Given a stack of N numbers and an array of numbers. Count the numbers of pop operations required to get each element of the array. Once an element is popped then its not pushed back again. Assume that the all the elements from the array present inside the stack initially.
Input: N = 5
Stack: 6 4 3 2 1
Array: 6 3 4 1 2
Output: 1 2 0 2 0
The 1st element of the stack is the same as the array elements. So to get 6, one pop is required i.e. pop 6 from the stack. To get 3, 2 elements will be popped i.e. 4 and 3. To get 4, 4 was already popped, thus we won’t pop more elements. Similarly, to get 1, we will pop 2 elements and for the 2, no pop count will be added.
Approach: This question can be solved easily by using a stack. We will keep popping the elements till we find the element we are searching. The only hurdle is that how to handle the case when the element is already popped and is not present in the stack. For that, we will maintain a hash map. As we pop an element from the stack we will insert that element in the hash map so that if the element comes later in the array we will first check if its present inside the hash maps or in other words has been popped out from the stack previously. Otherwise, we will know it’s present inside the stack and we will start popping the elements till we find the required number.
Below is the implementation of above approach:
1 2 0 2 0
Time Complexity: O(N)
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