Given binary string str of 0’s and 1’s only. The task is to count the total number of substrings of string str such that each substring has an equal number of consecutive 0’s and 1’s in it.
Examples:
Input: str = “010011”
Output: 4
Explanation:
The substrings with consecutive 0’s and 1’s are “01”, “10”, “0011”, “01”. Hence, the count is 4.
Note:
The two “01” are at different positions: [0, 1] and [3, 4].
“010011” has the same number of 0’s and 1’s but they are not consecutive.Input: str = “0001110010”
Output: 7
Explanation:
The substrings with consecutive 0’s and 1’s are “000111”, “0011”, “01”, “1100”, “10”, “01”, “10”.
Approach:
- Count the number of consecutive 0’s (or 1’s) from start of the string.
- Then count the number of consecutive 1’s (or 0’s) from the position in the string str where count of 0’s (or 1’s) ends.
- The total number of substrings with consecutive 0’s and 1’s is the minimum of the count of consecutive 0’s and 1’s found in above two steps.
- Repeat the above steps till the end of the string str.
Below is the implementation of the above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the count of substrings with equal no. // of consecutive 0's and 1's int countSubstring(string& S, int & n)
{ // To store the total count of substrings
int ans = 0;
int i = 0;
// Traversing the string
while (i < n) {
// Count of consecutive 0's & 1's
int cnt0 = 0, cnt1 = 0;
// Counting subarrays of type "01"
if (S[i] == '0' ) {
// Count the consecutive 0's
while (i < n && S[i] == '0' ) {
cnt0++;
i++;
}
// If consecutive 0's ends then check for
// consecutive 1's
int j = i;
// Counting consecutive 1's
while (j < n && S[j] == '1' ) {
cnt1++;
j++;
}
}
// Counting subarrays of type "10"
else {
// Count consecutive 1's
while (i < n && S[i] == '1' ) {
cnt1++;
i++;
}
// If consecutive 1's ends then check for
// consecutive 0's
int j = i;
// Count consecutive 0's
while (j < n && S[j] == '0' ) {
cnt0++;
j++;
}
}
// Update the total count of substrings with minimum
// of (cnt0, cnt1)
ans += min(cnt0, cnt1);
}
// Return answer
return ans;
} // Driver code int main()
{ string S = "0001110010" ;
int n = S.length();
// Function to print the count of substrings
cout << countSubstring(S, n);
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
// C implementation of the above approach #include <stdio.h> #include <string.h> // Find minimum between two numbers. int min( int num1, int num2)
{ return (num1 > num2) ? num2 : num1;
} // Function to find the count of substrings with equal no. // of consecutive 0's and 1's int countSubstring( char S[], int n)
{ // To store the total count of substrings
int ans = 0;
int i = 0;
// Traversing the string
while (i < n) {
// Count of consecutive 0's & 1's
int cnt0 = 0, cnt1 = 0;
// Counting subarrays of type "01"
if (S[i] == '0' ) {
// Count the consecutive 0's
while (i < n && S[i] == '0' ) {
cnt0++;
i++;
}
// If consecutive 0's ends then check for
// consecutive 1's
int j = i;
// Counting consecutive 1's
while (j < n && S[j] == '1' ) {
cnt1++;
j++;
}
}
// Counting subarrays of type "10"
else {
// Count consecutive 1's
while (i < n && S[i] == '1' ) {
cnt1++;
i++;
}
// If consecutive 1's ends then check for
// consecutive 0's
int j = i;
// Count consecutive 0's
while (j < n && S[j] == '0' ) {
cnt0++;
j++;
}
}
// Update the total count of substrings with minimum
// of (cnt0, cnt1)
ans += min(cnt0, cnt1);
}
// Return answer
return ans;
} // Driver code int main()
{ char S[] = "0001110010" ;
int n = strlen (S);
// Function to print the count of substrings
printf ( "%d" , countSubstring(S, n));
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
// Java implementation of the above approach class GFG {
// Function to find the count of substrings with equal
// no. of consecutive 0's and 1's
static int countSubstring(String S, int n)
{
// To store the total count of substrings
int ans = 0 ;
int i = 0 ;
// Traversing the string
while (i < n) {
// Count of consecutive 0's & 1's
int cnt0 = 0 , cnt1 = 0 ;
// Counting subarrays of type "01"
if (S.charAt(i) == '0' ) {
// Count the consecutive 0's
while (i < n && S.charAt(i) == '0' ) {
cnt0++;
i++;
}
// If consecutive 0's ends then check for
// consecutive 1's
int j = i;
// Counting consecutive 1's
while (j < n && S.charAt(j) == '1' ) {
cnt1++;
j++;
}
}
// Counting subarrays of type "10"
else {
// Count consecutive 1's
while (i < n && S.charAt(i) == '1' ) {
cnt1++;
i++;
}
// If consecutive 1's ends then check for
// consecutive 0's
int j = i;
// Count consecutive 0's
while (j < n && S.charAt(j) == '0' ) {
cnt0++;
j++;
}
}
// Update the total count of substrings with
// minimum of (cnt0, cnt1)
ans += Math.min(cnt0, cnt1);
}
// Return answer
return ans;
}
// Driver code
static public void main(String args[])
{
String S = "0001110010" ;
int n = S.length();
// Function to print the count of substrings
System.out.println(countSubstring(S, n));
}
} // This code is contributed by Aditya Kumar (adityakumar129) |
# Python3 implementation of the # above approach # Function to find the count # of substrings with equal no. # of consecutive 0's and 1's def countSubstring(S, n) :
# To store the total count
# of substrings
ans = 0 ;
i = 0 ;
# Traversing the string
while (i < n) :
# Count of consecutive
# 0's & 1's
cnt0 = 0 ; cnt1 = 0 ;
# Counting subarrays of
# type "01"
if (S[i] = = '0' ) :
# Count the consecutive
# 0's
while (i < n and S[i] = = '0' ) :
cnt0 + = 1 ;
i + = 1 ;
# If consecutive 0's
# ends then check for
# consecutive 1's
j = i;
# Counting consecutive 1's
while (j < n and S[j] = = '1' ) :
cnt1 + = 1 ;
j + = 1 ;
# Counting subarrays of
# type "10"
else :
# Count consecutive 1's
while (i < n and S[i] = = '1' ) :
cnt1 + = 1 ;
i + = 1 ;
# If consecutive 1's
# ends then check for
# consecutive 0's
j = i;
# Count consecutive 0's
while (j < n and S[j] = = '0' ) :
cnt0 + = 1 ;
j + = 1 ;
# Update the total count
# of substrings with
# minimum of (cnt0, cnt1)
ans + = min (cnt0, cnt1);
# Return answer
return ans;
# Driver code if __name__ = = "__main__" :
S = "0001110010" ;
n = len (S);
# Function to print the
# count of substrings
print (countSubstring(S, n));
# This code is contributed by Yash_R |
// C# implementation of the // above approach using System;
class GFG{
// Function to find the count
// of substrings with equal no.
// of consecutive 0's and 1's
static int countSubstring( string S, int n)
{
// To store the total count
// of substrings
int ans = 0;
int i = 0;
// Traversing the string
while (i < n) {
// Count of consecutive
// 0's & 1's
int cnt0 = 0, cnt1 = 0;
// Counting subarrays of
// type "01"
if (S[i] == '0' ) {
// Count the consecutive
// 0's
while (i < n && S[i] == '0' ) {
cnt0++;
i++;
}
// If consecutive 0's
// ends then check for
// consecutive 1's
int j = i;
// Counting consecutive 1's
while (j < n && S[j] == '1' ) {
cnt1++;
j++;
}
}
// Counting subarrays of
// type "10"
else {
// Count consecutive 1's
while (i < n && S[i] == '1' ) {
cnt1++;
i++;
}
// If consecutive 1's
// ends then check for
// consecutive 0's
int j = i;
// Count consecutive 0's
while (j < n && S[j] == '0' ) {
cnt0++;
j++;
}
}
// Update the total count
// of substrings with
// minimum of (cnt0, cnt1)
ans += Math.Min(cnt0, cnt1);
}
// Return answer
return ans;
}
// Driver code
static public void Main ()
{
string S = "0001110010" ;
int n = S.Length;
// Function to print the
// count of substrings
Console.WriteLine(countSubstring(S, n));
}
} // This code is contributed by Yash_R |
<script> // Javascript implementation of the // above approach // Function to find the count
// of substrings with equal no.
// of consecutive 0's and 1's
function countSubstring( S , n)
{
// To store the total count
// of substrings
var ans = 0;
var i = 0;
// Traversing the string
while (i < n) {
// Count of consecutive
// 0's & 1's
var cnt0 = 0, cnt1 = 0;
// Counting subarrays of
// type "01"
if (S.charAt(i) == '0' ) {
// Count the consecutive
// 0's
while (i < n && S.charAt(i) == '0 ')
{
cnt0++;
i++;
}
// If consecutive 0' s
// ends then check for
// consecutive 1's
var j = i;
// Counting consecutive 1's
while (j < n && S.charAt(j) == '1' )
{
cnt1++;
j++;
}
}
// Counting subarrays of
// type "10"
else {
// Count consecutive 1's
while (i < n && S.charAt(i) == '1 ')
{
cnt1++;
i++;
}
// If consecutive 1' s
// ends then check for
// consecutive 0's
var j = i;
// Count consecutive 0's
while (j < n && S.charAt(j) == '0' )
{
cnt0++;
j++;
}
}
// Update the total count
// of substrings with
// minimum of (cnt0, cnt1)
ans += Math.min(cnt0, cnt1);
}
// Return answer
return ans;
}
// Driver code
var S = "0001110010" ;
var n = S.length;
// Function to print the
// count of substrings
document.write(countSubstring(S, n));
// This code contributed by umadevi9616 </script> |
7
Time Complexity: O(N), where N = length of string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.