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Formatted question description: https://leetcode.ca/all/1318.html

1318. Minimum Flips to Make a OR b Equal to c (Medium)

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 <= 10^9
• 1 <= b <= 10^9
• 1 <= c <= 10^9

Related Topics:
Bit Manipulation

Solution 1.

• Java
• C++
• Python
• Go

• class Solution {
      public int minFlips(int a, int b, int c) {
          int flips = 0;
          StringBuffer aBinarySB = new StringBuffer(Integer.toBinaryString(a));
          StringBuffer bBinarySB = new StringBuffer(Integer.toBinaryString(b));
          StringBuffer cBinarySB = new StringBuffer(Integer.toBinaryString(c));
          while (aBinarySB.length() < 32)
              aBinarySB.insert(0, '0');
          while (bBinarySB.length() < 32)
              bBinarySB.insert(0, '0');
          while (cBinarySB.length() < 32)
              cBinarySB.insert(0, '0');
          for (int i = 0; i < 32; i++) {
              char aChar = aBinarySB.charAt(i);
              char bChar = bBinarySB.charAt(i);
              char cChar = cBinarySB.charAt(i);
              if (cChar == '0') {
                  if (aChar == '1')
                      flips++;
                  if (bChar == '1')
                      flips++;
              } else {
                  if (aChar == '0' && bChar == '0')
                      flips++;
              }
          }
          return flips;
      }
  }
  
  ############
  
  class Solution {
      public int minFlips(int a, int b, int c) {
          int ans = 0;
          for (int i = 0; i < 30; ++i) {
              int x = a >> i & 1, y = b >> i & 1, z = c >> i & 1;
              if ((x | y) != z) {
                  ans += x == 1 && y == 1 ? 2 : 1;
              }
          }
          return ans;
      }
  }
  

• // OJ: https://leetcode.com/problems/minimum-flips-to-make-a-or-b-equal-to-c/
  // Time: O(1)
  // Space: O(1)
  class Solution {
  public:
      int minFlips(int a, int b, int c) {
          int ans = 0;
          for (int i = 0; i < 31; ++i) {
              if ((c >> i) & 1) ans += ((a >> i) & 1) == 0 && ((b >> i) & 1) == 0;
              else ans += ((a >> i) & 1) + ((b >> i) & 1);
          }
          return ans;
      }
  };
  

• class Solution:
      def minFlips(self, a: int, b: int, c: int) -> int:
          ans = 0
          for i in range(31):
              x, y, z = (a >> i) & 1, (b >> i) & 1, (c >> i) & 1
              if (x | y) == z:
                  continue
              if x == 1 and y == 1 and z == 0:
                  ans += 2
              else:
                  ans += 1
          return ans
  
  
  

• func minFlips(a int, b int, c int) (ans int) {
  	for i := 0; i < 30; i++ {
  		x, y, z := a>>i&1, b>>i&1, c>>i&1
  		if (x | y) != z {
  			if x == 1 && y == 1 {
  				ans += 2
  			} else {
  				ans++
  			}
  		}
  	}
  	return
  }
  

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