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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 trueExample
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
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