# 44ms intuitive C++ solution

• Not a good solution but intuitive.

First elimination 0,1,2,3.....numbers from each array and store them in a[na] ,a[na-1],a[na-2].....(na is the size of nums1). Then do the same thing for second array.

After that, we could test combination of the two arrays of sizes from [0,k], [1,(k-1)], [2+(k-2)].....

When combining the two arrays, we choose the larger one at the beginning of two arrays and then 'delete' it. If the two numbers are equal, we compare the number behind them to find the one with a larger 'potential'.

Code:

``````class Solution {
public:
vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {

int na=nums1.size();
int nb=nums2.size();
vector<int> empty;
if(na+nb<k)return empty;
vector<vector<int>> a(na+1,empty);
vector<vector<int>> b(nb+1,empty);
a[na]=nums1;b[nb]=nums2;
int f=0;
for(int i=na-1;i>=0;i--){
a[i]=a[i+1];f=0;
for(int j=0;j<i;j++)if(a[i][j]<a[i][j+1]){a[i].erase(a[i].begin()+j);f=1;break;}
if(f==0)a[i].erase(a[i].begin()+i);
}

for(int i=nb-1;i>=0;i--){
b[i]=b[i+1];f=0;
for(int j=0;j<i;j++)if(b[i][j]<b[i][j+1]){b[i].erase(b[i].begin()+j);f=1;break;}
if(f==0)b[i].erase(b[i].begin()+i);
}
vector<int> max(k,0);
vector<int> tmp(k,0);
int at,bt;
for(int i=k-nb;i<=na;i++){
int j=k-i;
at=0;bt=0;
f=0;
while(at+bt<k){
if(at>=i){tmp[at+bt]=b[j][bt];bt++;}else{
if(bt>=j){tmp[at+bt]=a[i][at];at++;}else{
if(a[i][at]==b[j][bt]){
int aat=at,bbt=bt;
while(aat<i&&bbt<j&&a[i][aat]==b[j][bbt]){aat++;bbt++;}
if(aat>=i&&bbt>=j){tmp[at+bt]=a[i][at];at++;}else{
if(aat>=i){tmp[at+bt]=b[j][bt];bt++;}else{
if(bbt>=j){tmp[at+bt]=a[i][at];at++;}else{
if(a[i][aat]>b[j][bbt]){tmp[at+bt]=a[i][at];at++;} else {tmp[at+bt]=b[j][bt];bt++;}
}}}

}else{
if(a[i][at]>b[j][bt]){tmp[at+bt]=a[i][at];at++;} else {tmp[at+bt]=b[j][bt];bt++;}
}}}
if(f==0&&tmp[at+bt-1]>max[at+bt-1]){f=1;}
if(f==0&&tmp[at+bt-1]<max[at+bt-1]){break;}
if(f==1)max[at+bt-1]=tmp[at+bt-1];
}
}
return max;
}
};``````

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