Count of substrings that start and end with 1 in given Binary String
Last Updated :
03 Feb, 2023
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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:
// 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>
Output
3
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:
// 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>
Output
3
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:
// 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_
Output
3
Time Complexity: O(N), Traversing over the string of size N
Auxiliary Space: O(N), for recursion call stack
