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 : 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 zeroes. 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<iostream> 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<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 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<n; i++) { if (s.charAt(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.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
<?php // PHP program to count // substrings with odd // decimal value // function to count number // of substrings with odd // decimal representation function countSubstr( $s ) { $n = strlen ( $s ); // auxiliary array to // store count of 1's // before ith index $auxArr = array (); if ( $s [0] == '1' ) $auxArr [0] = 1; // store count of 1's // before i-th index for ( $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 $count = 0; // traverse the string // reversely to calculate // number of odd substrings // before i-th index for ( $i = $n - 1; $i >= 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
<script> // Javascript program to count substrings // with odd decimal value // Function to count number of substrings // with odd decimal representation function countSubstr(s) { let n = s.length; // auxiliary array to store count // of 1's before ith index let auxArr = new Array(n); if (s[0] == '1 ') auxArr[0] = 1; // Store count of 1' s before // i-th index for (let 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 let count = 0; // Traverse the string reversely to // calculate number of odd substrings // before i-th index for (let i = n - 1; i >= 0; i--) if (s[i] == '1' ) count += auxArr[i]; return count; } // Driver code let s = "1101" ; document.write(countSubstr(s)); // This code is contributed by rameshtravel07 </script> |
6
Time Complexity: O(n)
Auxiliary Space: O(n)
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