3133. Minimum Array End

C++

• class Solution {
   public:
    long long minEnd(int n, int x) {
      // Set x's 0s with (n - 1)'s LSb-to-MSb bits, preserving x's 1s. This
      // operation increase x for (n - 1) iterations while preserving x's 1s.
      const int kMaxBit = log2(n) + log2(x) + 2;
      const long k = n - 1;
      long ans = x;
      int kBinaryIndex = 0;
  
      for (int i = 0; i < kMaxBit; ++i) {
        if ((ans >> i & 1) == 0) {
          // Set x's 0 with k's bit if the running bit of k is 1.
          if (k >> kBinaryIndex & 1)
            ans |= 1L << i;
          ++kBinaryIndex;
        }
      }
  
      return ans;
    }
  };

JAVA

• import math
  
  class Solution:
      def minEnd(self, n: int, x: int) -> int:
          kMaxBit = math.log2(n) + math.log2(x) + 2
          k = n - 1
          ans = x
          kBinaryIndex = 0
  
          for i in range(int(kMaxBit)):
              if (ans >> i & 1) == 0:
                  if (k >> kBinaryIndex & 1):
                      ans |= 1 << i
                  kBinaryIndex += 1
  
          return ans

• 
  
• PYTHON

• public class Solution {
      public long minEnd(int n, int x) {
          int kMaxBit = (int)(Math.log(n) / Math.log(2) + Math.log(x) / Math.log(2) + 2);
          long k = n - 1;
          long ans = x;
          int kBinaryIndex = 0;
  
          for (int i = 0; i < kMaxBit; ++i) {
              if ((ans >> i & 1) == 0) {
                  if ((k >> kBinaryIndex & 1) == 1) {
                      ans |= 1L << i;
                  }
                  kBinaryIndex++;
              }
          }
  
          return ans;
      }
  }