diff --git a/Solution/1318. Minimum Flips to Make a OR b Equal to c/1318. Minimum Flips to Make a OR b Equal to c.py b/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
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diff --git a/Solution/1318. Minimum Flips to Make a OR b Equal to c/readme.md b/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
+---
+
+
+
+# [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
+
+
+
+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:
+
+
+ 1 <= a <= 10^91 <= b <= 10^91 <= c <= 10^9
+
+
+
+## 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 $O(1)$.
+
+
+
+#### 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;
+ }
+};
+```
+
+
+
+
+
+
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