# Minimum flips to make all 1s in left and 0s in right | Set 1 (Using Bitmask)

Given a binary array, we can flip all the 1 are in the left part and all the 0 to the right part.Calculate the minimum flips required to make all 1s in left and all 0s in right.

Examples:

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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

For solving this problem we use bitmasking. First, we convert this array to string, then we find the equivalent decimal number of that binary string. We try all masks with all possibilities of 1s in left and 0s in right. We iterate a loop till decimal number becomes zero. Each time we will do bitwise XOR of the number with mask and number of ones in XOR value will be the number of flips required. We decrease n by 1 and update the mask.

1-Take binary array as input
2-Convert array to string and then equivalent decimal number(num)
3-Take initial mask value and iterate till num <= 0
4-Find required flips using (num XOR mask)
5-Find minimum flips and decrease num and update mask
6-Return the minimum count

 `// Java program to find minimum flips to make ` `// all 1s in left ` `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `    ``// function to count minimum number of flips ` `    ``public` `static` `int` `findMiniFlip(``int``[] nums) ` `    ``{ ` `        ``int` `n = nums.length; ` `        ``String s = ``""``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``s += nums[i]; ` ` `  `        ``// This is converting string s into integer  ` `        ``// of base 2 (if s = '100' then num = 4) ` `        ``long` `num = Integer.parseInt(s, ``2``); ` ` `  `        ``// initialize minXor with n that can be maximum  ` `        ``// number of flips ` `        ``int` `minXor = n; ` ` `  `        ``// right shift 1 by (n-1) bits ` `        ``long` `mask = (``1` `<< (n-``1``)); ` `        ``while` `(n-``1` `> ``0``) { ` ` `  `            ``// calculate bitwise XOR of num and mask ` `            ``long` `temp = (num ^ mask); ` ` `  `            ``// Math.min(a, b) returns minimum of a and b ` `            ``// return minimum number of flips till that  ` `            ``// digit ` `            ``minXor = Math.min(minXor, countones(temp)); ` `            ``n--; ` ` `  `            ``mask = (mask | (``1` `<< n -``1``)); ` `        ``} ` `        ``return` `minXor; ` `    ``} ` ` `  `    ``// function to count number of 1s ` `    ``public` `static` `int` `countones(``long` `n) ` `    ``{ ` `        ``int` `c = ``0``; ` `        ``while` `(n > ``0``) { ` `            ``n = n & (n-``1``); ` `            ``c++; ` `        ``} ` `        ``return` `c; ` `    ``} ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int``[] nums = { ``1``, ``0``, ``1``, ``1``, ``0``, ``0``, ``0` `}; ` `        ``int` `n = findMiniFlip(nums); ` `        ``System.out.println(n); ` `    ``} ` `} `

Output:

```1
```

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Improved By : prateek dhawalia

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