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Q. Explain that how do we implement two stacks in one array A[1..n] in such a way that neither the stack overflows unless the total number of elements in both stacks together is n. The PUSH and POP operations should be running in O(1) time.
Ans:
Two stacks s1 and s2 can be implemented in one array A[1,2,...,N] as shown in the following figure
1 2 3 4 n-3 n-2 n-1 n S1 S2 Here we define A[1] as the bottom of stack S1 and let S1 "grow" to the right and we define A[n] as the bottom of the stack S2 and S2 "grow" to the left. In this particular case, overflow will occur, only S1 and S2 together have more than n elements. This technique or method will usually decrease the number of times overflow occurs. There will be separate push1, push2, pop1 and pop2 functions which are to be defined separately for the two stacks S1 and S2.
1 2 3 4 n-3 n-2 n-1 n
S1 S2
Here we define A[1] as the bottom of stack S1 and let S1 "grow" to the right and we define A[n] as the bottom of the stack S2 and S2 "grow" to the left. In this particular case, overflow will occur, only S1 and S2 together have more than n elements. This technique or method will usually decrease the number of times overflow occurs. There will be separate push1, push2, pop1 and pop2 functions which are to be defined separately for the two stacks S1 and S2.
* Initialise d & pi* for each vertex v within V( g ) g.d[v] := infinity g.pi[v] := nil g.d[s] := 0; * Set S to empty * S := { 0 } Q := V(g) * While (V-S)
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