Merge pull request #21 from iamAntimPal/Leetcode-75

Leetcode 75

Author: iamAntimPal

Solution/1318. Minimum Flips to Make a OR b Equal to c/1318. Minimum Flips to Make a OR b Equal to c.py

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+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

Solution/1318. Minimum Flips to Make a OR b Equal to c/readme.md

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+---
+comments: true
+difficulty: Medium
+edit_url: antim
+rating: 1382
+source: Weekly Contest 171 Q2
+tags:
+    - Bit Manipulation
+---
+
+<!-- problem:start -->
+
+# [1318. Minimum Flips to Make a OR b Equal to c](https://leetcode.com/problems/minimum-flips-to-make-a-or-b-equal-to-c)
+
+
+## Description
+
+<!-- description:start -->
+
+<p>Given 3 positives numbers <code>a</code>, <code>b</code> and <code>c</code>. Return the minimum flips required in some bits of <code>a</code> and <code>b</code> to make (&nbsp;<code>a</code> OR <code>b</code> == <code>c</code>&nbsp;). (bitwise OR operation).<br />
+Flip operation&nbsp;consists of change&nbsp;<strong>any</strong>&nbsp;single bit 1 to 0 or change the bit 0 to 1&nbsp;in their binary representation.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1318.Minimum%20Flips%20to%20Make%20a%20OR%20b%20Equal%20to%20c/images/sample_3_1676.png" style="width: 260px; height: 87px;" /></p>
+
+<pre>
+<strong>Input:</strong> a = 2, b = 6, c = 5
+<strong>Output:</strong> 3
+<strong>Explanation: </strong>After flips a = 1 , b = 4 , c = 5 such that (<code>a</code> OR <code>b</code> == <code>c</code>)</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> a = 4, b = 2, c = 7
+<strong>Output:</strong> 1
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> a = 1, b = 2, c = 3
+<strong>Output:</strong> 0
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<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>
+</ul>
+
+<!-- description:end -->
+
+## Solutions
+
+<!-- solution:start -->
+
+### 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 $O(1)$.
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+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
+```
+
+#### Java
+
+```java
+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;
+    }
+}
+```
+
+#### C++
+
+```cpp
+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;
+    }
+};
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->