Sum of Digits in the Minimum Number - Practice Coding Problems

Sum of Digits in the Minimum Number - Problem

Array Math Easy

Given an integer array nums, return 0 if the sum of the digits of the minimum integer in nums is odd, or 1 otherwise.

For example, if the minimum number is 23, the sum of digits is 2 + 3 = 5, which is odd, so return 0.

Input & Output

Example 1 — Basic Case

$ Input: nums = [99,77,33,44,11]

› Output: 1

💡 Note: Minimum is 11. Sum of digits: 1 + 1 = 2 (even), so return 1

Example 2 — Odd Sum

$ Input: nums = [18,38,58,23]

› Output: 0

💡 Note: Minimum is 18. Sum of digits: 1 + 8 = 9 (odd), so return 0

Example 3 — Single Digit

$ Input: nums = [7,4,9,6,2]

› Output: 1

💡 Note: Minimum is 2. Sum of digits: 2 (even), so return 1

Constraints

1 ≤ nums.length ≤ 100
1 ≤ nums[i] ≤ 99

Visualization

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Understanding the Visualization

Input Array

Array of positive integers

Find Minimum

Locate the smallest value in array

Sum Digits

Add up all digits of minimum number

Check Parity

Return 0 if odd, 1 if even

Key Takeaway

🎯 Key Insight: Find the minimum value, sum its digits, and check if the sum is odd (return 0) or even (return 1)

Asked in

a Amazon 15 G Google 8

The key insight is to find the minimum value in the array and calculate the sum of its digits. Best approach uses built-in min function followed by digit extraction. Time: O(n), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Optimized Single Pass	O(n + d)	O(1)	Find minimum and calculate digit sum in optimized way
Brute Force - Linear Search	O(n + d)	O(1)	Iterate through array to find minimum, then sum its digits

Optimized Single Pass — Algorithm Steps

Use built-in min() function to find minimum
Use modulo arithmetic to extract and sum digits
Return result based on parity of sum

Visualization

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Step-by-Step Walkthrough

Built-in Min

Use efficient built-in function to find minimum

String Conversion

Convert to string and process each digit

Sum & Check

Sum digits and check if odd or even

Code -

solution.c — C

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

int solution(int* nums, int numsSize) {
    // Find minimum value
    int minVal = INT_MAX;
    for (int i = 0; i < numsSize; i++) {
        if (nums[i] < minVal) {
            minVal = nums[i];
        }
    }
    
    // Calculate sum of digits using string approach
    char numStr[20];
    sprintf(numStr, "%d", abs(minVal));
    int digitSum = 0;
    for (int i = 0; numStr[i] != '\0'; i++) {
        digitSum += numStr[i] - '0';
    }
    
    // Return 0 if odd, 1 if even
    return digitSum % 2 == 1 ? 0 : 1;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int nums[1000];
    int count = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != ']') {
            nums[count++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    printf("%d\n", solution(nums, count));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + d)

O(n) for finding minimum + O(d) for digit processing where d ≤ 10

✓ Linear Growth

Space Complexity

O(1)

Only constant extra space for variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Optimized Single Pass — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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