weekly contest 116

961. N-Repeated Element in Size 2N Array

In a array A of size 2N, there are N+1 unique elements, and exactly one of these elements is repeated N times.

Return the element repeated N times.

Input: [1,2,3,3]
Output: 3

Input: [2,1,2,5,3,2]
Output: 2

Input: [5,1,5,2,5,3,5,4]
Output: 5

Approach #1: C++.

class Solution {
public:
    int repeatedNTimes(vector<int>& A) {
        int size = A.size();
        int repeatTime = size / 2;
        vector<int> temp(10005, 0);
        
        for (int a : A) {
            temp[a]++;
            if (temp[a] == repeatTime) 
                return a;
        }
        
    }
};

962. Maximum Width Ramp

Given an array A of integers, a ramp is a tuple (i, j) for which i < j and A[i] <= A[j].  The width of such a ramp is j - i.

Find the maximum width of a ramp in A.  If one doesn‘t exist, return 0.

Example 1:

Input: [6,0,8,2,1,5]
Output: 4
Explanation: 
The maximum width ramp is achieved at (i, j) = (1, 5): A[1] = 0 and A[5] = 5.

Input: [9,8,1,0,1,9,4,0,4,1]
Output: 7
Explanation: 
The maximum width ramp is achieved at (i, j) = (2, 9): A[2] = 1 and A[9] = 1.

Approach #2: C++.

typedef pair<int, int> pp;

class Solution {
public:
    int maxWidthRamp(vector<int>& A) {
        vector<pp> temp;
        int ans = 0;
        temp.push_back(make_pair(A[0], 0));
        for (int i = 1; i < A.size(); ++i) {
            if (A[i] < temp.back().first) {
                temp.push_back(make_pair(A[i], i));
            } else {
                int ramp = i - binarySearch(temp, A[i]);
                ans = max(ans, ramp);
            }
        }
        return ans;
    }
    
    int binarySearch(vector<pp> temp, int target) {
        int l = -1, r = temp.size()-1, mid;
        while (l+1 < r) {
            mid = (l + r) / 2;
            if (temp[mid].first > target) l = mid;
            else r = mid;
        }
        return temp[r].second;
    }
};

Analysis:

we use a vector to store the decrement elements from the begin to the end of A.

963. Minimum Area Rectangle II

Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the x and y axes.

If there isn‘t any rectangle, return 0.

Example 1:

Input: [[1,2],[2,1],[1,0],[0,1]]
Output: 2.00000
Explanation: The minimum area rectangle occurs at [1,2],[2,1],[1,0],[0,1], with an area of 2.

Input: [[0,1],[2,1],[1,1],[1,0],[2,0]]
Output: 1.00000
Explanation: The minimum area rectangle occurs at [1,0],[1,1],[2,1],[2,0], with an area of 1.

Input: [[0,3],[1,2],[3,1],[1,3],[2,1]]
Output: 0
Explanation: There is no possible rectangle to form from these points.

Input: [[3,1],[1,1],[0,1],[2,1],[3,3],[3,2],[0,2],[2,3]]
Output: 2.00000
Explanation: The minimum area rectangle occurs at [2,1],[2,3],[3,3],[3,1], with an area of 2.

Note:

1. 1 <= points.length <= 50
2. 0 <= points[i][0] <= 40000
3. 0 <= points[i][1] <= 40000
4. All points are distinct.
5. Answers within 10^-5 of the actual value will be accepted as correct.

Approach #1: C++.

Thinking..................................

964. Least Operators to Express Number

Given a single positive integer x, we will write an expression of the form x (op1) x (op2) x (op3) x ... where each operator op1, op2, etc. is either addition, subtraction, multiplication, or division (+, -, *, or /).  For example, with x = 3, we might write 3 * 3 / 3 + 3 - 3 which is a value of 3.

When writing such an expression, we adhere to the following conventions:

1. The division operator (/) returns rational numbers.
2. There are no parentheses placed anywhere.
3. We use the usual order of operations: multiplication and division happens before addition and subtraction.
4. It‘s not allowed to use the unary negation operator (-).  For example, "x - x" is a valid expression as it only uses subtraction, but "-x + x" is not because it uses negation.

We would like to write an expression with the least number of operators such that the expression equals the given target.  Return the least number of expressions used.

Input: x = 3, target = 19
Output: 5
Explanation: 3 * 3 + 3 * 3 + 3 / 3.  The expression contains 5 operations.

Input: x = 5, target = 501
Output: 8
Explanation: 5 * 5 * 5 * 5 - 5 * 5 * 5 + 5 / 5.  The expression contains 8 operations.

Input: x = 100, target = 100000000
Output: 3
Explanation: 100 * 100 * 100 * 100.  The expression contains 3 operations.

Note:

• 2 <= x <= 100
• 1 <= target <= 2 * 10^8

Approach #1: C++.

Thinking....................................

weekly contest 116

原文：https://www.cnblogs.com/ruruozhenhao/p/10163954.html