# [Solution] Flip to Invert solution codechef

Flip to Invert solution codechef – JJ has a binary string SS of length NN. JJ can perform the following operation on SS:

• Select an ii such that 1iN1≤i≤N, and flip SiSi (i.e. change 00 to 11 and 11 to 00)

## [Solution] Flip to Invert solution codechef

JJ wants to minimize the number of inversions in SS by performing the above operation at most KK times. Can you help JJ do so?

Recall that a pair of indices (i,j)(i,j) in SS is called an inversion if i<ji<j and Si>SjSi>Sj.

### Input Format

• The first line contains a single integer TT – the number of test cases. Then the test cases follow.
• The first line of each test case contains two integers NN and KK – the length of the binary string SS and the maximum number of times JJ can perform the given operation.
• The second line of each test case contains a binary string SS of length NN containing 00s and 11s only.

### Output Format

For each test case, output the minimum number of inversions possible after performing the given operation at most KK times.

## [Solution] Flip to Invert solution codechef

• 1T1051≤T≤105
• 1N1051≤N≤105
• 0KN0≤K≤N
• Sum of NN over all test cases does not exceed 21052⋅105

### Sample Input 1

3
8 3
00010111
5 1
10100
6 2
111000


### Sample Output 1

0
2
3


## Flip to Invert solution Explanation

Test case 1: We are allowed to perform at most 33 operations. We can perform the following operation on SS00010⎯⎯1110001111100010_111→00011111 which has 00 inversions. Therefore 00 is the answer.

Test case 2: We can perform the following operation on SS1⎯⎯0100001001_0100→00100 which has 22 inversions. It can be proven that this is the minimum number of inversions possible.

Test case 3: We can perform the following operations on SS11100⎯⎯0111010⎯⎯11101111100_0→111010_→111011 which has 33 inversions. It can be proven that this is the minimum number of inversions possible.