# 07.06.2023 [1318. Minimum Flips to Make a OR b Equal to c]

07.06.2023

1318. Minimum Flips to Make a OR b Equal to c medium
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Problem TLDR

Minimum a and b Int bit flips to make a or b == c.

Intuition

Naive implementation is to iterate over 32 bits and flip a or/and b bits to match c.
If we didn’t consider the case where a = 1 and b = 1 and c = 0, the result would be (a or b) xor c, as a or b gives us the left side of the equation, and xor c gives only bits that are needed to flip. For the corner case a = b = 1, c = 0, we must do additional flip to make 0, and we must make any other combinations 0:

a b c     a and b   c.inv()   (a and b) and c.inv()

0 0 1     0         0         0
0 1 0     0         1         0
0 1 1     0         0         0
1 0 0     0         1         0
1 0 1     0         0         0
1 1 0     1         1         1
1 1 1     1         0         0

Approach

Use Integer.bitCount.

Complexity

• Time complexity:
  O(1)

• Space complexity:
  O(1)

Code

fun minFlips(a: Int, b: Int, c: Int): Int =
Integer.bitCount((a or b) xor c) + Integer.bitCount((a and b) and c.inv())