Skip to content
Related Articles

Related Articles

Check if any circular rotation of String has at most X 1s between two adjacent 0s

Improve Article
Save Article
  • Last Updated : 04 Nov, 2022
Improve Article
Save Article

Given a binary string S of length N and an integer X, the task is to check if there exists a right-wise circular rotation of the string such that every 2 adjacent 0′s are separated by at most X 1′s.

Note: The first and the last 0s in the string are not considered to be adjacent 

Examples:

Input: S = “010110”, X = 1        
Output: Yes 
Explanation: The 6 circular rotations of the string 
S = {“010110”, “001011”, “100101”, “110010”, “011001”, “101100”} 
out of which second , third and fourth strings satisfy the criteria.
Hence there exist a binary string which satisfy above condition.

Input:  S = “0101”, X = 0
Output: No

An efficient approach: (No Addition of Auxiliary Space)

The input string could be thought of as a circle once we connect the last of the string to the first of it. The idea is to find out the first position of zero where the consecutive 1s between this element and the previous zero element is more than X. Next, if we need to rotate the string from that position to the left such that rest of elements in the string follows the constraints i.e. no of 1s between each pair of adjacent 0 is at most X. If we could not find any such arrangement after performing multiple rotation, then the Answer is No, else Yes. Only checking the rotation for the first incorrect position is sufficient because, there exists no right-wise circular rotation of the string such that every 2 adjacent 0′s are separated by at most X 1′s,  if there is multiple positions where the no of consecutive 1s between two consecutive 0s does not follow the strategy. This would been possible if we can rotate both clock wise (right rotation) and anti-clock wise (left rotation).

Follow the below steps to implement the above idea:

  1. Find out the first position of zero where it violates the rule (no of consecutive 1 between each pair of 0 is at most X), also track the first 0th position (it will help if any such position exists where this given rule is not satisfied)
  2. If any such position of zero is found, count the no of 1s from the next position and move circularly to the first of the string until the first 0th position). 
  3. If no of consecutive 1 found is more than X, Answer is NO, else YES

Below is the implementation of the above approach.

Java




import java.io.*;
 
public class GFG {
    public static void main(String[] args)
    {
        String str = "010110";
        int x = 1;
        boolean exist = checkIfSequeneceExist(str, x);
        if (exist)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
 
    private static boolean checkIfSequeneceExist(String str,
                                                 int x)
    {
        int i = 0, previousZeroPosition = -1,
            noOfConsequtive1 = 0, incorrectOthPosition = -1,
            firstOthPosition = -1;
        boolean exist = true;
 
        while (i < str.length()) {
            if (str.charAt(i)
                == '0') { // if the element is 0, check if
                          // any previous element=0 found
                if (previousZeroPosition != -1) {
                    if (noOfConsequtive1 <= x)
                        noOfConsequtive1
                            = 0; // valid condition
                    else {
                        incorrectOthPosition
                            = i; // violates the constraint
                        break;
                    }
                }
                else {
                    firstOthPosition = i;
                }
                previousZeroPosition
                    = i; // update previousZeroPosition to
                         // the latest 0th position
            }
            else if (firstOthPosition
                     != -1) // no need to track no of 1s if
                            // there is no element=0 found
                            // so far
                noOfConsequtive1++;
            i++;
        }
 
        if (incorrectOthPosition != -1) {
            noOfConsequtive1 = 0;
            i = (incorrectOthPosition + 1) % str.length();
            while (i != firstOthPosition) {
                i = (i + 1) % str.length();
                noOfConsequtive1++;
            }
            if (noOfConsequtive1 > x)
                exist = false;
        }
        return exist;
    }
}


C#




// C# code for the above approach
using System;
 
public class GFG {
 
    static bool checkIfSequeneceExist(string str, int x)
    {
        int i = 0, previousZeroPosition = -1,
            noOfConsequtive1 = 0, incorrectOthPosition = -1,
            firstOthPosition = -1;
        bool exist = true;
 
        while (i < str.Length) {
            if (str[i]
                == '0') { // if the element is 0, check if
                          // any previous element=0 found
                if (previousZeroPosition != -1) {
                    if (noOfConsequtive1 <= x)
                        noOfConsequtive1
                            = 0; // valid condition
                    else {
                        incorrectOthPosition
                            = i; // violates the constraint
                        break;
                    }
                }
                else {
                    firstOthPosition = i;
                }
                previousZeroPosition
                    = i; // update previousZeroPosition to
                         // the latest 0th position
            }
            else if (firstOthPosition
                     != -1) // no need to track no of 1s if
                            // there is no element=0 found
                            // so far
                noOfConsequtive1++;
            i++;
        }
 
        if (incorrectOthPosition != -1) {
            noOfConsequtive1 = 0;
            i = (incorrectOthPosition + 1) % str.Length;
            while (i != firstOthPosition) {
                i = (i + 1) % str.Length;
                noOfConsequtive1++;
            }
            if (noOfConsequtive1 > x)
                exist = false;
        }
        return exist;
    }
 
    static public void Main()
    {
 
        // Code
        string str = "010110";
        int x = 1;
        bool exist = checkIfSequeneceExist(str, x);
        if (exist)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by lokeshmvs21.


Output

YES

Approach: The problem can be solved based on the following observation: 

Assuming the last character to be adjacent to first, we can find the number of ones between each pair of adjacent ones in a list. Now, the rotation of binary string is equivalent to deleting at most one element of this list. So if rest of the elements are up to X, then the answer is YES.

Follow the below steps to implement the above idea:

  • Find the positions of all 0’s in the array.
  • Find the number of 1s between any two adjacent 0s.
  • If there is at most 1 such segment of contiguous 1s having length X or more then that segment can be partitioned to be in the start or last. So print “Yes”;
  • Otherwise, it prints “No”.

Below is the implementation of the above approach.

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check whether
// such a rotation exists
void check(string s, int n, int x)
{
    vector<int> pos;
    for (int i = 0; i < n; i++) {
        if (s[i] == '0')
            pos.push_back(i);
    }
    int cnt = 0;
    for (int i = 0; i + 1 < pos.size(); i++) {
        if ((pos[i + 1] - pos[i] - 1) > x)
            cnt++;
    }
    if (!pos.empty() and n - pos.back() - 1 + pos[0] > x)
        cnt++;
    if (cnt <= 1)
        cout << "Yes\n";
    else
        cout << "No\n";
}
 
// Driver Code
int main()
{
    string S = "010110";
    int N = S.length();
    int X = 1;
 
    // Function call
    check(S, N, X);
 
    return 0;
}


Java




// Java code to implement the approach
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function to check whether
    // such a rotation exists
    public static void check(String s, int n, int x)
    {
        int i = 0, previousZeroPosition=-1, noOfConsequtive1 =0, incorrectOthPosition = -1, firstOthPosition = -1;
        boolean exist=true;
         
        while(i< n) {
            if(str.charAt(i) == '0') {
                if(previousZeroPosition != -1) {
                    if(noOfConsequtive1 <= x)
                        noOfConsequtive1 = 0;
                    else{
                        incorrectOthPosition = i;
                        break;
                    }
                }else {
                  firstOthPosition = i;
                }
                previousZeroPosition = i;
 
            }else if(firstOthPosition != -1) {
                noOfConsequtive1++;
            }
            i++;
        }
         
        if(incorrectOthPosition != -1) {
            noOfConsequtive1 = 0; i = (incorrectOthPosition + 1)%n;
            while(i != firstOthPosition) {
                i = (i + 1)%n;
                noOfConsequtive1++;
            }
            if(noOfConsequtive1 > x)
                exist = false;
                 
        }
        return exist;
    }
    // Driver Code
    public static void main(String[] args)
    {
        String S = "010110";
        int N = S.length();
        int X = 1;
 
        // Function call
        check(S, N, X);
    }
}


Python3




# Python code to implement the approach
 
# Function to check whether
# such a rotation exists
def check(s, n, x):
    pos = []
    for i in range(n):
        if (s[i] == '0'):
            pos.append(i)
 
    cnt = 0
    for i in range(len(pos)-1):
        if ((pos[i + 1] - pos[i] - 1) > x):
            cnt += 1
    if (len(pos) != 0 and n - pos[len(pos)-1] - 1 + pos[0] > x):
        cnt += 1
    if (cnt <= 1):
        print("Yes")
    else:
        print("No")
 
# Driver Code
if __name__ == "__main__":
    S = "010110"
    N = len(S)
    X = 1
     
    # Function call
    check(S, N, X)
 
# This code is contributed by Rohit Pradhan


C#




using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
 
public class GFG{
 
      // Function to check whether
    // such a rotation exists
    public static void check(String s, int n, int x)
    {
        ArrayList pos = new ArrayList();
        for (int i = 0; i < n; i++) {
            if (s[i] == '0')
                pos.Add(i);
        }
        int cnt = 0;
        for (int i = 0; i + 1 < pos.size(); i++) {
            if (((int)pos[i + 1] - (int)pos[i] - 1) > x)
                cnt++;
        }
        if (!pos.Any()
            && (n - pos[pos.Count - 1] - 1 + pos[0]
                > x))
            cnt++;
        if (cnt <= 1)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
   
    static public void Main (){
 
        // Code
          String S = "010110";
        int N = S.Length;
        int X = 1;
  
        // Function call
        check(S, N, X);
    }
}


Javascript




<script>
 
    function check(s, n, x){
        let pos = [];
        for(let i = 0; i < n; i++){
            if(s.charAt(i) == '0'){
                pos.push(i);
            }
        }
         
        var cnt = 0;
        for(let i = 0; i + 1 < pos.length; i++){
            if((pos[i + 1] - pos[i] - 1) > x){
                cnt += 1;
            }
        }
        if((pos.length != 0) && (n - pos[(pos.length) - 1] + pos[0] > x))
        {
            cnt += 1;
        }
        if(cnt <= 1){
            document.write("Yes");
        }
        else{
            document.write("No");
        }
    }
 
    let S = "010110";
    let N = S.length;
    let X = 1;
     
    // Function call
    check(S, N, X);
     
    // This code is contributed by lokeshmvs21.
</script>


Output

Yes

Time Complexity: O(N) 
Auxiliary Space: O(N)


My Personal Notes arrow_drop_up
Related Articles

Start Your Coding Journey Now!