Today I have done Minimum Flips to Make a OR b Equal to c on leetcode and leetcode's November LeetCoding Challenge with cpp.

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

Solution

nothing to say

class Solution {
public:
    int minFlips(int a, int b, int c) {
        int answer = 0;
        for(int i = 0; i < 32; ++i) {
            int cur = 1 << i;
            if(cur & c) {
                if(!(cur&(a|b))) answer += 1;
            } else {
                if(cur&a) answer += 1;
                if(cur&b) answer += 1;
            }
        }
        return answer;
    }
};

November LeetCoding Challenge 25

Description

Smallest Integer Divisible by K

Given a positive integer K, you need to find the length of the smallest positive integer N such that N is divisible by K, and N only contains the digit 1.

Return the length of N. If there is no such N, return -1.

Note: N may not fit in a 64-bit signed integer.

Example 1:

Input: K = 1
Output: 1
Explanation: The smallest answer is N = 1, which has length 1.

Example 2:

Input: K = 2
Output: -1
Explanation: There is no such positive integer N divisible by 2.

Example 3:

Input: K = 3
Output: 3
Explanation: The smallest answer is N = 111, which has length 3.

Constraints:

• 1 <= K <= 105

Solution

not so elegant

class Solution {
public:
    int smallestRepunitDivByK(int K) {
        if(K & 1 == 0) return -1;
        int cur = 1;
        int sum = 1;
        for(int i = 1; i <= K; ++i) {
            if(sum % K == 0) return i;
            sum %= K;
            cur = cur * 10;
            cur %= K;
            sum += cur;
        }
        return -1;
    }
};