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Minimum flips to make all 1s in right and 0s in left

  • Difficulty Level : Medium
  • Last Updated : 22 Apr, 2021

Given binary string str, the task is to find the minimum number of characters required to be flipped to make all 1s in right and 0s in left.

Examples:

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Input: str = “100101”
Output: 2
Explanation:
Flipping str[0] and str[4] modifies str = “000111”. Therefore, the required output is 2.

Input: S = “00101001”
Output: 2
Explanation:
Flipping str[2] and str[4] modifies str = “00000001”. Therefore, the required output is 2.



Approach: The idea is to count the number of 0s on the right side of each index of the string and count the number of 1s on the left side of each index of the string. Follow the steps below to solve the problem:

  • Initialize a variable, say cntzero, to store the total count of 0s in the given string.
  • Traverse the string and count the total number of 0s in the given string.
  • If cntzero in the given string is 0, or equal to the length of the string, the result will be 0.
  • Traverse the string and for each index, find the sum of the count of 1s on the left side of the index and the count of 0s on the right side of the index.
  • Find the minimum value from all sums obtained.

Below is the implementation of the above approach:

C++




// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum count
// of flips required to make all 1s on
// the right and all 0s on the left of
// the given string
int minimumCntOfFlipsRequired(string str)
{
    // Stores length of str
    int n = str.length();
 
    // Store count of 0s in the string
    int zeros = 0;
 
    // Traverse the string
    for (int i = 0; i < n; i++) {
 
        // If current character is 0
        if (str[i] == '0') {
 
            // Update zeros
            zeros++;
        }
    }
 
    // If count of 0s in the string
    // is 0 or n
    if (zeros == 0 || zeros == n) {
        return 0;
    }
 
    // Store minimum count of flips
    // required to make all 0s on
    // the left and all 1s on the right
    int minFlips = INT_MAX;
 
    // Stores count of 1s on the left
    // of each index
    int currOnes = 0;
 
    // Stores count of flips required to make
    // string monotonically increasing
    int flips;
 
    // Traverse the string
    for (int i = 0; i < n; i++) {
 
        // If current character
        // is 1
        if (str[i] == '1') {
 
            // Update currOnes
            currOnes++;
        }
 
        // Update flips
        flips = currOnes + (zeros - (i + 1 - currOnes));
 
        // Update the minimum
        // count of flips
        minFlips = min(minFlips, flips);
    }
 
    return minFlips;
}
 
// Driver Code
int main()
{
    string str = "100101";
    cout << minimumCntOfFlipsRequired(str);
    return 0;
}


Java




// Java program to implement
// the above approach
 
import java.io.*;
 
class GFG {
 
    // Function to find the minimum count of flips
    // required to make all 1s on the right and
    // all 0s on the left of the given string
    public static int minimumCntOfFlipsRequired(
        String str)
    {
        // Stores length of str
        int n = str.length();
 
        // Store count of 0s in the string
        int zeros = 0;
 
        // Traverse the string
        for (int i = 0; i < n; i++) {
 
            // If current character is 0
            if (str.charAt(i) == '0') {
 
                // Update zeros
                zeros++;
            }
        }
 
        // If count of 0s in the string
        // is 0 or n
        if (zeros == 0 || zeros == n) {
            return 0;
        }
 
        // Store minimum count of flips
        // required to make all 0s on
        // the left and 1s on the right
        int minFlips = Integer.MAX_VALUE;
 
        // Stores count of 1s on the left
        // side of each index
        int currOnes = 0;
 
        // Stores count of flips required to make
        // all 0s on the left and 1s on the right
        int flips;
 
        // Traverse the string
        for (int i = 0; i < n; i++) {
 
            // If current character is 1
            if (str.charAt(i) == '1') {
 
                // Update currOnes
                currOnes++;
            }
 
            // Update flips
            flips = currOnes + (zeros - (i + 1 - currOnes));
 
            // Update the minimum
            // count of flips
            minFlips = Math.min(minFlips, flips);
        }
 
        return minFlips;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String s1 = "100101";
        System.out.println(minimumCntOfFlipsRequired(s1));
    }
}


Python3




# Python3 program to implement
# the above approach
 
# Function to find the minimum count
# of flips required to make all 1s on
# the right and all 0s on the left of
# the given string
def minimumCntOfFlipsRequired(str):
     
    # Stores length of str
    n = len(str);
 
 
    # Store count of 0s in the string
    zeros = 0;
 
 
    # Traverse the string
    for i in range(n):
         
         
        # If current character
        # is 0
        if (str[i] == '0'):
             
             
            # Update zeros
            zeros += 1;
 
 
    # If count of 0s in the string
    # is 0 or n
    if (zeros == 0 or zeros == n):
        return 0;
 
 
    # Store minimum count of flips
    # required to make all 0s on the
    # left and all 1s on the right
    minFlips = 10000001;
 
 
    # Stores count of 1s on the left
    # of each index
    currOnes = 0;
     
     
    # Stores count of flips required to make
    # all 0s on the left and all 1s on the right
    flips = 0;
 
 
    # Traverse the string
    for i in range(n):
 
        # If current character is 1
        if (str[i] == '1'):
 
            # Update currOnes
            currOnes += 1;
 
        # Update flips
        flips = currOnes + (zeros - (i + 1 - currOnes));
 
 
        # Update the minimum
        # count of flips
        minFlips = min(minFlips, flips);
 
    return minFlips;
 
 
 
# Driver Code
if __name__ == '__main__':
    str = "100101";
    print(minimumCntOfFlipsRequired(str));


C#




// C# program to implement
// the above approach 
using System;
  
class GFG{
  
// Function to find the minimum count of flips
// required to make all 1s on the right and
// all 0s on the left of the given string
public static int minimumCntOfFlipsRequired(string str)
{
     
    // Stores length of str
    int n = str.Length;
     
    // Store count of 0s in the string
    int zeros = 0;
 
    // Traverse the string
    for(int i = 0; i < n; i++)
    {
         
        // If current character is 0
        if (str[i] == '0')
        {
             
            // Update zeros
            zeros++;
        }
    }
 
    // If count of 0s in the string
    // is 0 or n
    if (zeros == 0 || zeros == n)
    {
        return 0;
    }
 
    // Store minimum count of flips
    // required to make all 0s on
    // the left and 1s on the right
    int minFlips = Int32.MaxValue;
 
    // Stores count of 1s on the left
    // side of each index
    int currOnes = 0;
 
    // Stores count of flips required
    // to make all 0s on the left and
    // 1s on the right
    int flips;
 
    // Traverse the string
    for(int i = 0; i < n; i++)
    {
         
        // If current character is 1
        if (str[i] == '1')
        {
             
            // Update currOnes
            currOnes++;
        }
 
        // Update flips
        flips = currOnes +
             (zeros - (i + 1 - currOnes));
 
        // Update the minimum
        // count of flips
        minFlips = Math.Min(minFlips, flips);
    }
    return minFlips;
}
 
// Driver code
public static void Main()
{
    string s1 = "100101";
     
    Console.WriteLine(minimumCntOfFlipsRequired(s1));
}
}
 
// This code is contributed by sanjoy_62


Javascript




<script>
// Javascript program to implement
// the above approach
 
    // Function to find the minimum count of flips
    // required to make all 1s on the right and
    // all 0s on the left of the given string
   function minimumCntOfFlipsRequired(str)
    {
        // Stores length of str
        let n = str.length;
  
        // Store count of 0s in the string
        let zeros = 0;
  
        // Traverse the string
        for (let i = 0; i < n; i++) {
  
            // If current character is 0
            if (str[i] == '0') {
  
                // Update zeros
                zeros++;
            }
        }
  
        // If count of 0s in the string
        // is 0 or n
        if (zeros == 0 || zeros == n) {
            return 0;
        }
  
        // Store minimum count of flips
        // required to make all 0s on
        // the left and 1s on the right
        let minFlips = Number.MAX_VALUE;
  
        // Stores count of 1s on the left
        // side of each index
        let currOnes = 0;
  
        // Stores count of flips required to make
        // all 0s on the left and 1s on the right
        let flips;
  
        // Traverse the string
        for (let i = 0; i < n; i++) {
  
            // If current character is 1
            if (str[i] == '1') {
  
                // Update currOnes
                currOnes++;
            }
  
            // Update flips
            flips = currOnes + (zeros - (i + 1 - currOnes));
  
            // Update the minimum
            // count of flips
            minFlips = Math.min(minFlips, flips);
        }
  
        return minFlips;
    }
 
    // Driver Code
     
    let s1 = "100101";
    document.write(minimumCntOfFlipsRequired(s1));
 
</script>


Output

2

Time Complexity: O(N), where N is the length of the string
Auxiliary Space: O(1)

 




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