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

• Difficulty Level : Easy
• Last Updated : 16 Jul, 2021

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 :

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

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

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 == ``'1'``)``        ``auxArr = 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 == ``'1'``)``        ``auxArr = 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``?>`

## Javascript

 ``

Output :

`6`

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

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