1237. Find Positive Integer Solution for a Given Equation*

Posted 2022-08-01 珍妮的选择

1237. Find Positive Integer Solution for a Given Equation*

​​https://leetcode.com/problems/find-positive-integer-solution-for-a-given-equation/​​

题目描述

Given a function ​​f(x, y)​​​ and a value ​​z​​​, return all positive integer pairs ​​x​​​ and ​​y​​​ where ​​f(x,y) == z​​.

The function is constantly increasing, i.e.:

• ​​f(x, y) < f(x + 1, y)​​
• ​​f(x, y) < f(x, y + 1)​​

The function interface is defined like this:

interface CustomFunction 
public:
  // Returns positive integer f(x, y) for any given positive integer x and y.
  int f(int x, int y);
;

For custom testing purposes you’re given an integer ​​function_id​​​ and a target ​​z​​​ as input, where ​​function_id​​ represent one function from an secret internal list, on the examples you’ll know only two functions from the list.

You may return the solutions in any order.

Example 1:

Input: function_id = 1, z = 5
Output: [[1,4],[2,3],[3,2],[4,1]]
Explanation: function_id = 1 means that f(x, y) =

Example 2:

Input: function_id = 2, z = 5
Output: [[1,5],[5,1]]
Explanation: function_id = 2 means that f(x, y) =

Constraints:

• ​​1 <= function_id <= 9​​
• ​​1 <= z <= 100​​
• It’s guaranteed that the solutions of​​f(x, y) == z​​​ will be on the range​​1 <= x, y <= 1000​​
• It’s also guaranteed that​​f(x, y)​​​ will fit in 32 bit signed integer if​​1 <= x, y <= 1000​​

C++ 实现 1

这道题的目的是考察二分法. 我一开始没有考虑到… 用的暴力的方法, beats 100%.

题目描述中, 说 ​​f(x, y)​​​ 具体的形式未知, 只知道它递增. 另外, 注意约束中 ​​1 <= x, y <= 1000​​, 那么写两个 for 循环一个值一个值尝试即可.

/*
 * // This is the custom function interface.
 * // You should not implement it, or speculate about its implementation
 * class CustomFunction 
 * public:
 *     // Returns f(x, y) for any given positive integers x and y.
 *     // Note that f(x, y) is increasing with respect to both x and y.
 *     // i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
 *     int f(int x, int y);
 * ;
 */

class Solution 
public:
    vector<vector<int>> findSolution(CustomFunction& customfunction, int z) 
        vector<vector<int>> res;
        for (int x = 1; x <= 1000; ++ x) 
            for (int y = 1; y <= 1000; ++ y) 
                if (customfunction.f(x, y) > z) break;
                if (customfunction.f(x, y) == z) res.push_back(x, y);
            
        
        return res;
    
;

C++ 实现 2

实际上, 本题的目的是考察二分法. 因为题目中有递增的字眼, 要对这样的条件敏感一些.

来自 LeetCode Submission.

/*
 * // This is the custom function interface.
 * // You should not implement it, or speculate about its implementation
 * class CustomFunction 
 * public:
 *     // Returns f(x, y) for any given positive integers x and y.
 *     // Note that f(x, y) is increasing with respect to both x and y.
 *     // i.e. f(x, y) < f(x + 1, y), f(x, y) < f(x, y + 1)
 *     int f(int x, int y);
 * ;
 */
/*
    1. for all value of x, apply binary search to find the value of y
    2. Time Complexity : O(x logy)
*/
class Solution 
public:
    vector<vector<int>> findSolution(CustomFunction& cf, int z) 
        vector<vector<int>> res;
        for( int x = 1; x <= 1000; ++x ) 
            int lo = 1, hi = 1000;
            while( lo < hi ) 
                int mid = lo + ( hi - lo )/2;
                if( z > cf.f(x, mid) )
                    lo = mid + 1;
                else
                    hi = mid;
            
            if( cf.f(x, lo) == z )
                res.push_back(x, lo);
        
        return res;
    
;