Count of substrings that start and end with 1 in given Binary String
Last Updated : 03 Feb, 2023
Given a binary string, count the number of substrings that start and end with 1.
Examples:
Input: "00100101"
Output: 3
Explanation: three substrings are "1001", "100101" and "101"
Input: "1001"
Output: 1
Explanation: one substring "1001"
Count of substrings that start and end with 1 in given Binary String using Nested Loop:
A Simple Solution is to run two loops. Outer loops pick every 1 as a starting point and the inner loop searches for ending 1 and increments count whenever it finds 1.
Below is the implementation of above approach:
C++ // A simple C++ program to count number of // substrings starting and ending with 1 #include <iostream> using namespace std; int countSubStr(char str[]) { int res = 0; // Initialize result // Pick a starting point for (int i = 0; str[i] != '\0'; i++) { if (str[i] == '1') { // Search for all possible ending point for (int j = i + 1; str[j] != '\0'; j++) if (str[j] == '1') res++; } } return res; } // Driver program to test above function int main() { char str[] = "00100101"; cout << countSubStr(str); return 0; } // A simple Java program to count number of // substrings starting and ending with 1 class CountSubString { int countSubStr(char str[], int n) { int res = 0; // Initialize result // Pick a starting point for (int i = 0; i < n; i++) { if (str[i] == '1') { // Search for all possible ending point for (int j = i + 1; j < n; j++) { if (str[j] == '1') res++; } } } return res; } // Driver program to test the above function public static void main(String[] args) { CountSubString count = new CountSubString(); String string = "00100101"; char str[] = string.toCharArray(); int n = str.length; System.out.println(count.countSubStr(str, n)); } } # A simple Python 3 program to count number of # substrings starting and ending with 1 def countSubStr(st, n): # Initialize result res = 0 # Pick a starting point for i in range(0, n): if (st[i] == '1'): # Search for all possible ending point for j in range(i+1, n): if (st[j] == '1'): res = res + 1 return res # Driver program to test above function st = "00100101" list(st) n = len(st) print(countSubStr(st, n), end="") # This code is contributed # by Nikita Tiwari.
// A simple C# program to count number of // substrings starting and ending with 1 using System; class GFG { public virtual int countSubStr(char[] str, int n) { int res = 0; // Initialize result // Pick a starting point for (int i = 0; i < n; i++) { if (str[i] == '1') { // Search for all possible // ending point for (int j = i + 1; j < n; j++) { if (str[j] == '1') { res++; } } } } return res; } // Driver Code public static void Main(string[] args) { GFG count = new GFG(); string s = "00100101"; char[] str = s.ToCharArray(); int n = str.Length; Console.WriteLine(count.countSubStr(str, n)); } } // This code is contributed by Shrikant13 <?php // A simple PHP program to count number of // substrings starting and ending with 1 function countSubStr($str) { $res = 0; // Initialize result // Pick a starting point for ($i = 0; $i < strlen($str); $i++) { if ($str[$i] == '1') { // Search for all possible // ending point for ($j = $i + 1; $j < strlen($str); $j++) if ($str[$j] == '1') $res++; } } return $res; } // Driver Code $str = "00100101"; echo countSubStr($str); // This code is contributed by ita_c ?> <script> // A simple javascript program to count number of // substrings starting and ending with 1 function countSubStr(str,n) { let res = 0; // Initialize result // Pick a starting point for (let i = 0; i<n; i++) { if (str[i] == '1') { // Search for all possible ending point for (let j = i + 1; j< n; j++) { if (str[j] == '1') res++; } } } return res; } // Driver program to test the above function let string = "00100101"; let n=string.length; document.write(countSubStr(string,n)); // This code is contributed by rag2127 </script> Time Complexity: O(N2),
Auxiliary Space: O(1)
Count of substrings that start and end with 1 in a given Binary String using Subarray count:
We know that if count of 1's is m, then there will be m * (m - 1) / 2 possible subarrays.
Follow the steps to solve the problem:
- Count the number of 1's. Let the count of 1's be m.
- Return m(m-1)/2
Below is the implementation of above approach:
C++ // A O(n) C++ program to count number of // substrings starting and ending with 1 #include <iostream> using namespace std; int countSubStr(char str[]) { int m = 0; // Count of 1's in input string // Traverse input string and count of 1's in it for (int i = 0; str[i] != '\0'; i++) { if (str[i] == '1') m++; } // Return count of possible pairs among m 1's return m * (m - 1) / 2; } // Driver program to test above function int main() { char str[] = "00100101"; cout << countSubStr(str); return 0; } // A O(n) Java program to count number of substrings // starting and ending with 1 class CountSubString { int countSubStr(char str[], int n) { int m = 0; // Count of 1's in input string // Traverse input string and count of 1's in it for (int i = 0; i < n; i++) { if (str[i] == '1') m++; } // Return count of possible pairs among m 1's return m * (m - 1) / 2; } // Driver program to test the above function public static void main(String[] args) { CountSubString count = new CountSubString(); String string = "00100101"; char str[] = string.toCharArray(); int n = str.length; System.out.println(count.countSubStr(str, n)); } } # A Python3 program to count number of # substrings starting and ending with 1 def countSubStr(st, n): # Count of 1's in input string m = 0 # Traverse input string and # count of 1's in it for i in range(0, n): if (st[i] == '1'): m = m + 1 # Return count of possible # pairs among m 1's return m * (m - 1) // 2 # Driver program to test above function st = "00100101" list(st) n = len(st) print(countSubStr(st, n), end="") # This code is contributed # by Nikita Tiwari.
// A O(n) C# program to count // number of substrings starting // and ending with 1 using System; class GFG { int countSubStr(char[] str, int n) { int m = 0; // Count of 1's in // input string // Traverse input string and // count of 1's in it for (int i = 0; i < n; i++) { if (str[i] == '1') m++; } // Return count of possible // pairs among m 1's return m * (m - 1) / 2; } // Driver Code public static void Main(String[] args) { GFG count = new GFG(); String strings = "00100101"; char[] str = strings.ToCharArray(); int n = str.Length; Console.Write(count.countSubStr(str, n)); } } // This code is contributed by princiraj <?php // A simple PHP program to count number of // substrings starting and ending with 1 function countSubStr($str) { $m = 0; // Initialize result // Pick a starting point for ($i = 0; $i < strlen($str); $i++) { if ($str[$i] == '1') { $m++; } } // Return count of possible // pairs among m 1's return $m * ($m - 1) / 2; } // Driver Code $str = "00100101"; echo countSubStr($str); // This code is contributed // by Akanksha Rai ?> <script> // A O(n) javascript program to count number of substrings //starting and ending with 1 function countSubStr(str,n) { let m = 0; // Count of 1's in input string // Traverse input string and count of 1's in it for (let i = 0; i < n; i++) { if (str[i] == '1') m++; } // Return count of possible pairs among m 1's return m * Math.floor((m - 1) / 2); } // Driver program to test the above function let str = "00100101"; let n = str.length; document.write(countSubStr(str, n)); // This code is contributed by avanitrachhadiya2155 </script> Time Complexity: O(N), where n is the length of the string.
Auxiliary Space: O(1).
Count of substrings that start and end with 1 in given Binary String using Recursion:
This approach is the same as the above approach but here to calculate the count of 1s we use recursion.
Follow the steps to solve the problem:
- Count the number of 1's using recursion. Let the count of 1's be m.
- Return m(m-1)/2
Below is the implementation of above approach:
C++ // A O(n) C++ program to count number of // substrings starting and ending with 1 #include <bits/stdc++.h> using namespace std; int helper(int n, char str[], int i) { // if 'i' is on the last index if (i == n - 1) return (str[i] == '1') ? 1 : 0; // if current char is 1 // add 1 to the answer if (str[i] == '1') return 1 + helper(n, str, i + 1); // if it is zero else return helper(n, str, i + 1); } int countSubStr(char str[]) { int n = strlen(str); // counting the number of 1's in the string int count = helper(n, str, 0); // return the number of combinations return (count * (count - 1)) / 2; } // Driver program to test above function int main() { char str[] = "00100101"; cout << countSubStr(str); return 0; } // this code is contributed by rajdeep999 /*package whatever //do not write package name here */ import java.io.*; class GFG { static int helper(int n, char str[], int i) { // if 'i' is on the last index if (i == n - 1) return (str[i] == '1') ? 1 : 0; // if current char is 1 // add 1 to the answer if (str[i] == '1') return 1 + helper(n, str, i + 1); // if it is zero else return helper(n, str, i + 1); } static int countSubStr(char str[]) { int n = str.length; // counting the number of 1's in the string int count = helper(n, str, 0); // return the number of combinations return (count * (count - 1)) / 2; } public static void main (String[] args) { char str[] = "00100101".toCharArray(); System.out.println(countSubStr(str)); } } // This code is contributed by aadityaburujwale. class GFG : @staticmethod def helper( n, str, i) : # if 'i' is on the last index if (i == n - 1) : return 1 if (str[i] == '1') else 0 # if current char is 1 # add 1 to the answer if (str[i] == '1') : return 1 + GFG.helper(n, str, i + 1) else : return GFG.helper(n, str, i + 1) @staticmethod def countSubStr( str) : n = len(str) # counting the number of 1's in the string count = GFG.helper(n, str, 0) # return the number of combinations return int((count * (count - 1)) / 2) @staticmethod def main( args) : str = list("00100101") print(GFG.countSubStr(str)) if __name__=="__main__": GFG.main([]) # This code is contributed by aadityaburujwale. // Include namespace system using System; public class GFG { public static int helper(int n, char[] str, int i) { // if 'i' is on the last index if (i == n - 1) { return (str[i] == '1') ? 1 : 0; } // if current char is 1 // add 1 to the answer if (str[i] == '1') { return 1 + GFG.helper(n, str, i + 1); } else { return GFG.helper(n, str, i + 1); } } public static int countSubStr(char[] str) { var n = str.Length; // counting the number of 1's in the string var count = GFG.helper(n, str, 0); // return the number of combinations return (int)((count * (count - 1)) / 2); } public static void Main(String[] args) { char[] str = "00100101".ToCharArray(); Console.WriteLine(GFG.countSubStr(str)); } } // This code is contributed by aadityaburujwale. // A O(n) JS program to count number of // substrings starting and ending with 1 function helper(n, str, i) { // if 'i' is on the last index if (i == n - 1) { return (str[i] == '1') ? 1 : 0; } // if current char is 1 // add 1 to the answer if (str[i] == '1') { return 1 + helper(n, str, i + 1); } // if it is zero else { return helper(n, str, i + 1); } } function countSubStr(str) { let n = str.length; // counting the number of 1's in the string let count = helper(n, str, 0); // return the number of combinations return (count * (count - 1)) / 2; } // Driver program to test above function console.log(countSubStr("00100101")); // This code is contributed by akashish_ Time Complexity: O(N), Traversing over the string of size N
Auxiliary Space: O(N), for recursion call stack
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