Weekly Contest 129

1020. Partition Array Into Three Parts With Equal Sum

Given an array A of integers, return true if and only if we can partition the array into three non-empty parts with equal sums.

Formally, we can partition the array if we can find indexes i+1 < j with (A[0] + A[1] + ... + A[i] == A[i+1] + A[i+2] + ... + A[j-1] == A[j] + A[j-1] + ... + A[A.length - 1])

 

Example 1:

Input: [0,2,1,-6,6,-7,9,1,2,0,1]
Output: true
Explanation: 0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1

Example 2:

Input: [0,2,1,-6,6,7,9,-1,2,0,1]
Output: false

Example 3:

Input: [3,3,6,5,-2,2,5,1,-9,4]
Output: true
Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4

 

Note:

1. 3 <= A.length <= 50000
2. -10000 <= A[i] <= 10000

 

Approach #1:

class Solution {
public:
    bool canThreePartsEqualSum(vector<int>& A) {
        int len = A.size();
        vector<int> sums(len, 0);
        unordered_map<int, int> m_;
        sums[0] = A[0];
        m_[sums[0]] = 0;
        for (int i = 1; i < len; ++i) {
            sums[i] = sums[i-1] + A[i];
            m_[sums[i]] = i;
        }
        for (int i = 0; i < len-2; ++i) {
            int s, m, e;
            
            if (m_.find(2*sums[i]) != m_.end() && m_.find(3*sums[i]) != m_.end()) {
                s = i, m = m_[2*sums[i]], e = m_[3*sums[i]];
                if (s < m && m < e && e == len-1) return true;
            }
                
        }
        
        return false;
    }
};

　　

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1022. Smallest Integer Divisible by K

Given a positive integer K, you need find 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.

 

Example 1:

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

Example 2:

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

Example 3:

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

 

Note:

• 1 <= K <= 10^5

 

Approach #1: 

class Solution {
public:
    int smallestRepunitDivByK(int K) {
        if (K % 2 == 0) return -1;
        int last = 0;
        int temp = 0;
        for (int i = 1; i <= K; ++i) {
            temp = last;   
            temp = temp * 10 + 1;
            temp = temp % K;
            if (temp % K == 0) return i;
            last = temp;
        }
        return -1;
    }
};

　　

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1021. Best Sightseeing Pair

Given an array A of positive integers, A[i] represents the value of the i-th sightseeing spot, and two sightseeing spots i and j have distance j - i between them.

The score of a pair (i < j) of sightseeing spots is (A[i] + A[j] + i - j) : the sum of the values of the sightseeing spots, minus the distance between them.

Return the maximum score of a pair of sightseeing spots.

 

Example 1:

Input: [8,1,5,2,6]
Output: 11
Explanation: i = 0, j = 2, A[i] + A[j] + i - j = 8 + 5 + 0 - 2 = 11

 

Note:

1. 2 <= A.length <= 50000
2. 1 <= A[i] <= 1000

 

Approach #1:

class Solution {
public:
    int maxScoreSightseeingPair(vector<int>& A) {
        int res = 0, cur = 0;
        
        for (int a : A) {
            res = max(res, cur + a);
            cur = max(cur, a) - 1;
        }
        
        return res;
    }
};

　　

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1023. Binary String With Substrings Representing 1 To N

Given a binary string S (a string consisting only of '0' and '1's) and a positive integer N, return true if and only if for every integer X from 1 to N, the binary representation of X is a substring of S.

 

Example 1:

Input: S = "0110", N = 3
Output: true

Example 2:

Input: S = "0110", N = 4
Output: false

 

Note:

1. 1 <= S.length <= 1000
2. 1 <= N <= 10^9

 

Approach #1:

class Solution {
public:
    bool queryString(string S, int N) {
        vector<bool> seen(N, false);
        for (int i = 0; i < S.length(); ++i) {
            for (auto j = i, num = 0; num <= N && j < S.length(); ++j) {
                num = (num << 1) + S[j] - '0';
                if (num > 0 && num <= N) seen[num-1] = true;
            }
        }
        
        return all_of(seen.begin(), seen.end(), [](bool s) { return s; });
    }
};

　　

 

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来源： https://www.cnblogs.com/ruruozhenhao/p/10588154.html