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.
Example 1:
Input: [1,2,3,3]
Output: 3
Example 2:
Input: [2,1,2,5,3,2]
Output: 2
Example 3:
Input: [5,1,5,2,5,3,5,4]
Output: 5
Note:
4 <= A.length <= 100000 <= A[i] < 10000A.lengthis even
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.
Example 2:
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.
Note:
2 <= A.length <= 500000 <= A[i] <= 50000
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.
Example 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.
Example 3:
Input: [[0,3],[1,2],[3,1],[1,3],[2,1]]
Output: 0
Explanation: There is no possible rectangle to form from these points.
Example 4:
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 <= points.length <= 500 <= points[i][0] <= 400000 <= points[i][1] <= 40000- All points are distinct.
- Answers within
10^-5of the actual value will be accepted as correct.
Approach #1: C++.
Thinking..................................
Approach #2: Java.
import java.awt.Point;
class Solution {
public double minAreaFreeRect(int[][] points) {
int N = points.length;
Point[] A = new Point[N];
for (int i = 0; i < N; ++i)
A[i] = new Point(points[i][0], points[i][1]);
Map<Integer, Map<Point, List<Point>>> seen = new HashMap();
for (int i = 0; i < N; ++i)
for (int j = i+1; j < N; ++j) {
// center is twice actual to keep integer precision
Point center = new Point(A[i].x + A[j].x, A[i].y + A[j].y);
int r2 = (A[i].x - A[j].x) * (A[i].x - A[j].x);
r2 += (A[i].y - A[j].y) * (A[i].y - A[j].y);
if (!seen.containsKey(r2))
seen.put(r2, new HashMap<Point, List<Point>>());
if (!seen.get(r2).containsKey(center))
seen.get(r2).put(center, new ArrayList<Point>());
seen.get(r2).get(center).add(A[i]);
}
double ans = 0.0;
for (Map<Point, List<Point>> info: seen.values()) {
for (Point center: info.keySet()) { // center is twice actual
List<Point> candidates = info.get(center);
int clen = candidates.size();
for (int i = 0; i < clen; ++i)
for (int j = i+1; j < clen; ++j) {
Point P = candidates.get(i);
Point Q = candidates.get(j);
Point Q2 = new Point(center);
Q2.translate(-Q.x, -Q.y);
double area = P.distance(Q) * P.distance(Q2);
if (area > ans)
ans = area;
}
}
}
return ans;
}
}
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:
- The division operator (
/) returns rational numbers. - There are no parentheses placed anywhere.
- We use the usual order of operations: multiplication and division happens before addition and subtraction.
- 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.
Example 1:
Input: x = 3, target = 19
Output: 5
Explanation: 3 * 3 + 3 * 3 + 3 / 3. The expression contains 5 operations.
Example 2:
Input: x = 5, target = 501
Output: 8
Explanation: 5 * 5 * 5 * 5 - 5 * 5 * 5 + 5 / 5. The expression contains 8 operations.
Example 3:
Input: x = 100, target = 100000000
Output: 3
Explanation: 100 * 100 * 100 * 100. The expression contains 3 operations.
Note:
2 <= x <= 1001 <= target <= 2 * 10^8
Approach #1: C++.
Thinking....................................
Approach #2: Python. (can't understand)
class Solution(object):
def leastOpsExpressTarget(self, x, target):
"""
:type x: int
:type target: int
:rtype: int
"""
# 由于不能有括号，所以每一位（x进制）必须是由自己组成或者由下一位减自己余数组成,所以首先短除法求出每一位的余数
rl = list()
while target:
rl.append(target%x)
target/=x
n = len(rl)
# 在个位的时候，必须用x/x表示1，所以*2，但其它位不用，所以单独先提出
pos = rl[0] * 2
neg = (x-rl[0]) * 2
for i in range(1,n):
# 正数表示时，可用自己+剩下的正数表示或者多加一个本位然后减去上一位的余数表示
# 负数表示时，可用自己的余数加上剩下的正数表示或者用自己的余数+剩下的余数，但此时可以合并同级项减少运算符
# 如在10进制下，86可表示为
# 80 + 6 （6 为下一位正数表示
# 80 + 10 - 4 （4 为下一位负数表示）
# 100 - 20 + 6 （100-20为本位余数表示，6为下一位正数表示
# 100 - 20 + 10 - 4 （100-20为本位余数表示，10 -4 为下一位余数表示
# 在此时， 20 和 10注定为同一位且符号相反，可以省去两个符号（一个符号在该位用i 个符号表示（如100为第二位，所以表示为+10 * 10，用两个符号，在此时所有符号均带自己的正负号
pos, neg = min(rl[i] * i + pos, rl[i] * i + i + neg), min((x - rl[i]) * i + pos, (x-rl[i]) * i + i + neg - 2 * i)
# 因为在前面算法中所有位都带自己的正负号，所以第一位应该减去自己的符号，所以总量减1
# 或者在余数表示法中，加上一个更高位的减去自己表示本位余数
# 所以此题归根结底还是考察对进制的理解而不是简单的理解bfs, dfs，那样复杂度远远高于此，但是是对惯性思维者的一种挑战
return min(pos-1, n+neg-1)
come from: https://leetcode.com/problems/least-operators-to-express-number/discuss/208376/python2-O(log-target)-chinese