Smallest Index With Equal Value

Smallest Index With Equal Value - Problem

Array Easy

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.

Input & Output

Example 1 — Found Match

$ Input: nums = [0,1,2,3,4,5,6,7,8,9]

› Output: 0

💡 Note: At index 0: 0 % 10 = 0 and nums[0] = 0, so return 0

Example 2 — Later Match

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

› Output: -1

💡 Note: No index i satisfies i % 10 == nums[i]: 0≠4, 1≠3, 2≠2 is false, 3≠1

Example 3 — Double Digit Index

$ Input: nums = [1,2,3,4,5,6,7,8,9,0,1,2,3]

› Output: 10

💡 Note: At index 10: 10 % 10 = 0 but nums[10] = 1, continue. At index 12: 12 % 10 = 2 but nums[12] = 3. Actually at index 10: 10 % 10 = 0 and nums[10] = 1, so no match until later.

Constraints

1 ≤ nums.length ≤ 100
0 ≤ nums[i] ≤ 9

Visualization

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Asked in

G Google 15 a Amazon 12

The key insight is to iterate through the array and check if each index modulo 10 equals the value at that index. Since we want the smallest index, we return the first match. Best approach is linear scan with Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Linear Scan	O(n)	O(1)	Check each index to see if i mod 10 equals nums[i]

Linear Scan — Algorithm Steps

Iterate through array indices from 0 to n-1
For each index i, calculate i mod 10
Check if i mod 10 equals nums[i]
Return first matching index, or -1 if none found

Visualization

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Step-by-Step Walkthrough

Start at index 0

Check if 0 % 10 == nums[0]

Continue scanning

Check each subsequent index

Return first match

Return index when condition is met

Code -

solution.c — C

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

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

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[100];
    int numsSize = 0;
    
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (token[0] >= '0' && token[0] <= '9' || token[0] == '-') {
            nums[numsSize++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, numsSize);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array of n elements

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Linear Scan — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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