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.
Below is the implementation of the above approach:
<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> |
2
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:
- Check if the frequency of ‘1’ in the string, S is equal to 0 or not. If found to be true, print the value of N as the result.
- Otherwise, perform the following steps:
- Store the maximum number of consecutive 0s in the given string in a variable, say X.
- Initialize two variables, start as 0 and end as N-1.
- Increment the value of cnt and start by 1 while S[start] is not equal to ‘1‘.
- Increment the value of cnt and decrement end by 1 while S[end] is not equal to ‘1‘.
- Print the maximum of X and cnt as a result.
Below is the implementation of the above approach:
<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> |
2
Time Complexity: O(N)
Auxiliary Space: O(1)
Please refer complete article on Maximum number of 0s placed consecutively at the start and end in any rotation of a Binary String for more details!