leetcode 1187. Make Array Strictly Increasing

Given two integer arrays arr1 and arr2, return the minimum number of operations (possibly zero) needed to make arr1 strictly increasing.

In one operation, you can choose two indices 0 <= i < arr1.length and 0 <= j < arr2.length and do the assignment arr1[i] = arr2[j].

If there is no way to make arr1 strictly increasing, return -1.

 

Example 1:

Input: arr1 = [1,5,3,6,7], arr2 = [1,3,2,4]
Output: 1
Explanation: Replace 5 with 2, then arr1 = [1, 2, 3, 6, 7].

Example 2:

Input: arr1 = [1,5,3,6,7], arr2 = [4,3,1]
Output: 2
Explanation: Replace 5 with 3 and then replace 3 with 4. arr1 = [1, 3, 4, 6, 7].

Example 3:

Input: arr1 = [1,5,3,6,7], arr2 = [1,6,3,3]
Output: -1
Explanation: You can't make arr1 strictly increasing.

 

Constraints:

• 1 <= arr1.length, arr2.length <= 2000
• 0 <= arr1[i], arr2[i] <= 10^9

解题思路：

对于arr1的第 i - 1 位，这位可以改，也可以不改。但是都要保证前i- 1位严格递增 ；

设置了一个vector> cnt_max ；

cnt_max[i]中存的是<修改的次数 ， 修改后的最大值> 。

第i位，遍历cnt_max[i - 1]中的每对元素

for(auto ele : cnt_max[i - 1])

{

         //不改

         if(arr1[i - 1] > ele.second) //可以不改；

         //要改

         auto it = arr2_num.upper_bound(ele.second) ;

         if(it != arr2_num.end()）//在arr2_num中可以找到一个比ele.second大的数，可以改

}

如果cnt_max[i].empty()，就直接返回-1 ；

最后返回cnt_max[arr1.size()].begin()->first ;//这个是判断arr1[arr1.size()-1]位改还是不改后的最少修改次数 ；

class Solution {
public:
    int makeArrayIncreasing(vector& arr1, vector& arr2) 
    {
        int n1 = arr1.size() ;
        vector> cnt_max(n1 + 1) ;
        cnt_max[0][0] = -1 ;
        set arr2_num ;
        for(int i = 0 ; i < arr2.size() ; ++i)
            arr2_num.insert(arr2[i]) ;
        for(int i = 1 ; i <= n1 ; ++i)
        {
            for(auto ele : cnt_max[i - 1])
            {
                if(arr1[i - 1] > ele.second)
                {
                    if(cnt_max[i].count(ele.first))
                    {
                        cnt_max[i][ele.first] = min(cnt_max[i][ele.first] , arr1[i - 1]) ;
                    }
                    else
                    {
                        cnt_max[i][ele.first] = arr1[i - 1] ;
                    }
                }
                auto it = arr2_num.upper_bound(ele.second) ;
                if(it != arr2_num.end())
                {
                    if(cnt_max[i].count(ele.first + 1))
                    {
                        cnt_max[i][ele.first + 1] = min(cnt_max[i][ele.first + 1] , *it) ;
                    }
                    else
                    {
                        cnt_max[i][ele.first + 1] = *it ;
                    }
                }
            }
            if(cnt_max[i].empty()) return -1 ;
        }
        
        return cnt_max[n1].begin()->first ;
    }
};