# C++ Program to check if two stacks of letters can be emptied or not

Suppose, there are 2n number of letters and each of them has an integer number between 1 to n written on them. There are exactly two letters that have the same number written on them. These letters are arranged into m stacks and stack i has letters stack[i] on it. Our task is to empty all the stacks in the following manner

• We have to choose any two stacks and remove the top letter from both of them.

• The letters that we have removed must have the same number on both of them.

If we can empty the m stacks in this manner, we print true or otherwise we return false.

So, if the input is like n = 3, m = 2, stacks = {{2, 1, 3}, {2, 1, 3}}, then the output will be true.

There are two stacks and each of the stacks has letters that have the numbers 2, 1, 3 written on them respectively. So, we can remove from the two stacks and empty them in the given manner.

To solve this, we will follow these steps −

Define one 2D array dp
Define an array tvec
for initialize i := 0, when i < m, update (increase i by 1), do:
k := size of stacks[i]
for initialize j := 0, when j < k, update (increase j by 1), do:
if j < 0, then:
insert p at the end of dp[stacks[i, j]]
(increase tvec[p] by 1)
p := stacks[i, j]
Define an array tp
for initialize i := 1, when i <= n, update (increase i by 1), do:
Define one queue q
insert i into q
while (q is not empty), do:
if not tp[first element of q] and tvec[first element of q] is same as 0, then:
for each element next in dp[first element of q], do:
(decrease tvec[next] by 1)
insert next into q
tp[first element of q] := true
delete first element from q
for initialize i := 1, when i <= n, update (increase i by 1), do:
if tvec[i] is not equal to 0, then:
return false
return true

## Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;

bool solve(int n, int m, vector<vector<int>> stacks){
vector<vector<int>> dp(n + 1);
vector<int> tvec(n + 1);
for(int i = 0; i < m; i++){
int k = stacks[i].size();
int p;
for(int j = 0; j < k; j++){
if(j > 0){
dp[stacks[i][j]].push_back(p);
tvec[p]++;
}
p = stacks[i][j];
}
}
vector<bool> tp(n + 1);
for(int i = 1; i <= n; i++){
queue<int> q;
q.push(i);
while(!q.empty()){
if(!tp[q.front()] && tvec[q.front()] == 0){
for(auto next: dp[q.front()]){
tvec[next]--;
q.push(next);
}
tp[q.front()]=true;
}
q.pop();
}
}
for(int i = 1; i <= n; i++){
if(tvec[i] != 0){
return false;
}
}
return true;
}
int main() {
int n = 3, m = 2;
vector<vector<int>> stacks = {{2, 1, 3}, {2, 1, 3}};
cout<< solve(n, m, stacks);
return 0;
}

## Input

3, 2, {{2, 1, 3}, {2, 1, 3}}

## Output

1