## Problem Statement

In the “Even Substring Count” problem we have given an input string which is formed by digits. Write a program or code to find the count of substrings which when converting into integer form even.

## Input Format

The first and only one line containing a string “s”.

## Output Format

The first line contains an integer value that represents the number of substrings of “s” which when converting into integer form even.

## Constraints

- 1<=|s|<=10^6
- s[i] is in the range from 0 to 9

## Example

1234

6

**Explanation:** Here we simply check the position of the even digit and add the position of this digit(1-base indexing) into the answer. Like here the position of ‘2’ is 2 so we add 2 to our answer. Now moving ahead and see the next digit is ‘4’ and we add 4 to our answer. So, we got our answer which is 6. And if we check manually 2, 4, 12, 34, 234, 1234 are the substrings that are even and their_count is also equal to 6.

## Algorithm for Even Substring Count

**1.** Declare one variable ans and set it to zero.

**2.** Start traversing from the beginning of the string and check for the digit.

**3**. If the digit is even then add the position(1-based indexing) of this number into the answer.

**4.** Print ans which containing the count of the substrings.

## Implementation

### C++ Program for Even Substring Count

#include <bits/stdc++.h> using namespace std; int main() { string s; cin>>s; int n=s.length(); int ans=0; for(int i=0;i<n;i++) { if((s[i]-'0')%2==0) { ans+=(i+1); } } cout<<ans<<endl; return 0; }

### Java Program for Even Substring Count

import java.util.Scanner; class sum { public static void main(String[] args) { Scanner sr = new Scanner(System.in); String s = sr.next(); int n = s.length(); int ans=0; for(int i=0;i<n;i++) { int x = s.charAt(i)-48; if(x%2==0) { ans+=(i+1); } }ven dig System.out.println(ans); } }

213578

7

## Complexity Analysis for Even Substring Count

### Time Complexity

**O(n)** where **n** is the size of the given string. Here we simply traverse the given string and add the index of the even digit into our answer.

### Space Complexity

**O(1)** because we don’t use any auxiliary space here. Here we only declare one variable to store the answer and print it.