Minimum flips to make all 1s in left and 0s in right | Set 2

Last Updated : 5 Sep, 2022

Given a binary string, we can flip any bit in the given string i.e., convert 1 to 0 or vice-versa. Calculate the minimum flips required to make all 1s on the left and all 0s on the right.

Examples : 

Input: 1011000  
Output: 1
Explanation: 1 flip is required to make it 1111000.

Input: 00001 
Output: 2
Explanation: 2 flips required to make it 10000.

We have discussed a bitmask based solution in below post. Minimum flips to make all 1s in left and 0s in right | Set 1 (Using Bitmask)
It can be done with O(N) time complexity (where N - number of bits) and O(N) extra memory 

  1. Calculate number of flips of '0' needed to be done while moving from left to right to have all '1' in bits.
  2. Calculate number of flips of '1' needed to be done while moving from right to left to have all '0' in bits.
  3. Traversing through all positions between bits and find minimal sum of '0'-flips+'1'-flips from both arrays.

Implementation:

C++ 
// CPP program to find minimum flips required
// to make all 1s in left and 0s in right.
#include <bits/stdc++.h>
using namespace std;

int minimalFilps(string bits)
{
    int n = bits.length();

    // two arrays will keep the count for number
    // of 0s' and 1s' to be flipped while 
    // traversing from left to right and right to
    // left respectively
    int flipsFromLeft[n];
    int flipsFromRight[n];

    // Fill flipsFromLeft[]
    int flips = 0; 
    for (int i = 0; i < n; i++) {
        if (bits[i] == '0') 
            flips++;         
        flipsFromLeft[i] = flips; 
    }

    // Fill flipsFromRight[] 
    flips = 0;
    for (int i = n - 1; i >= 0; i--) {
        if (bits[i] == '1') 
            flips++;         
        flipsFromRight[i] = flips; 
    }

    // initialize minFlip to highest int value. If sum 
    // of leftflip and rightFlip is smaller than minflips, 
    // then update minFlips
    int minFlips = INT_MAX; 
    for (int i = 1; i < n; i++) {
        if (flipsFromLeft[i - 1] + flipsFromRight[i] < minFlips)
            minFlips = flipsFromLeft[i - 1] + flipsFromRight[i];
    }

    return minFlips;
}

// Driver code
int main()
{
    string bits = "00001";
    cout << minimalFilps(bits) << endl;
    return 0;
}
Java 
// Java program to find minimum flips required
// to make all 1s in left and 0s in right.
import java.io.*;

class GFG 
{
    static int minimalFilps(String bits)
    {
        int n = bits.length();
    
        // two arrays will keep the count 
        // for number of 0s' and 1s' to be
        // flipped while traversing from 
        // left to right and right to
        // left respectively
        int flipsFromLeft[] = new int[n];
        int flipsFromRight[] =new int[n] ;
    
        // Fill flipsFromLeft[]
        int flips = 0; 
        for (int i = 0; i < n; i++) 
        {
            if (bits.charAt(i) == '0') 
                flips++;         
            flipsFromLeft[i] = flips; 
        }
    
        // Fill flipsFromRight[] 
        flips = 0;
        for (int i = n - 1; i >= 0; i--) 
        {
            if (bits.charAt(i) == '1') 
                flips++;         
            flipsFromRight[i] = flips; 
        }
    
        // initialize minFlip to highest int value. If sum 
        // of leftflip and rightFlip is smaller than minflips, 
        // then update minFlips
        int minFlips = Integer.MAX_VALUE; 
        for (int i = 1; i < n; i++) 
        {
            if (flipsFromLeft[i - 1] + flipsFromRight[i] 
                                              < minFlips)
                minFlips = flipsFromLeft[i - 1] 
                           + flipsFromRight[i];
        }
    
        return minFlips;
    }
    
    // Driver code
    public static void main (String[] args) 
    {
        String bits = "00001";
        System.out.println(minimalFilps(bits)); 
        
    }
}

// This code is contributed by vt_m.
Python3 
# Python 3 program to find minimum flips required
# to make all 1s in left and 0s in right.
import sys

def minimalFilps(bits):
    n = len(bits)

    # two arrays will keep the count for number
    # of 0s' and 1s' to be flipped while 
    # traversing from left to right and right to
    # left respectively
    flipsFromLeft = [0 for i in range(n)]
    flipsFromRight = [0 for i in range(n)]

    # Fill flipsFromLeft[]
    flips = 0
    for i in range(0, n, 1):
        if (bits[i] == '0'):
            flips = flips + 1    
        flipsFromLeft[i] = flips 
    
    # Fill flipsFromRight[] 
    flips = 0
    i = n - 1
    while(i >= 0):
        if (bits[i] == '1'):
            flips = flips + 1
        i = i - 1
        flipsFromRight[i] = flips 
    
    # initialize minFlip to highest int value. 
    # If sum of leftflip and rightFlip is smaller 
    # than minflips, then update minFlips
    minFlips = sys.maxsize 
    for i in range(1, n, 1):
        if (flipsFromLeft[i - 1] + 
            flipsFromRight[i] < minFlips):
            minFlips = (flipsFromLeft[i - 1] + 
                        flipsFromRight[i])
    
    return minFlips

# Driver code
if __name__ == '__main__':
    bits = "00001"
    print(minimalFilps(bits))

# This code is contributed by
# Surendra_Gangwar
C# 
// C# program to find minimum flips required
// to make all 1s in left and 0s in right.
using System;

class GFG 
{
    static int minimalFilps(String bits)
    {
        int n = bits.Length;
    
        // two arrays will keep the count 
        // for number of 0s' and 1s' to be
        // flipped while traversing from 
        // left to right and right to
        // left respectively
        int []flipsFromLeft = new int[n];
        int []flipsFromRight =new int[n] ;
    
        // Fill flipsFromLeft[]
        int flips = 0; 
        for (int i = 0; i < n; i++) 
        {
            if (bits[i] == '0') 
                flips++;         
            flipsFromLeft[i] = flips; 
        }
    
        // Fill flipsFromRight[] 
        flips = 0;
        for (int i = n - 1; i >= 0; i--) 
        {
            if (bits[i] == '1') 
                flips++;         
            flipsFromRight[i] = flips; 
        }
    
        // initialize minFlip to highest int value.
        // If sum of leftflip and rightFlip is smaller
        // than minflips, then update minFlips
        int minFlips = int.MaxValue; 
        for (int i = 1; i < n; i++) 
        {
            if (flipsFromLeft[i - 1] + flipsFromRight[i] < minFlips)
            minFlips = flipsFromLeft[i - 1] + flipsFromRight[i];
        }
    
        return minFlips;
    }
    
    // Driver code
    public static void Main () 
    {
        string bits = "00001";
        Console.WriteLine(minimalFilps(bits)); 
        
    }
}

// This code is contributed by vt_m.
PHP 
<?php
// PHP program to find minimum 
// flips required to make all 
// 1s in left and 0s in right.

function minimalFilps($bits)
{
    $n = strlen($bits);

    // two arrays will keep the
    // count for number of 0s' 
    // and 1s' to be flipped 
    // while traversing from 
    // left to right and right 
    // to left respectively
    $flipsFromLeft[$n] = 0;
    $flipsFromRight[$n] = 0;

    // Fill flipsFromLeft[]
    $flips = 0; 
    for ($i = 0; $i < $n; $i++) {
        if ($bits[$i] == '0') 
            $flips++;         
        $flipsFromLeft[$i] = $flips; 
    }

    // Fill flipsFromRight[] 
    $flips = 0;
    for ($i = $n - 1; $i >= 0; $i--) 
    {
        if ($bits[$i] == '1') 
            $flips++;         
        $flipsFromRight[$i] = $flips; 
    }

    // initialize minFlip to 
    // highest int value. If sum 
    // of leftflip and rightFlip 
    // is smaller than minflips, 
    // then update minFlips
    $INT_MAX=2147483647;
    $minFlips = $INT_MAX; 
    for ($i = 1; $i < $n; $i++) 
    {
        if ($flipsFromLeft[$i - 1] + 
            $flipsFromRight[$i] < $minFlips)
            $minFlips = $flipsFromLeft[$i - 1] + 
                        $flipsFromRight[$i];
    }

    return $minFlips;
}

    // Driver Code
    $bits = "00001";
    echo minimalFilps($bits) ;

// This code is contributed by nitin mittal.
?>
JavaScript 
<script>
    // Javascript program to find minimum flips required
    // to make all 1s in left and 0s in right.
    
    function minimalFilps(bits)
    {
        let n = bits.length;
      
        // two arrays will keep the count 
        // for number of 0s' and 1s' to be
        // flipped while traversing from 
        // left to right and right to
        // left respectively
        let flipsFromLeft = new Array(n);
        flipsFromLeft.fill(0);
        let flipsFromRight =new Array(n);
        flipsFromRight.fill(0);
      
        // Fill flipsFromLeft[]
        let flips = 0; 
        for (let i = 0; i < n; i++) 
        {
            if (bits[i] == '0') 
                flips++;         
            flipsFromLeft[i] = flips; 
        }
      
        // Fill flipsFromRight[] 
        flips = 0;
        for (let i = n - 1; i >= 0; i--) 
        {
            if (bits[i] == '1') 
                flips++;         
            flipsFromRight[i] = flips; 
        }
      
        // initialize minFlip to highest int value.
        // If sum of leftflip and rightFlip is smaller
        // than minflips, then update minFlips
        let minFlips = Number.MAX_VALUE; 
        for (let i = 1; i < n; i++) 
        {
            if (flipsFromLeft[i - 1] + flipsFromRight[i] < minFlips)
                minFlips = flipsFromLeft[i - 1] + flipsFromRight[i];
        }
      
        return minFlips;
    }
    
    let bits = "00001";
      document.write(minimalFilps(bits)); 
 
 // This code is contributed by divyesh072019.
</script>

Output

 2

Time complexity: O(N) where N is the length of the given binary string.
Auxiliary space: O(N), for creating arrays flipsFromLeft and flipsFromRight of size N.

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Article Tags:
DSA
Arrays
binary-string

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