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Minimum shifts of substrings of 1s required to group all 1s together in a given Binary string

  • Last Updated : 21 Apr, 2021

Given a binary string S of length N, the task is to print the minimum number of indices, substrings consisting only of 1s are required to be shifted such that all 1s present in the string are grouped together.

Examples:

Input: S = “00110111011”
Output: 2
Explanation: 
Operation 1: Shift substring {S[2], S[3]} (0-based indexing) to right and place between S[4] and S[5]. Now, S is modified to “00011111011”. 
Operation 2: Shift substring {S[10], S[11]} to left and place between S[7] and S[8]. Now, S is modified to “00011111110”. 
Therefore, 2 substrings are required to be shifted.

Input: S = “1001001”
Output: 4
Explanation: 
Operation 1: Shift ‘1’ at S[0] to right by two indices and place between S[2] and S[3]. Now, S is modified to “0011001”. 
Operation 2: Shift ‘1’ at S[6] to left by two indices and place between S[3] and S[4]. Now, S is modified to “0011100”. 
Therefore, 4 substrings are required to be shifted.

Approach: The problem can be solved by observing that the minimum number of operations required is equal to the number of 0s present between any pair of consecutive 1s. Follow the steps below to solve the problem:



  1. Iterate over the characters of the string and search for the first and last occurrences of the character ‘1’ and store it in variable, say firstOne and lastOne respectively.
  2. Traverse the range [firstOne, lastOne] and count the number of ‘0’s present in the substring {S[firstOne], .. , S[lastOne]} and print it as the required output.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count indices substrings
// of 1s need to be shifted such that
// all 1s in the string are grouped together
void countShifts(string str)
{
    // Stores first occurrence of '1'
    int firstOne = -1;
 
    // Stores last occurrence of '1'
    int lastOne = -1;
 
    // Count of 0s between firstOne and lastOne
    int count = 0;
 
    // Traverse the string to find the
    // first and last occurrences of '1'
    for (int i = 0; i < str.length(); i++) {
        if (str[i] == '1') {
 
            if (firstOne == -1)
                firstOne = i;
 
            lastOne = i;
        }
    }
 
    if ((firstOne == -1) || (firstOne == lastOne)) {
        cout << 0;
        return;
    }
 
    // Count number of 0s present between
    // firstOne and lastOne
    for (int i = firstOne; i <= lastOne; i++) {
 
        if (str[i] == '0') {
 
            count++;
        }
    }
 
    // Print minimum operations
    cout << count << endl;
}
 
// Driver Code
int main()
{
 
    // Given string
    string str = "00110111011";
 
    countShifts(str);
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
      
// Function to count indices substrings
// of 1s need to be shifted such that
// all 1s in the string are grouped together
static void countShifts(String str)
{
     
    // Stores first occurrence of '1'
    int firstOne = -1;
    
    // Stores last occurrence of '1'
    int lastOne = -1;
    
    // Count of 0s between firstOne and lastOne
    int count = 0;
     
    // Traverse the string to find the
    // first and last occurrences of '1'
    for(int i = 0; i < str.length(); i++)
    {
        if (str.charAt(i) == '1')
        {
            if (firstOne == -1)
                firstOne = i;
    
            lastOne = i;
        }
    }
    
    if ((firstOne == -1) || (firstOne == lastOne))
    {
        System.out.print(0);
        return;
    }
    
    // Count number of 0s present between
    // firstOne and lastOne
    for(int i = firstOne; i <= lastOne; i++)
    {
        if (str.charAt(i) == '0')
        {      
            count++;
        }
    }
    
    // Print minimum operations
    System.out.println(count);
}
 
// Driver code
public static void main (String[] args)
{
     
    // Given string
    String str = "00110111011";
    
    countShifts(str); 
}
}
 
// This code is contributed by code_hunt

Python3




# Python3 program for the above approach
  
# Function to count indices substrings
# of 1s need to be shifted such that
# all 1s in the string are grouped
# together
def countShifts(str):
     
    # Stores first occurrence of '1'
    firstOne = -1
  
    # Stores last occurrence of '1'
    lastOne = -1
  
    # Count of 0s between firstOne
    # and lastOne
    count = 0
  
    # Traverse the string to find the
    # first and last occurrences of '1'
    for i in range(len(str)):
        if (str[i] == '1'):
  
            if (firstOne == -1):
                firstOne = i
  
            lastOne = i
             
    if ((firstOne == -1) or
        (firstOne == lastOne)):
        print(0)
        return
     
    # Count number of 0s present between
    # firstOne and lastOne
    for i in range(firstOne, lastOne + 1, 1):
        if (str[i] == '0'):
            count += 1
             
    # Print minimum operations
    print(count)
 
# Driver Code
 
# Given string
str = "00110111011"
 
countShifts(str)
 
# This code is contributed by sanjoy_62

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
     
    // Function to count indices substrings
    // of 1s need to be shifted such that
    // all 1s in the string are grouped together
    static void countShifts(string str)
    {
        // Stores first occurrence of '1'
        int firstOne = -1;
       
        // Stores last occurrence of '1'
        int lastOne = -1;
       
        // Count of 0s between firstOne and lastOne
        int count = 0;
       
        // Traverse the string to find the
        // first and last occurrences of '1'
        for (int i = 0; i < str.Length; i++)
        {
            if (str[i] == '1')
            {
       
                if (firstOne == -1)
                    firstOne = i;
       
                lastOne = i;
            }
        }
       
        if ((firstOne == -1) || (firstOne == lastOne))
        {
            Console.Write(0);
            return;
        }
       
        // Count number of 0s present between
        // firstOne and lastOne
        for (int i = firstOne; i <= lastOne; i++)
        {
       
            if (str[i] == '0')
            {      
                count++;
            }
        }
       
        // Print minimum operations
        Console.WriteLine(count);
    }
 
  // Driver code
  static void Main()
  {
     
        // Given string
        string str = "00110111011";
       
        countShifts(str); 
  }
}
 
// This code is contributed by divyeshrabadiya07

Javascript




<script>
// javascript program to implement
// the above approach
 
    // Function to count indices substrings
    // of 1s need to be shifted such that
    // all 1s in the string are grouped together
    function countShifts(str)
    {
     
        // Stores first occurrence of '1'
        let firstOne = -1;
        
        // Stores last occurrence of '1'
        let lastOne = -1;
        
        // Count of 0s between firstOne and lastOne
        let count = 0;
        
        // Traverse the string to find the
        // first and last occurrences of '1'
        for (let i = 0; i < str.length; i++)
        {
            if (str[i] == '1')
            {
        
                if (firstOne == -1)
                    firstOne = i;
        
                lastOne = i;
            }
        }
        
        if ((firstOne == -1) || (firstOne == lastOne))
        {
            Console.Write(0);
            return;
        }
        
        // Count number of 0s present between
        // firstOne and lastOne
        for (let i = firstOne; i <= lastOne; i++)
        {
        
            if (str[i] == '0')
            {     
                count++;
            }
        }
        
        // Prlet minimum operations
        document.write(count);
    }
     
// Driver code
 
      // Given string
        let str = "00110111011";
        
        countShifts(str);
   
  // This code is contributed by splevel62.
</script>
Output: 
2

 

Time Complexity: O(N)
Auxiliary Space: O(1)

 

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