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Minimum Number of K Consecutive Bit Flips in C++
Suppose we have an array A. This is containing only 0s and 1s, here a K-bit flip consists of choosing a (contiguous) subarray of length K and simultaneously inverting the bits n the subarray. We have to find the minimum number of K-bit flips required so that there is no 0 in the array. If there is no such possibility, then return -1.
So, if the input is like [0,0,0,1,0,1,1,0] and K = 3, then the output will be 3, as we need to perform the operations three times, on the first try flip A[0] to A[3], the array will be [1,1,1,1,0,1,1,0], on the second run flip A[4] to A[6], the array will be [1,1,1,1,1,0,0,0], and in the 3rd move change A[5] to A[7], the array will be [1,1,1,1,1,1,1,1].
To solve this, we will follow these steps −
n := size of A
Define an array moves of size n
counter := 0
for initialize i := 0, when i + K <= n, update (increase i by 1), do −
if i > 0, then −
moves[i] := moves[i] + moves[i - 1]
if not ((moves[i] mod 2) + A[i]) mod 2 is non-zero, then −
moves[i] := moves[i] + 1
if i + K < n, then −
moves[i + K] - = 1
(increase counter by 1)
for i < n, update (increase i by 1), do −
if i > 0, then −
moves[i] := moves[i] + moves[i - 1]
if not ((moves[i] mod 2) + A[i]) mod 2 is non-zero, then −
return -1
return counter
Let us see the following implementation to get better understanding −
Output
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int minKBitFlips(vector<int>& A, int K){
int n = A.size();
vector<int> moves(n);
int i;
int counter = 0;
for (i = 0; i + K <= n; i++) {
if (i > 0)
moves[i] += moves[i - 1];
if (!(((moves[i] % 2) + A[i]) % 2)) {
moves[i] += 1;
if (i + K < n)
moves[i + K] -= 1;
counter++;
}
}
for (; i < n; i++) {
if (i > 0)
moves[i] += moves[i - 1];
if (!(((moves[i] % 2) + A[i]) % 2))
return -1;
}
return counter;
}
};
main(){
Solution ob;
vector<int> v = {0,0,0,1,0,1,1,0};
cout << (ob.minKBitFlips(v, 3));
}Input
{0,0,0,1,0,1,1,0}, 3Output
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