with Digit Sum Calculation	O(n * d)	O(1)	Iterate through array and calculate digit sum for each element until we find a match

Linear Scan with Digit Sum Calculation — Algorithm Steps

Start from index 0 and iterate through the array
For each element at index i, calculate the sum of its digits
Compare the digit sum with the current index i
If they match, return the current index
If no match is found after checking all elements, return -1

Visualization

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Step-by-Step Walkthrough

Start at index 0

Begin scanning from the first element

Calculate digit sum

Add up all digits of the current number

Compare with index

Check if digit sum equals current index

Return if match found

Return index immediately when condition is met

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int digitSum(int n) {
    int total = 0;
    n = abs(n); // Handle negative numbers
    while (n > 0) {
        total += n % 10;
        n /= 10;
    }
    return total;
}

int smallestIndexWithDigitSum(int* nums, int numsSize) {
    for (int i = 0; i < numsSize; i++) {
        if (digitSum(nums[i]) == i) {
            return i;
        }
    }
    return -1;
}

int main() {
    int nums[1000];
    int size = 0;
    char ch;
    int num = 0;
    int sign = 1;
    int reading = 0;
    
    // Simple parser for [1,2,3] format
    while ((ch = getchar()) != '\n' && ch != EOF) {
        if (ch >= '0' && ch <= '9') {
            num = num * 10 + (ch - '0');
            reading = 1;
        } else if (ch == '-') {
            sign = -1;
        } else if ((ch == ',' || ch == ']') && reading) {
            nums[size++] = num * sign;
            num = 0;
            sign = 1;
            reading = 0;
        }
    }
    
    printf("%d\n", smallestIndexWithDigitSum(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n * d)

O(n) to iterate through array, O(d) to calculate digit sum where d is average number of digits

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for calculations

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Linear Scan with Digit Sum Calculation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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