Is there a test case where this code can break? - c++

Question

This is the problem I am trying to solve.
Given two arrays a and b, check whether they are similar.
Two arrays are called similar if one can be obtained from another by swapping at most one pair of elements in one of the arrays.
I wrote the following code to solve the problem. I tried to make some optimizations but it fails on a hidden test. Can anyone help me find where the code could break?
bool is_similar(vector<int> a, vector<int> b, int start){
return memcmp(a.data() + start, b.data() + start, (a.size() - start) * sizeof(int)) == 0;
}
bool solution(vector<int> a, vector<int> b) {
int sizea = a.size(), sizeb = b.size();
if(sizea != sizeb) return false;
for(int i = 0; i < sizea; i++){
if(a[i] != b[i]){
auto it = find(b.begin() + i + 1, b.end(), a[i]);
if(it != b.end()){
swap(b[i], *it);
return is_similar(a, b, i + 1);
}
return false;
}
}
return true;
}

Answer 1

I started from scratch again and came up with a much better solution. Thank you all for your input. I'm loving this platform
The main problem was assuming the inputs were unique. This is the general flow of the initial code. A good test case to demonstrate the fault is when a = [0,1,1,3,4], b = [0,4,1,3,1]
Loop for the length of one of the arrays (they are of equal length)
Do nothing if the values are equal at the same index
If the values are not equal (now we are at index 1), search for 1 in array b and swap it with the current index. b becomes [0,1,4,3,1]
Now if we go on and call is_similar(a,b), it returns false.
Clearly if we swapped the contents at index 1 and 4, in array b, the arrays would be similar. That is why it fails.
A better approach is to collect the values at the indices where the values are different and determine if one swap is enough to make the two arrays similar. This is the code snippet.
bool solution(vector<int> a, vector<int> b) {
int sizea = a.size(), sizeb = b.size();
if(sizea != sizeb) return false;
int i= 0;
vector<int> a1;
vector<int> b1;
for(i = 0; i < sizea; i++){
if(a[i] != b[i]){
a1.emplace_back(a[i]);
b1.emplace_back(b[i]);
}
}
if(a1.size() == 0) return true;
if(a1.size() == 2) return a1[0] == b1[1] && a1[1] == b1[0];
return false;
}