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

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 ``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

 `= 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

 ``

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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