Maximum number of 0s placed consecutively at the start and end in any rotation of a Binary String
Last Updated :
23 Jul, 2025
Given a binary string S of size N, the task is to maximize the sum of the count of consecutive 0s present at the start and end of any of the rotations of the given string S.
Examples:
Input: S = "1001"
Output: 2
Explanation:
All possible rotations of the string are:
"1001": Count of 0s at the start = 0; at the end = 0. Sum= 0 + 0 = 0.
"0011": Count of 0s at the start = 2; at the end = 0. Sum = 2 + 0=2
"0110": Count of 0s at the start = 1; at the end = 1. Sum= 1 + 1 = 2.
"1100": Count of 0s at the start = 0; at the end = 2. Sum = 0 + 2 = 2
Therefore, the maximum sum possible is 2.
Input: S = "01010"
Output: 2
Explanation:
All possible rotations of the string are:
"01010": Count of 0s at the start = 1; at the end = 1. Sum= 1+1=1
"10100": Count of 0s at the start = 0; at the end = 2. Sum= 0+2=2
"01001": Count of 0s at the start = 1; at the end = 0. Sum= 1+0=1
"10010": Count of 0s at the start = 0; at the end = 1. Sum= 0+1=1
"00101": Count of 0s at the start = 2; at the end = 0. Sum= 2+0=2
Therefore, the maximum sum possible is 2.
Naive Approach: The simplest idea is to generate all rotations of the given string and for each rotation, count the number of 0s present at the beginning and end of the string and calculate their sum. Finally, print the maximum sum obtained.
Algorithm:
- Initialize a counter variable c0 to 0 to count the frequency of 0s in the given string.
- Traverse the string and for each character, if it is 0, increment the value of c0 by 1.
- If the value of c0 is equal to the length of the string n, it means that all the characters in the string are 0, so the maximum sum of consecutive 0s present at the start and end of the string will be n. Print n and return.
- Concatenate the string with itself and store it in a new string s.
- Initialize a variable mx to 0 to store the maximum sum of consecutive 0s present at the start and end of a string in any rotation of the given string.
- Generate all rotations of the string s using a loop that iterates from 0 to n-1.
- For each rotation, initialize two variables cs and ce to 0 to store the number of consecutive 0s at the start and end of the string, respectively.
- Traverse the rotated string from the current index i to i+n-1, and for each character, if it is 0, increment the value of cs by 1, else break out of the loop.
- Traverse the rotated string from the current index i+n-1 to i, and for each character, if it is 0, increment the value of ce by 1, else break out of the loop.
- Calculate the sum of cs and ce and store it in a variable val.
- Update the value of mx to the maximum of its current value and val.
- After the loop ends, the value of mx will contain the maximum sum of consecutive 0s present at the start and end of a string in any rotation of the given string. Print mx.
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of a string present in any
// of the rotations of the given string
void findMaximumZeros(string str, int n)
{
// Check if all the characters
// in the string are 0
int c0 = 0;
// Iterate over characters
// of the string
for (int i = 0; i < n; ++i) {
if (str[i] == '0')
c0++;
}
// If the frequency of '1' is 0
if (c0 == n) {
// Print n as the result
cout << n;
return;
}
// Concatenate the string
// with itself
string s = str + str;
// Stores the required result
int mx = 0;
// Generate all rotations of the string
for (int i = 0; i < n; ++i) {
// Store the number of consecutive 0s
// at the start and end of the string
int cs = 0;
int ce = 0;
// Count 0s present at the start
for (int j = i; j < i + n; ++j) {
if (s[j] == '0')
cs++;
else
break;
}
// Count 0s present at the end
for (int j = i + n - 1; j >= i; --j) {
if (s[j] == '0')
ce++;
else
break;
}
// Calculate the sum
int val = cs + ce;
// Update the overall
// maximum sum
mx = max(val, mx);
}
// Print the result
cout << mx;
}
// Driver Code
int main()
{
// Given string
string s = "1001";
// Store the size of the string
int n = s.size();
findMaximumZeros(s, n);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of a string present in any
// of the rotations of the given string
static void findMaximumZeros(String str, int n)
{
// Check if all the characters
// in the string are 0
int c0 = 0;
// Iterate over characters
// of the string
for(int i = 0; i < n; ++i)
{
if (str.charAt(i) == '0')
c0++;
}
// If the frequency of '1' is 0
if (c0 == n)
{
// Print n as the result
System.out.print(n);
return;
}
// Concatenate the string
// with itself
String s = str + str;
// Stores the required result
int mx = 0;
// Generate all rotations of the string
for(int i = 0; i < n; ++i)
{
// Store the number of consecutive 0s
// at the start and end of the string
int cs = 0;
int ce = 0;
// Count 0s present at the start
for(int j = i; j < i + n; ++j)
{
if (s.charAt(j) == '0')
cs++;
else
break;
}
// Count 0s present at the end
for(int j = i + n - 1; j >= i; --j)
{
if (s.charAt(j) == '0')
ce++;
else
break;
}
// Calculate the sum
int val = cs + ce;
// Update the overall
// maximum sum
mx = Math.max(val, mx);
}
// Print the result
System.out.print(mx);
}
// Driver Code
public static void main(String[] args)
{
// Given string
String s = "1001";
// Store the size of the string
int n = s.length();
findMaximumZeros(s, n);
}
}
// This code is contributed by susmitakundugoaldanga
# Python3 program for the above approach
# Function to find the maximum sum of
# consecutive 0s present at the start
# and end of a string present in any
# of the rotations of the given string
def findMaximumZeros(st, n):
# Check if all the characters
# in the string are 0
c0 = 0
# Iterate over characters
# of the string
for i in range(n):
if (st[i] == '0'):
c0 += 1
# If the frequency of '1' is 0
if (c0 == n):
# Print n as the result
print(n)
return
# Concatenate the string
# with itself
s = st + st
# Stores the required result
mx = 0
# Generate all rotations of the string
for i in range(n):
# Store the number of consecutive 0s
# at the start and end of the string
cs = 0
ce = 0
# Count 0s present at the start
for j in range(i, i + n):
if (s[j] == '0'):
cs += 1
else:
break
# Count 0s present at the end
for j in range(i + n - 1, i - 1, -1):
if (s[j] == '0'):
ce += 1
else:
break
# Calculate the sum
val = cs + ce
# Update the overall
# maximum sum
mx = max(val, mx)
# Print the result
print(mx)
# Driver Code
if __name__ == "__main__":
# Given string
s = "1001"
# Store the size of the string
n = len(s)
findMaximumZeros(s, n)
# This code is contributed by ukasp.
// C# program for the above approach
using System;
class GFG{
// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of a string present in any
// of the rotations of the given string
static void findMaximumZeros(string str, int n)
{
// Check if all the characters
// in the string are 0
int c0 = 0;
// Iterate over characters
// of the string
for(int i = 0; i < n; ++i)
{
if (str[i] == '0')
c0++;
}
// If the frequency of '1' is 0
if (c0 == n)
{
// Print n as the result
Console.Write(n);
return;
}
// Concatenate the string
// with itself
string s = str + str;
// Stores the required result
int mx = 0;
// Generate all rotations of the string
for(int i = 0; i < n; ++i)
{
// Store the number of consecutive 0s
// at the start and end of the string
int cs = 0;
int ce = 0;
// Count 0s present at the start
for(int j = i; j < i + n; ++j)
{
if (s[j] == '0')
cs++;
else
break;
}
// Count 0s present at the end
for(int j = i + n - 1; j >= i; --j)
{
if (s[j] == '0')
ce++;
else
break;
}
// Calculate the sum
int val = cs + ce;
// Update the overall
// maximum sum
mx = Math.Max(val, mx);
}
// Print the result
Console.Write(mx);
}
// Driver Code
public static void Main(string[] args)
{
// Given string
string s = "1001";
// Store the size of the string
int n = s.Length;
findMaximumZeros(s, n);
}
}
// This code is contributed by AnkThon
<script>
// Javascript program for the above approach
// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of a string present in any
// of the rotations of the given string
function findMaximumZeros(str, n)
{
// Check if all the characters
// in the string are 0
var c0 = 0;
var i;
// Iterate over characters
// of the string
for (i = 0; i < n; ++i) {
if (str[i] == '0')
c0++;
}
// If the frequency of '1' is 0
if (c0 == n) {
// Print n as the result
document.write(n);
return;
}
// Concatenate the string
// with itself
var s = str + str;
// Stores the required result
var mx = 0;
var j;
// Generate all rotations of the string
for (i = 0; i < n; ++i) {
// Store the number of consecutive 0s
// at the start and end of the string
var cs = 0;
var ce = 0;
// Count 0s present at the start
for (j = i; j < i + n; ++j) {
if (s[j] == '0')
cs++;
else
break;
}
// Count 0s present at the end
for (j = i + n - 1; j >= i; --j) {
if (s[j] == '0')
ce++;
else
break;
}
// Calculate the sum
var val = cs + ce;
// Update the overall
// maximum sum
mx = Math.max(val, mx);
}
// Print the result
document.write(mx);
}
// Driver Code
// Given string
var s = "1001";
// Store the size of the string
var n = s.length;
findMaximumZeros(s, n);
</script>
Time Complexity: O(N2)
Auxiliary Space: O(N)
Efficient Approach: The idea is to find the maximum number of consecutive 0s in the given string. Also, find the sum of consecutive 0s at the start and the end of the string, and then print the maximum out of them.
Follow the steps below to solve the problem:
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of any rotation of the string str
void findMaximumZeros(string str, int n)
{
// Stores the count of 0s
int c0 = 0;
for (int i = 0; i < n; ++i) {
if (str[i] == '0')
c0++;
}
// If the frequency of '1' is 0
if (c0 == n) {
// Print n as the result
cout << n;
return;
}
// Stores the required sum
int mx = 0;
// Find the maximum consecutive
// length of 0s present in the string
int cnt = 0;
for (int i = 0; i < n; i++) {
if (str[i] == '0')
cnt++;
else {
mx = max(mx, cnt);
cnt = 0;
}
}
// Update the overall maximum sum
mx = max(mx, cnt);
// Find the number of 0s present at
// the start and end of the string
int start = 0, end = n - 1;
cnt = 0;
// Update the count of 0s at the start
while (str[start] != '1' && start < n) {
cnt++;
start++;
}
// Update the count of 0s at the end
while (str[end] != '1' && end >= 0) {
cnt++;
end--;
}
// Update the maximum sum
mx = max(mx, cnt);
// Print the result
cout << mx;
}
// Driver Code
int main()
{
// Given string
string s = "1001";
// Store the size of the string
int n = s.size();
findMaximumZeros(s, n);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of any rotation of the string str
static void findMaximumZeros(String str, int n)
{
// Stores the count of 0s
int c0 = 0;
for(int i = 0; i < n; ++i)
{
if (str.charAt(i) == '0')
c0++;
}
// If the frequency of '1' is 0
if (c0 == n)
{
// Print n as the result
System.out.print(n);
return;
}
// Stores the required sum
int mx = 0;
// Find the maximum consecutive
// length of 0s present in the string
int cnt = 0;
for(int i = 0; i < n; i++)
{
if (str.charAt(i) == '0')
cnt++;
else
{
mx = Math.max(mx, cnt);
cnt = 0;
}
}
// Update the overall maximum sum
mx = Math.max(mx, cnt);
// Find the number of 0s present at
// the start and end of the string
int start = 0, end = n - 1;
cnt = 0;
// Update the count of 0s at the start
while (str.charAt(start) != '1' && start < n)
{
cnt++;
start++;
}
// Update the count of 0s at the end
while (str.charAt(end) != '1' && end >= 0)
{
cnt++;
end--;
}
// Update the maximum sum
mx = Math.max(mx, cnt);
// Print the result
System.out.println(mx);
}
// Driver Code
public static void main (String[] args)
{
// Given string
String s = "1001";
// Store the size of the string
int n = s.length();
findMaximumZeros(s, n);
}
}
// This code is contributed by sanjoy_62
# Python3 program for the above approach
# Function to find the maximum sum of
# consecutive 0s present at the start
# and end of any rotation of the string str
def findMaximumZeros(string, n):
# Stores the count of 0s
c0 = 0
for i in range(n):
if (string[i] == '0'):
c0 += 1
# If the frequency of '1' is 0
if (c0 == n):
# Print n as the result
print(n, end = "")
return
# Stores the required sum
mx = 0
# Find the maximum consecutive
# length of 0s present in the string
cnt = 0
for i in range(n):
if (string[i] == '0'):
cnt += 1
else:
mx = max(mx, cnt)
cnt = 0
# Update the overall maximum sum
mx = max(mx, cnt)
# Find the number of 0s present at
# the start and end of the string
start = 0
end = n - 1
cnt = 0
# Update the count of 0s at the start
while (string[start] != '1' and start < n):
cnt += 1
start += 1
# Update the count of 0s at the end
while (string[end] != '1' and end >= 0):
cnt += 1
end -= 1
# Update the maximum sum
mx = max(mx, cnt)
# Print the result
print(mx, end = "")
# Driver Code
if __name__ == "__main__":
# Given string
s = "1001"
# Store the size of the string
n = len(s)
findMaximumZeros(s, n)
# This code is contributed by AnkThon
// C# program for the above approach
using System;
class GFG{
// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of any rotation of the string str
static void findMaximumZeros(string str, int n)
{
// Stores the count of 0s
int c0 = 0;
for(int i = 0; i < n; ++i)
{
if (str[i] == '0')
c0++;
}
// If the frequency of '1' is 0
if (c0 == n)
{
// Print n as the result
Console.Write(n);
return;
}
// Stores the required sum
int mx = 0;
// Find the maximum consecutive
// length of 0s present in the string
int cnt = 0;
for(int i = 0; i < n; i++)
{
if (str[i] == '0')
cnt++;
else
{
mx = Math.Max(mx, cnt);
cnt = 0;
}
}
// Update the overall maximum sum
mx = Math.Max(mx, cnt);
// Find the number of 0s present at
// the start and end of the string
int start = 0, end = n - 1;
cnt = 0;
// Update the count of 0s at the start
while (str[start] != '1' && start < n)
{
cnt++;
start++;
}
// Update the count of 0s at the end
while (str[end] != '1' && end >= 0)
{
cnt++;
end--;
}
// Update the maximum sum
mx = Math.Max(mx, cnt);
// Print the result
Console.Write(mx);
}
// Driver Code
static public void Main ()
{
// Given string
string s = "1001";
// Store the size of the string
int n = s.Length;
findMaximumZeros(s, n);
}
}
// This code is contributed by avijitmondal1998
<script>
//Javascript program for
//the above approach
// Function to find the maximum sum of
// consecutive 0s present at the start
// and end of any rotation of the string str
function findMaximumZeros(str, n)
{
// Stores the count of 0s
var c0 = 0;
for (var i = 0; i < n; ++i) {
if (str[i] == '0')
c0++;
}
// If the frequency of '1' is 0
if (c0 == n) {
// Print n as the result
document.write( n);
return;
}
// Stores the required sum
var mx = 0;
// Find the maximum consecutive
// length of 0s present in the string
var cnt = 0;
for (var i = 0; i < n; i++) {
if (str[i] == '0')
cnt++;
else {
mx = Math.max(mx, cnt);
cnt = 0;
}
}
// Update the overall maximum sum
mx = Math.max(mx, cnt);
// Find the number of 0s present at
// the start and end of the string
var start = 0, end = n - 1;
cnt = 0;
// Update the count of 0s at the start
while (str[start] != '1' && start < n) {
cnt++;
start++;
}
// Update the count of 0s at the end
while (str[end] != '1' && end >= 0) {
cnt++;
end--;
}
// Update the maximum sum
mx = Math.max(mx, cnt);
// Print the result
document.write( mx);
}
var s = "1001";
// Store the size of the string
var n = s.length;
findMaximumZeros(s, n);
// This code is contributed by SoumikMondal
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
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