Count of substrings of a Binary string containing only 1s
Last Updated :
16 Nov, 2023
Given a binary string of length N, we need to find out how many substrings of this string contain only 1s.
Examples:
Input: S = "0110111"
Output: 9
Explanation:
There are 9 substring with only 1's characters.
"1" comes 5 times.
"11" comes 3 times.
"111" comes 1 time.
Input: S = "000"
Output: 0
The_Approach: the approach simple we will going to traverse over array and count one follow the steps.
- initialize total=0 and currone=0.
- then traverse over string and count ones in string till zero occur.
- if zero occur then just add the number of possible substring that are possible from currone as.
- total+=(currone*(currone+1))/2. and after addition currone=0.
- do this till n-1.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
#include <iostream>
using namespace std;
int main()
{
// Given string.
string s = "0110111";
// to avoid int overflow.
int mod = 1000000007;
long long int total = 0;
long long int one = 0;
// traverse over string.
for (int i = 0; i < s.size(); i++) {
// count the ones.
if (s[i] == '1') {
++one;
}
else {
long long x = (one * (one + 1)) % mod;
total += x / 2;
one = 0;
}
}
// count if reamin.
if (one != 0)
total += (long long)(one * (one + 1)) / 2;
// printing the output.
cout << total << endl;
return 0;
}
import java.util.*;
public class GFG {
public static void main(String[] args) {
// Given string.
String s = "0110111";
// To avoid integer overflow.
long mod = 1000000007;
long total = 0;
long one = 0;
// Traverse over the string.
for (int i = 0; i < s.length(); i++) {
// Count the ones.
if (s.charAt(i) == '1') {
one++;
} else {
long x = (one * (one + 1)) % mod;
total += x / 2;
one = 0;
}
}
// Count if remaining ones.
if (one != 0) {
total += (long) (one * (one + 1)) / 2;
}
// Printing the output.
System.out.println(total);
}
}
# Given string
s = "0110111"
# to avoid int overflow
mod = 1000000007
total = 0
one = 0
# traverse over string
for i in range(len(s)):
# count the ones
if s[i] == '1':
one += 1
else:
x = (one * (one + 1)) % mod
total += x // 2
one = 0
# count if remain
if one != 0:
total += (one * (one + 1)) // 2
# printing the output
print(total)
using System;
class Program {
static void Main()
{
// Given string.
string s = "0110111";
// to avoid int overflow.
int mod = 1000000007;
long total = 0;
long one = 0;
// traverse over string.
for (int i = 0; i < s.Length; i++)
{
// count the ones.
if (s[i] == '1') {
++one;
}
else {
long x = (one * (one + 1)) % mod;
total += x / 2;
one = 0;
}
}
// count if reamin.
if (one != 0)
total += (long)(one * (one + 1)) / 2;
// printing the output.
Console.WriteLine(total);
}
}
// This code is contributed by user_dtewbxkn77n
// Given string.
let s = "0110111";
// to avoid int overflow.
let mod = 1000000007;
let total = 0;
let one = 0;
// traverse over string.
for (let i = 0; i < s.length; i++) {
// count the ones.
if (s[i] == '1') {
++one;
}
else {
let x = (one * (one + 1)) % mod;
total += x / 2;
one = 0;
}
}
// count if remain.
if (one != 0)
total += (one * (one + 1)) / 2;
// printing the output.
console.log(total);
Complexity Analysis:
Time Complexity : O(n).
Auxiliary Space : O(1).
Approach: The idea is to traverse the binary string and count the consecutive ones in the string. Below is the illustration of the approach:
- Traverse the given binary string from index 0 to length - 1
- Count the number of consecutive "1" till index i.
- For each new character str[i], there will be more substring with all character's as "1"
Below is the implementation of the above approach:
C++
// C++ implementation to find
// count of substring containing
// only ones
#include <bits/stdc++.h>
using namespace std;
// Function to find the total number
// of substring having only ones
int countOfSubstringWithOnlyOnes(string s)
{
int res = 0, count = 0;
for (int i = 0; i < s.length(); i++) {
count = s[i] == '1' ? count + 1 : 0;
res = (res + count);
}
return res;
}
// Driver Code
int main()
{
string s = "0110111";
cout << countOfSubstringWithOnlyOnes(s)
<< endl;
return 0;
}
// Java implementation to find
// count of substring containing
// only ones
class GFG{
// Function to find the total number
// of substring having only ones
static int countOfSubstringWithOnlyOnes(String s)
{
int res = 0, count = 0;
for(int i = 0; i < s.length(); i++)
{
count = s.charAt(i) == '1' ? count + 1 : 0;
res = (res + count);
}
return res;
}
// Driver code
public static void main(String[] args)
{
String s = "0110111";
System.out.println(countOfSubstringWithOnlyOnes(s));
}
}
// This code is contributed by dewantipandeydp
# Python3 implementation to find
# count of substring containing
# only ones
# Function to find the total number
# of substring having only ones
def countOfSubstringWithOnlyOnes(s):
count = 0
res = 0
for i in range(0,len(s)):
if s[i] == '1':
count = count + 1
else:
count = 0;
res = res + count
return res
# Driver Code
s = "0110111"
print(countOfSubstringWithOnlyOnes(s))
# This code is contributed by jojo9911
// C# implementation to find count
// of substring containing only ones
using System;
class GFG{
// Function to find the total number
// of substring having only ones
static int countOfSubstringWithOnlyOnes(string s)
{
int res = 0, count = 0;
for(int i = 0; i < s.Length; i++)
{
count = s[i] == '1' ? count + 1 : 0;
res = (res + count);
}
return res;
}
// Driver code
public static void Main(string[] args)
{
string s = "0110111";
Console.Write(countOfSubstringWithOnlyOnes(s));
}
}
// This code is contributed by rutvik_56
<script>
// Javascript implementation to find
// count of substring containing
// only ones
// Function to find the total number
// of substring having only ones
function countOfSubstringWithOnlyOnes(s)
{
var res = 0, count = 0;
for (var i = 0; i < s.length; i++) {
count = s[i] == '1' ? count + 1 : 0;
res = (res + count);
}
return res;
}
// Driver Code
var s = "0110111";
document.write( countOfSubstringWithOnlyOnes(s));
</script>
Time Complexity: O(n), where n is the size of the given string
Auxiliary Space: O(1)
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