Smallest Index With Equal Value - Practice Coding Problems

Smallest Index With Equal Value - Problem

Array Easy

You are given a 0-indexed integer array nums. Your task is to find the smallest index i where a special condition is met: the index's last digit equals the value at that position.

More formally, find the smallest index i such that i mod 10 == nums[i]. If no such index exists, return -1.

What does x mod y mean? It's the remainder when x is divided by y. For example, 13 mod 10 = 3 because 13 ÷ 10 = 1 remainder 3.

Example: In array [0,1,2], index 0 has value 0, and 0 mod 10 = 0 ✓. Index 1 has value 1, and 1 mod 10 = 1 ✓. Since we want the smallest index, the answer is 0.

Input & Output

example_1.py — Basic Case

$ Input: [0,1,2]

› Output: 0

💡 Note: Index 0: 0 mod 10 = 0, nums[0] = 0 → Match! Return 0.

example_2.py — Later Match

$ Input: [4,3,2,1]

› Output: 2

💡 Note: Index 0: 0 mod 10 = 0, nums[0] = 4 → No match. Index 1: 1 mod 10 = 1, nums[1] = 3 → No match. Index 2: 2 mod 10 = 2, nums[2] = 2 → Match! Return 2.

example_3.py — No Match

$ Input: [1,2,3,4,5,6,7,8,9]

› Output: -1

💡 Note: No index i satisfies the condition i mod 10 == nums[i], so return -1.

Constraints

1 ≤ nums.length ≤ 100
0 ≤ nums[i] ≤ 9
Array is 0-indexed
Return smallest valid index or -1 if none exists

Visualization

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Understanding the Visualization

Initialize

Start scanning from index 0

Check Condition

For each index i, check if i % 10 equals nums[i]

Early Return

Return immediately when first match is found

No Match

If entire array is scanned without match, return -1

Key Takeaway

🎯 Key Insight: Since we need the smallest index, we must scan from left to right and return the first match immediately. No preprocessing can help us avoid this sequential check.

Asked in

a Amazon 8 G Google 5 ⊞ Microsoft 3

This problem requires a linear scan approach since we need the smallest index where i mod 10 == nums[i]. The optimal solution iterates through the array once, checking each index against its value, with O(n) time and O(1) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Linear Scan (Optimal)	O(n)	O(1)	Iterate through array and check each index against its value

Linear Scan (Optimal) — Algorithm Steps

Start from index 0 and iterate through the array
For each index i, calculate i mod 10 (last digit of index)
Check if i mod 10 equals nums[i]
If condition is met, return i immediately
If no index satisfies the condition, return -1

Visualization

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Step-by-Step Walkthrough

Start at Index 0

Check if 0 % 10 equals nums[0]

Check Index 1

Check if 1 % 10 equals nums[1]

Continue Until Match

Keep checking until condition is satisfied

Return Result

Return first matching index or -1

Code -

solution.c — C

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

int smallestEqual(int* nums, int numsSize) {
    for (int i = 0; i < numsSize; i++) {
        if (i % 10 == nums[i]) {
            return i;
        }
    }
    return -1;
}

int main() {
    int nums[1000];
    int size = 0;
    char c;
    int num = 0;
    int reading = 0;
    
    while ((c = getchar()) != '\n' && c != EOF) {
        if (c >= '0' && c <= '9') {
            num = num * 10 + (c - '0');
            reading = 1;
        } else if (reading) {
            nums[size++] = num;
            num = 0;
            reading = 0;
        }
    }
    if (reading) nums[size++] = num;
    
    printf("%d\n", smallestEqual(nums, size));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

We may need to check each element once in the worst case

✓ Linear Growth

Space Complexity

O(1)

Only using a constant amount of extra space for loop variables

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Linear Scan (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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