Count Integers With Even Digit Sum

Problem Description

Given a positive integer num, return the number of positive integers less than or equal to num whose digit sums are even.

The digit sum of a positive integer is the sum of all its digits.

Example

Example 1:

Input: num = 4
Output: 2
Explanation:
The only integers less than or equal to 4 whose digit sums are even are 2 and 4.  

Example 2:

Input: num = 30
Output: 14
Explanation:
The 14 integers less than or equal to 30 whose digit sums are even are
2, 4, 6, 8, 11, 13, 15, 17, 19, 20, 22, 24, 26, and 28.

Constraints

• 1 <= num <= 1000

Solution Approach

Intuition:

To efficiently Count Integers With Even Digit Sum

Solution Implementation

Code In Different Languages:

JavaScript

Written by @Ishitamukherjee2004

 
class Solution {
countEven(num) {
 let cnt = 0;
 for(let i = 1; i <= num; i++){
   let sum = 0;
   if(i < 10){
     if(i % 2 == 0) cnt++;
   } else{
     let temp = i;
     while(temp > 0){
       let rev = temp % 10;
       sum += rev;
       temp = Math.floor(temp / 10);
     }
     if(sum % 2 == 0) cnt++;
   }
 }
 return cnt;
}
}

TypeScript

Written by @Ishitamukherjee2004

 
class Solution {
countEven(num: number): number {
 let cnt = 0;
 for(let i = 1; i <= num; i++){
   let sum = 0;
   if(i < 10){
     if(i % 2 == 0) cnt++;
   } else{
     let temp = i;
     while(temp > 0){
       let rev = temp % 10;
       sum += rev;
       temp = Math.floor(temp / 10);
     }
     if(sum % 2 == 0) cnt++;
   }
 }
 return cnt;
}
}

Python

Written by @Ishitamukherjee2004

 
class Solution:
 def countEven(self, num):
     cnt = 0
     for i in range(1, num + 1):
         sum = 0
         if i < 10:
             if i % 2 == 0:
                 cnt += 1
         else:
             temp = i
             while temp > 0:
                 rev = temp % 10
                 sum += rev
                 temp = temp // 10
             if sum % 2 == 0:
                 cnt += 1
     return cnt

Java

Written by @Ishitamukherjee2004

public class Solution {
 public int countEven(int num) {
     int cnt = 0;
     for(int i = 1; i <= num; i++){
         int sum = 0;
         if(i < 10){
             if(i % 2 == 0) cnt++;
         } else{
             int temp = i;
             while(temp > 0){
                 int rev = temp % 10;
                 sum += rev;
                 temp = temp / 10;
             }
             if(sum % 2 == 0) cnt++;
         }
     }
     return cnt;
 }
}

C++

Written by @Ishitamukherjee2004

class Solution {
public:
 int countEven(int num) {
     int cnt = 0;
     for(int i=1; i<=num; i++){
         int sum = 0;
         if(i<10){
             if(i%2==0) cnt++;
         }
         else{
             int temp = i;
             while(temp>0){
                 int rev = temp%10;
                 sum+=rev;
                 temp/=10;
             }
             if(sum%2==0) cnt++;

         }
     }
     return cnt;
 }
};

Complexity Analysis

• Time Complexity: $O(n)$
• Space Complexity: $O(1)$
• The time complexity is $O(n)$,where n is the input number num. This is because the algorithm uses a single loop that iterates from 1 to num.
• The space complexity is $O(1)$ because we are not using any extra space.