# Leetcode [1085] - Sum of Digits in the Minimum Number (Java Solution)

**Leetcode [1085]** - Sum of Digits in the Minimum Number (Java Solution)

**Category:** Easy

**What is used**: Arrays, Linear Search

## Question

Given an **array A** of positive integers, let **S** be the *sum of the digits of the minimal element* of **A.**

Return **0** if S is *odd, otherwise* return **1.**

### Example 1:

**Input**: [34,23,1,24,75,33,54,8]

**Output**: 0

**Explanation**: The minimal element is 1, and the sum of those digits is S = 1 which is odd, so the answer is 0.

### Example 2:

**Input**: [99,77,33,66,55]

**Output**: 1

**Explanation**: The minimal element is 33, and the sum of those digits is S = 3 + 3 = 6 which is even, so the answer is 1.

#### Note:

- 1 <=
**A.length**<= 100 - 1 <=
**A[i].length**<= 100

## Solution

This is a easy problem. The 3 steps to solve this problem should be:

1. Find the minimum of the array - I used Linear Search **O(n)**

2. Find the sum of the minimum number - The standard way of modulo 10 and divide by 10. **O(n)**

3. Check if sum is odd - Simple if-else statement

```
```
class Solution {
public int sumOfDigits(int[] A) {
//Find minimum
int min = Integer.MAX_VALUE;
for(int i = 0; i < A.length; i++){
if(A[i] < min){
min = A[i];
}
}
int n = 0;
int sum = 0;
//Sum of minimum
while(min > 0){
n = min % 10;
sum = sum + n;
min = min / 10;
}
//Odd-Even Check
if(sum % 2 != 0){
return 0;
}else{
return 1;
}
}
```

#### Time complexity:

The time complexity of this code is **O(n)**

#### Submission Details:

**Runtime**: 0 ms, faster than 100.00% of Java online submissions for Sum of Digits in the Minimum Number.

**Memory Usage**: 34.3 MB, less than 100.00% of Java online submissions for Sum of Digits in the Minimum Number.

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