# Number of substrings with odd decimal value in a binary string

Given a binary string containing only 0’s and 1’s. Write a program to find number of sub-strings of this string whose decimal representation is odd.

Examples :

Input : 101
Output : 3
Explanation : Substrigs with odd decimal
representation are:
{1, 1, 101}

Input : 1101
Output : 6
Explanation : Substrigs with odd decimal
representation are:
{1, 1, 1, 11, 101, 1011}

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Brute force Approach: The simplest approach to solve above problem is to generate all possible substrings of the given string and convert them to decimal and check if the decimal representation is odd or not. You may refer to this article for binary to decimal conversion.

Time Complexity: O(n*n)

Efficient approach: An efficient approach is to observe that if the last digit of a binary number is 1 then it is odd otherwise it is even. So our problem now is reduced to check all substrings with value at last index as 1. We can easily solve this problem in single traversal by traversing from the end. If the value of i-th index in the string is 1 then there is i odd substrings before this index. But this also includes strings with leading zero’s. So to handle this we can take an auxiliary array to keep count of number of 1’s before ith index. We count only pairs of 1s.

Below is the implementation of this approach:

## C++

 // CPP program to count substrings // with odd decimal value #include using namespace std;    // function to count number of substrings  // with odd decimal representation int countSubstr(string s) {        int n = s.length();            // auxiliary array to store count      // of 1's before ith index     int auxArr[n] = {0};            if (s[0] == '1')         auxArr[0] = 1;            // store  count of 1's before      // i-th  index     for (int i=1; i=0; i--)             if (s[i] == '1')             count += auxArr[i];                return count; }    // Driver code int main() {     string s = "1101";         cout << countSubstr(s);         return 0; }

## Java

 // Java program to count substrings // with odd decimal value import java.io.*; import java.util.*;    class GFG {       // function to count number of substrings  // with odd decimal representation static int countSubstr(String s) {      int n = s.length();            // auxiliary array to store count      // of 1's before ith index     int[] auxArr=new int[n];            if (s.charAt(0) == '1')         auxArr[0] = 1;            // store count of 1's before      // i-th index     for (int i=1; i=0; i--)          if (s.charAt(i) == '1')             count += auxArr[i];             return count; }    public static void main (String[] args) {      String s = "1101";      System.out.println(countSubstr(s));            } }    // This code is contributed by Gitanjali.

## Python3

 # python program to count substrings # with odd decimal value import math     # function to count number of substrings  # with odd decimal representation def countSubstr( s):        n = len(s)            # auxiliary array to store count      # of 1's before ith index     auxArr= [0 for i in range(n)]            if (s[0] == '1'):         auxArr[0] = 1            # store count of 1's before      # i-th index     for i in range(0,n):                  if (s[i] == '1'):             auxArr[i] = auxArr[i-1]+1       else:             auxArr[i] = auxArr[i-1]                   # variable to store answer     count = 0            # traverse the string reversely to      # calculate number of odd substrings      # before i-th index     for i in range(n-1,-1,-1):          if (s[i] == '1'):             count += auxArr[i]             return count # Driver method s = "1101" print (countSubstr(s))    # This code is contributed by Gitanjali.

## C#

 // C# program to count substrings // with odd decimal value using System;    class GFG {        // Function to count number of substrings  // with odd decimal representation static int countSubstr(string s) {      int n = s.Length;            // auxiliary array to store count      // of 1's before ith index     int[] auxArr = new int[n];            if (s[0] == '1')         auxArr[0] = 1;            // store count of 1's before      // i-th index     for (int i = 1; i < n; i++)     {         if (s[i] == '1')             auxArr[i] = auxArr[i - 1] + 1;         else             auxArr[i] = auxArr[i - 1];     }            // variable to store answer     int count = 0;            // Traverse the string reversely to      // calculate number of odd substrings      // before i-th index     for (int i = n - 1; i >= 0; i--)          if (s[i] == '1')             count += auxArr[i];             return count; }    // Driver Code public static void Main () {     string s = "1101";      Console.WriteLine(countSubstr(s));     } }    // This code is contributed by vt_m.

## PHP

 = 0; \$i--)          if (\$s[\$i] == '1')             \$count += \$auxArr[\$i];             return \$count; }    // Driver code \$s = "1101";  echo countSubstr(\$s);     // This code is contributed by aj_36 ?>

Output :

6

Time Complexity: O(n)
Auxiliary Space: O(n)

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