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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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

✓ Brute Force - Linear Search

⏱️ Time: O(n + d) Space: O(1)

First pass through the array to find the minimum value. Then extract each digit of the minimum number and sum them up. Finally check if the sum is odd or even.

Optimized Single Pass

⏱️ Time: O(n + d) Space: O(1)

Use built-in functions to find minimum efficiently, then use mathematical operations to sum digits without string conversion. This approach is cleaner and slightly more efficient.

Brute Force - Linear Search — Algorithm Steps

Iterate through array to find minimum value
Extract digits of minimum number and sum them
Return 0 if sum is odd, 1 if even

Visualization

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Step-by-Step Walkthrough

Find Minimum

Scan array to find minimum value

Sum Digits

Extract and sum digits of minimum number

Check Parity

Return 0 if odd sum, 1 if even sum

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
    int digitSum = 0;
    int num = abs(minVal);  // Handle negative numbers
    while (num > 0) {
        digitSum += num % 10;
        num /= 10;
    }
    
    // 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) to find minimum + O(d) to sum digits where d is number of digits

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Sum of Digits in the Minimum Number - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Linear Search — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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