Optimal Division – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
You are given an integer array nums. The adjacent integers in nums will perform the float division.
- For example, for
nums = [2,3,4], we will evaluate the expression"2/3/4".
However, you can add any number of parenthesis at any position to change the priority of operations. You want to add these parentheses such the value of the expression after the evaluation is maximum.
Return the corresponding expression that has the maximum value in string format.
Note: your expression should not contain redundant parenthesis.
Example 1:
Input: nums = [1000,100,10,2] Output: "1000/(100/10/2)" Explanation: 1000/(100/10/2) = 1000/((100/10)/2) = 200 However, the bold parenthesis in "1000/((100/10)/2)" are redundant since they do not influence the operation priority. So you should return "1000/(100/10/2)". Other cases: 1000/(100/10)/2 = 50 1000/(100/(10/2)) = 50 1000/100/10/2 = 0.5 1000/100/(10/2) = 2
Example 2:
Input: nums = [2,3,4] Output: "2/(3/4)" Explanation: (2/(3/4)) = 8/3 = 2.667 It can be shown that after trying all possibilities, we cannot get an expression with evaluation greater than 2.667
Constraints:
1 <= nums.length <= 102 <= nums[i] <= 1000- There is only one optimal division for the given input.
C++ Optimal DivisionLeetCode Solution
class Solution {
public:
string optimalDivision(vector<int>& nums) {
int n = nums.size();
string expr;
for (int i = 0; i < n; i++) {
if (i > 0) {
expr += "/";
}
if (i == 1 && n > 2) {
expr += "(";
}
expr += to_string(nums[i]);
if (i == n - 1 && n > 2) {
expr += ")";
}
}
return expr;
}
};
Java Optimal Division LeetCode Solution
public class Solution {
class Result {
String str;
double val;
}
public String optimalDivision(int[] nums) {
int len = nums.length;
return getMax(nums, 0, len - 1).str;
}
private Result getMax(int[] nums, int start, int end) {
Result r = new Result();
r.val = -1.0;
if (start == end) {
r.str = nums[start] + "";
r.val = (double)nums[start];
}
else if (start + 1 == end) {
r.str = nums[start] + "/" + nums[end];
r.val = (double)nums[start] / (double)nums[end];
}
else {
for (int i = start; i < end; i++) {
Result r1 = getMax(nums, start, i);
Result r2 = getMin(nums, i + 1, end);
if (r1.val / r2.val > r.val) {
r.str = r1.str + "/" + (end - i >= 2 ? "(" + r2.str + ")" : r2.str);
r.val = r1.val / r2.val;
}
}
}
//System.out.println("getMax " + start + " " + end + "->" + r.str + ":" + r.val);
return r;
}
private Result getMin(int[] nums, int start, int end) {
Result r = new Result();
r.val = Double.MAX_VALUE;
if (start == end) {
r.str = nums[start] + "";
r.val = (double)nums[start];
}
else if (start + 1 == end) {
r.str = nums[start] + "/" + nums[end];
r.val = (double)nums[start] / (double)nums[end];
}
else {
for (int i = start; i < end; i++) {
Result r1 = getMin(nums, start, i);
Result r2 = getMax(nums, i + 1, end);
if (r1.val / r2.val < r.val) {
r.str = r1.str + "/" + (end - i >= 2 ? "(" + r2.str + ")" : r2.str);
r.val = r1.val / r2.val;
}
}
}
//System.out.println("getMin " + start + " " + end + "->" + r.str + ":" + r.val);
return r;
}
}
Python 3 Optimal Division LeetCode Solution
def optimalDivision(self, A):
A = map(str, A)
if len(A) <= 2: return '/'.join(A)
return '{}/({})'.format(A[0], '/'.join(A[1:]))
Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1
