Weekly Contest 128

1012. Complement of Base 10 Integer

Every non-negative integer N has a binary representation.  For example, 5 can be represented as "101" in binary, 11 as "1011" in binary, and so on.  Note that except for N = 0, there are no leading zeroes in any binary representation.

The complement of a binary representation is the number in binary you get when changing every 1 to a 0 and 0 to a 1.  For example, the complement of "101" in binary is "010" in binary.

For a given number N in base-10, return the complement of it's binary representation as a base-10 integer.

Example 1:

Input: 5
Output: 2
Explanation: 5 is "101" in binary, with complement "010" in binary, which is 2 in base-10.

Example 2:

Input: 7
Output: 0
Explanation: 7 is "111" in binary, with complement "000" in binary, which is 0 in base-10.

Example 3:

Input: 10
Output: 5
Explanation: 10 is "1010" in binary, with complement "0101" in binary, which is 5 in base-10.

Note:

1. 0 <= N < 10^9

Approach #1: Math. [C++]

class Solution {
public:
    int bitwiseComplement(int N) {
        int X = 1;
        while (N > X) X = X * 2 + 1;
        return X ^ N;
    }
};

　　

Analysis:

Claim ----- The XOR operation evaluates the difference in the individual bits, i.e it gives information about whether the bits are identical or not.

Proof ----- It's easy once youknow the definition of XOR.  0^0 = 1^1 = 0 (as the bits don't differ), whereas 0^1 = 1^0 = 1 (as the bits are difference).

Claim ----- XOR of identical numbers is zero.

Proof ----- As argued above, the bits of identical numbers do not differ at any position. Hence, XOR is zero.

Claim ----- 0 XOR any number is the number itself.

Proof ----- XOR gives us the bit difference. Since all the bits in 0 are unset, therefore the difference in bits is the number itself.

Claim ----- XOR of a number with its complement results in a number with all set bits.

Proof ----- This is trivial,  since bits of a number and its complement differ at every position(according to the definition of complement).

So, number ^ complement = all_set_bits ==> number ^ number ^ complement = number ^ all_set_bits ===> 0 ^ complement = number ^ all_set_bits

So, complement = number ^ all_set_bits.

So, we find out the number containing all the set bits and XOR with the original number to get the answer.

Reference:

https://leetcode.com/problems/complement-of-base-10-integer/discuss/256734/Detailed-Explanation-using-XOR-C%2B%2BJavaScript

1013. Pairs of Songs With Total Durations Divisible by 60

In a list of songs, the i-th song has a duration of time[i] seconds. 

Return the number of pairs of songs for which their total duration in seconds is divisible by 60.  Formally, we want the number of indices i < j with (time[i] + time[j]) % 60 == 0.

Example 1:

Input: [30,20,150,100,40]
Output: 3
Explanation: Three pairs have a total duration divisible by 60:
(time[0] = 30, time[2] = 150): total duration 180
(time[1] = 20, time[3] = 100): total duration 120
(time[1] = 20, time[4] = 40): total duration 60

Example 2:

Input: [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.

Note:

1. 1 <= time.length <= 60000
2. 1 <= time[i] <= 500

 

Approach #1: Brute force + Map. [C++]

class Solution {
public:
    int numPairsDivisibleBy60(vector<int>& time) {
        int ans = 0;
        int len = time.size();
        map<int, vector<int>> m;
        vector<int> duration = {60, 120, 180, 240, 300, 360, 420, 480, 540, 600,
                                660, 720, 780, 840, 900, 960, 1020};
        for (int i = 0; i < len; ++i) m[time[i]].push_back(i);
        for (int i = 0; i < len; ++i) {
            for (int j = 0; j < duration.size(); ++j) {
                if (duration[j] - time[i] > 0) {
                    int tmp = duration[j] - time[i];
                    if (m.count(tmp)) {
                        int count = m[tmp].end() - upper_bound(m[tmp].begin(), m[tmp].end(), i);
                        ans += count;
                    }
                }
            }
        }
        
        return ans;
    }
};

　　

Approach #2: Orz.

    int numPairsDivisibleBy60(vector<int>& time) {
        vector<int> c(60);
        int res = 0;
        for (int t : time) {
            res += c[(60 - t % 60) % 60];
            c[t % 60] += 1;
        }
        return res;
    }

　　

Analysis:

Calculate the time%60 then it will be exactly same as two sum problem.

Reference:

https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/discuss/256738/JavaC%2B%2BPython-Two-Sum-with-K-60

1014. Capacity To Ship Packages Within D Days

A conveyor belt has packages that must be shipped from one port to another within Ddays.

The i-th package on the conveyor belt has a weight of weights[i].  Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship.

Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within Ddays.

Example 1:

Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5
Output: 15
Explanation: 
A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
1st day: 1, 2, 3, 4, 5
2nd day: 6, 7
3rd day: 8
4th day: 9
5th day: 10

Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed. 

Example 2:

Input: weights = [3,2,2,4,1,4], D = 3
Output: 6
Explanation: 
A ship capacity of 6 is the minimum to ship all the packages in 3 days like this:
1st day: 3, 2
2nd day: 2, 4
3rd day: 1, 4

Example 3:

Input: weights = [1,2,3,1,1], D = 4
Output: 3
Explanation: 
1st day: 1
2nd day: 2
3rd day: 3
4th day: 1, 1

Note:

1. 1 <= D <= weights.length <= 50000
2. 1 <= weights[i] <= 500

Approach #1: Binary search. [C++]

class Solution {
public:
    int shipWithinDays(vector<int>& weights, int D) {
        int left = *max_element(weights.begin(), weights.end());
        int right = 25000000;
        while (left < right) {
            int mid = (right + left) / 2;
            int need = 1, cur = 0;
            for (int i = 0; i < weights.size() && need <= D; cur += weights[i++]) {
                if (cur + weights[i] > mid)
                    cur = 0, need++;
            }
            if (need > D) left = mid + 1;
            else right = mid;
        }
        return left;
    }
};

　　

Analysis:

Given the number of bags, return the minimum capacity of each bag, so that we can put items one by one into all bags.

Reference:

https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/discuss/256729/JavaC%2B%2BPython-Binary-Search

1015. Numbers With Repeated Digits

Given a positive integer N, return the number of positive integers less than or equal to N that have at least 1 repeated digit.

Example 1:

Input: 20
Output: 1
Explanation: The only positive number (<= 20) with at least 1 repeated digit is 11.

Example 2:

Input: 100
Output: 10
Explanation: The positive numbers (<= 100) with atleast 1 repeated digit are 11, 22, 33, 44, 55, 66, 77, 88, 99, and 100.

Example 3:

Input: 1000
Output: 262

Note:

1. 1 <= N <= 10^9

 

 Approach #1: 

class Solution {
public:
    int numDupDigitsAtMostN(int N) {
        int invalid = 0;
        
        int c = floor(log10(N+1)) + 1;
        for (int i = 0; i < c-1; ++i) {
            invalid += 9 * perm(9, i);
        }
        
        int digits = 0;
        for (int i = 0; i < c; ++i) {
            int digit = ((N+1) / (int)pow(10, c-i-1)) % 10;
            
            for (int j = (i > 0 ? 0 : 1); j < digit; ++j) {
                if (((digits >> j) & 1) == 0) {
                    invalid += perm(9 - i, c - i - 1);
                }
            }
            
            if ((digits >> digit) & 1) break;
            digits |= 1 << digit;
        }
        
        return N - invalid;
    }
    
    int perm(int m, int n) {
        int out = 1;
        while (m > 1 && n > 0) {
            out *= m;
            m--;
            n--;
        }
        
        return out;
    }
};

　　

Analysis:

For example, with the number 350, we have 3 digits, meaning we can start by finding all invalid numbers from 0 to 99 (e.g. the first two digits). To start, let's assuming we only have 1 digit available. In this case, we can't vary any other digits in the number since there are none, and because there is only 1 digit they are all invalid. Thus, since there are 9 total numbers with 1 digit, we have 9 invalid permutations for this digit. Similary, for 2 digits, we have 1 digit we can vary (e.g. 1x has x that can be varied, 2y has y that can be varied, so on and so forth). Plugging that into our formula, we have perm(9,1) which results in 9. Because there are 9 possible digits for the first digit, we can multiply the result by 9 (perm(9, 1) * 9) which gives us 81 invalid digits. Adding that onto our first result of 9, and we get 90 invalid for a number range of 1-99 (meaning we have 9 valid digits in that range).

At this point, for the number 350, we know that thare are at least 90 invalid digits from 1 - 100 as a result (since 100 is valid). Now however we need to count the number of invalid digits from 100 - 350. This can be done by varying each of the digits in 351 (e.g. N+1), and finding the valid permutations of that as a result. For example:

3XX -> perm(9-0, 3-0-1) -> perm(9, 2)
X5X -> perm(9-1, 3-1-1) -> perm(8, 1)
XX1 -> perm(9-2, 3-2-1) -> perm(7, 0)

We then add this number of invalid permutations to our count based on the number we have. However, if we have previously seen a number in that range. we ignore it. For example, when we get to the 5 in 351, we will only add perm(8, 1)'s result 4 time, since the third time has alredy been accounted for when we went over the 3 in 351. Once we've done all of this, we can simply subtract our number of nvalid numbers from our original number N to get our result.

Here is what this process looks like in action:

350 -> 351
invalid digits -> 0

1 digit -> X -> perm(9, 0) * 9 -> 9 invalid digits
2 digits -> YX -> perm(9, 1) * 9 -> 81 invalid digits
invalid digits -> 90

0XX -> invalid so don't count the invalid digits.
1XX -> perm(9, 2) -> 72 invalid digits
2XX -> perm(9, 2) -> 72 invalid digits
3XX -> stop counting invalid numbers for the first digit.
X0X -> perm(8, 1) -> 8 invalid digits
X1X -> perm(8, 1) -> 8 invalid digits
X2X -> perm(8, 1) -> 8 invalid digits
X3X -> perm(8, 1) -> 8 invalid digits -> but because we've already looked at the digit 3 previously we can skip this.
X4X -> perm(8, 1) -> 8 invalid digits
X5X -> stop counting invalid numbers for the second digit.
XX0 -> perm(7, 0) -> 1 invalid digit
XX1 -> stop counting invalid numbers for the third and final digit.
invalid digits -> 267

result -> 350 - 267 = 83

Reference:

https://leetcode.com/problems/numbers-with-repeated-digits/discuss/257241/C%2B%2B-with-Alternative-Explanation