Design a Stack With Increment Operation | Better Approach | O(1) | Leetcode 1381 | codestorywithMIK
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In this video we will try to solve a good Design DSA Problem : Design a Stack With Increment Operation | Better Approach | O(1) Time | Leetcode 1381 | codestorywithMIK
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Problem Name : Design a Stack With Increment Operation | Better Approach | O(1) Time | Leetcode 1381 | codestorywithMIK
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My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Design/Design a Stack With Increment Operation.cpp
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Summary :
Initialization (CustomStack(int maxSize)):
The constructor initializes a maximum stack size (n) and two vectors:
st: Holds the elements in the stack.
increments: Keeps track of incremental values for each element, which will be added during the pop operation.
Push Operation (void push(int x)):
If the stack has not reached its maximum size (n), the new element x is pushed onto st, and a corresponding 0 is added to increments.
Pop Operation (int pop()):
The pop operation removes and returns the top element from the stack. If the stack is empty, it returns -1.
It adds the current top's increment value to the next-to-top element’s increment, effectively propagating the increment downwards.
Returns the value of the top element adjusted by its corresponding increment.
Increment Operation (void increment(int k, int val)):
This method adds val to the first k elements in the stack (or the entire stack if k is larger than the stack size).
The increment is added to the increments vector at index min(k-1, st.size()-1), meaning the effect is propagated when elements are popped.
Key Design Points:
The increments vector enables efficient management of the increment operation, as it avoids incrementing all elements individually and instead defers the update until a pop occurs.
The time complexity for push, pop, and increment operations is constant (O(1)), making this approach very efficient for stack operations with increments.
✨ Timelines✨
00:00 - Introduction
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