How do you implement 2 stacks using one array.

# Implementing 2 stacks in one array?

**Mukul_d**#2

Algorithm:

- Start with two indexes, one at the left end and other at the right end
- The left index simulates the first stack and the right index simulates the second stack.
- If we want to push an element into the first stack then put the element at left index.
- Similarly, if we want to push an element into the second stack then put the element at the right index.
- First stack grows towards the right and the second stack grows towards left.

**Can we extend to three ? like 3 stack using one array ?**

yes we can do have to jsut use 3 way partitioning but for k arrays there is other algorithm which is 0(n) complexity in space along with 0(1) in pusha n pop

**satish**#4

Method 1 (Divide the space in two halves)

A simple way to implement two stacks is to divide the array in two halves and assign the half half space to two stacks, i.e., use arr[0] to arr[n/2] for stack1, and arr[n/2+1] to arr[n-1] for stack2 where arr[] is the array to be used to implement two stacks and size of array be n.

The problem with this method is inefficient use of array space. A stack push operation may result in stack overflow even if there is space available in arr[]. For example, say the array size is 6 and we push 3 elements to stack1 and do not push anything to second stack2. When we push 4th element to stack1, there will be overflow even if we have space for 3 more elements in array.

Method 2 (A space efficient implementation)

This method efficiently utilizes the available space. It doesnâ€™t cause an overflow if there is space available in arr[]. The idea is to start two stacks from two extreme corners of arr[]. stack1 starts from the leftmost element, the first element in stack1 is pushed at index 0. The stack2 starts from the rightmost corner, the first element in stack2 is pushed at index (n-1). Both stacks grow (or shrink) in opposite direction. To check for overflow, all we need to check is for space between top elements of both stacks. This check is highlighted in the below code.