Assume we desire to store multiple stacks using physical locations Min through Max. Stacks grow from left to right only in the diagram (non-circular implementation). A stack is considered empty anytime B[J]- T[J] (the base of queue J is equal to the top of queue J). Overflow occurs anytime T[J] > B[J+1]. For convenience, given N queues, we will utilize N + 1 base pointers with BIN + 1] = Max to detect overflow out of the last stack. As an initial condition, we always set B[1] = T[11-Min-1 for the first queue to maximize memory utilization where Min is the first location that can actually hold data develop an algorithm to provide equal initial space allocation for each of N stacks prior to application execution. Assume we wish to utilize 3 stacks. If the value of min is 22 and the value of Max is 104 calculate the initial base addresses allocating equal space to all 3 stacks using the formula developed in class. The sample diagram below shows 3 stacks. Stacks 1 and 2 are currently empty. Stack 3 is full and contains two items. work clearly show your Bob Tom Min-1 Min Min+1 o Max-2 Max-1 Max B2 B Assume we desire to store multiple stacks using physical locations Min through Max. Stacks grow from left to right only in the diagram (non-circular implementation). A stack is considered empty anytime B[J]- T[J] (the base of queue J is equal to the top of queue J). Overflow occurs anytime T[J] > B[J+1]. For convenience, given N queues, we will utilize N + 1 base pointers with BIN + 1] = Max to detect overflow out of the last stack. As an initial condition, we always set B[1] = T[11-Min-1 for the first queue to maximize memory utilization where Min is the first location that can actually hold data develop an algorithm to provide equal initial space allocation for each of N stacks prior to application execution. Assume we wish to utilize 3 stacks. If the value of min is 22 and the value of Max is 104 calculate the initial base addresses allocating equal space to all 3 stacks using the formula developed in class. The sample diagram below shows 3 stacks. Stacks 1 and 2 are currently empty. Stack 3 is full and contains two items. work clearly show your Bob Tom Min-1 Min Min+1 o Max-2 Max-1 Max B2 B