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Javascript Program to Find Maximum number of 0s placed consecutively at the start and end in any rotation of a Binary String

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  • Last Updated : 27 Jan, 2022

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:

Javascript




<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>
Output: 
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:

Below is the implementation of the above approach: 

Javascript




<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>
Output: 
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!


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