LeetCode 1318. Minimum Flips to Make a OR b Equal to c Solution in Java, C++, Python & More | Explanation + Code

Description

Given 3 positives numbers a, b and c. Return the minimum flips required in some bits of a and b to make ( a OR b == c ). (bitwise OR operation).

Flip operation consists of change any single bit 1 to 0 or change the bit 0 to 1 in their binary representation.

 

Example 1:

Input: a = 2, b = 6, c = 5

Output: 3

Explanation: After flips a = 1 , b = 4 , c = 5 such that (a OR b == c)

Example 2:

Input: a = 4, b = 2, c = 7

Output: 1

Example 3:

Input: a = 1, b = 2, c = 3

Output: 0

 

Constraints:

<li><code>1 &lt;= a &lt;= 10^9</code></li>

<li><code>1 &lt;= b&nbsp;&lt;= 10^9</code></li>

<li><code>1 &lt;= c&nbsp;&lt;= 10^9</code></li>

Solutions

Solution 1: Bit Manipulation

We can enumerate each bit of the binary representation of a, b, and c, denoted as x, y, and z respectively. If the bitwise OR operation result of x and y is different from z, we then check if both x and y are 1. If so, we need to flip twice, otherwise, we only need to flip once. We accumulate all the required flip times.

The time complexity is O(log M), where M is the maximum value of the numbers in the problem. The space complexity is O(1).

PythonJavaC++GoTypeScript

class Solution: def minFlips(self, a: int, b: int, c: int) -> int: ans = 0 for i in range(32): x, y, z = a >> i & 1, b >> i & 1, c >> i & 1 ans += x + y if z == 0 else int(x == 0 and y == 0) return ans(code-box)

class Solution { public int minFlips(int a, int b, int c) { int ans = 0; for (int i = 0; i < 32; ++i) { int x = a >> i & 1, y = b >> i & 1, z = c >> i & 1; ans += z == 0 ? x + y : (x == 0 && y == 0 ? 1 : 0); } return ans; } }(code-box)

class Solution { public: int minFlips(int a, int b, int c) { int ans = 0; for (int i = 0; i < 32; ++i) { int x = a >> i & 1, y = b >> i & 1, z = c >> i & 1; ans += z == 0 ? x + y : (x == 0 && y == 0 ? 1 : 0); } return ans; } };(code-box)

func minFlips(a int, b int, c int) (ans int) { for i := 0; i < 32; i++ { x, y, z := a>>i&1, b>>i&1, c>>i&1 if z == 0 { ans += x + y } else if x == 0 && y == 0 { ans++ } } return }(code-box)

function minFlips(a: number, b: number, c: number): number { let ans = 0; for (let i = 0; i < 32; ++i) { const [x, y, z] = [(a >> i) & 1, (b >> i) & 1, (c >> i) & 1]; ans += z === 0 ? x + y : x + y === 0 ? 1 : 0; } return ans; }(code-box)