# Daily Coding Challenge #63

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.

/*
Reverse Bits

Reverse bits of a given 32 bits unsigned integer.

Example 1:

Input: 00000010100101000001111010011100
Output: 00111001011110000010100101000000
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.
Example 2:

Input: 11111111111111111111111111111101
Output: 10111111111111111111111111111111
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10111111111111111111111111111111.

Note:

Note that in some languages such as Java, there is no unsigned integer type. In this case, both input and output will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 2 above the input represents the signed integer -3 and the output represents the signed integer -1073741825.

If this function is called many times, how would you optimize it?
*/

class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t ans=0;
for(int i=0;i<32;i++){
// use n&1 to get the least bit
// shift 1 bit to the left for ans and apply OR to it and the least bit from n
ans=(ans<<1|n&1);
// shift 1 bit to the right
n>>=1;
}
return ans;
}
};

static const auto io_sync_off = []() {std::ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();


/*
LeetCode - Number of Good Pairs

Given an array of integers nums.

A pair (i,j) is called good if nums[i] == nums[j] and i < j.

Return the number of good pairs.

Example 1:

Input: nums = [1,2,3,1,1,3]
Output: 4
Explanation: There are 4 good pairs (0,3), (0,4), (3,4), (2,5) 0-indexed.
Example 2:

Input: nums = [1,1,1,1]
Output: 6
Explanation: Each pair in the array are good.
Example 3:

Input: nums = [1,2,3]
Output: 0

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100
*/

class Solution {
public:
int numIdenticalPairs(vector<int>& nums) {
int ans=0;
// find all nums[i] == nums[j] and i < j
for(int i=0;i<nums.size();i++){
for(int j=i+1;j<nums.size();j++){
if(nums[i]==nums[j]) ans++;
}
}
return ans;
}
};


/*
LeetCode - Number of Substrings With Only 1s

Given a binary string s (a string consisting only of '0' and '1's).

Return the number of substrings with all characters 1's.

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: s = "0110111"
Output: 9
Explanation: There are 9 substring in total with only 1's characters.
"1" -> 5 times.
"11" -> 3 times.
"111" -> 1 time.
Example 2:

Input: s = "101"
Output: 2
Explanation: Substring "1" is shown 2 times in s.
Example 3:

Input: s = "111111"
Output: 21
Explanation: Each substring contains only 1's characters.
Example 4:

Input: s = "000"
Output: 0

Constraints:

s[i] == '0' or s[i] == '1'
1 <= s.length <= 10^5
*/

class Solution {
public:
int numSub(string s) {
// count approach
int ans=0,cnt=0,m=1e9+7;
for(char c:s){
if(c=='1'){
cnt++;
ans=(ans+cnt)%m;
}else {
cnt=0;
}
}
return ans;
}
};


The source code is available in corresponding repo below. Star and watch for timely updates!

## wingkwong / atcoder

### Discussion   