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Validate Stack Sequences in C++
Suppose we have two sequences pushed and popped with distinct values, we have to find true if and only if this could have been the result of a sequence of the push and pop operations on an initially empty stack. So if the input is push = [1,2,3,4,5], and pop = [4,5,3,2,1], then the output will be true. We can use push(1), push(2), push(3), push(4), pop() : 4, push(5), pop() : 5, pop() : 3, pop() : 2, pop() : 1
To solve this, we will follow these steps −
Create one method called solve(). This will take pushed and popped arrays
define a stack st, set index := 0
for i in range 0 to size of pushed array
push pushed[i] into st
if popped[index] = stack top element, then
index := index + 1
pop from stack
while st is not empty, and popped[index] = top of st
index := index + 1
delete from st
while index < size of popped
if popped[index] = stack top, then
increase index by 1
delete from stack
otherwise come out from the loop
return true, when stack is empty
this solve method will be called from the main section like below −
return solve(pushed, popped)
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool solve(vector<int>& pushed, vector<int>& popped){
stack <int> st;
int currentIndexOfPopped = 0;
for(int i =0;i<pushed.size();i++){
st.push(pushed[i]);
if(popped[currentIndexOfPopped] == st.top()){
currentIndexOfPopped++;
st.pop();
while(!st.empty() && popped[currentIndexOfPopped]==st.top()){
currentIndexOfPopped++;
st.pop();
}
}
}
while(currentIndexOfPopped <popped.size()){
if (popped[currentIndexOfPopped]==st.top()){
currentIndexOfPopped++;
st.pop();
}else{
break;
}
}
return st.empty();
}
bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
Solution s;
bool flag = s.solve(pushed, popped);
return flag;
}
};
main(){
vector<int> v = {1,2,3,4,5};
vector<int> v1 = {4,5,3,2,1};
Solution ob;
cout << (ob.validateStackSequences(v, v1));
}Input
[1,2,3,4,5] [4,5,3,2,1]
Output
1
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