problem1172

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1172. Dinner Plate Stacks

You have an infinite number of stacks arranged in a row and numbered (left to right) from 0, each of the stacks has the same maximum capacity.

Implement the DinnerPlates class:

• DinnerPlates(int capacity) Initializes the object with the maximum capacity of the stacks.
• void push(int val) pushes the given positive integer val into the leftmost stack with size less than capacity.
• int pop() returns the value at the top of the rightmost non-empty stack and removes it from that stack, and returns -1 if all stacks are empty.
• int popAtStack(int index) returns the value at the top of the stack with the given index and removes it from that stack, and returns -1 if the stack with that given index is empty.

Example:

Input:
["DinnerPlates","push","push","push","push","push","popAtStack","push","push","popAtStack","popAtStack","pop","pop","pop","pop","pop"]
[[2],[1],[2],[3],[4],[5],[0],[20],[21],[0],[2],[],[],[],[],[]]
Output:
[null,null,null,null,null,null,2,null,null,20,21,5,4,3,1,-1]

Explanation:
DinnerPlates D = DinnerPlates(2);  // Initialize with capacity = 2
D.push(1);
D.push(2);
D.push(3);
D.push(4);
D.push(5);         // The stacks are now:  2  4
                                           1  3  5
                                           ﹈ ﹈ ﹈
D.popAtStack(0);   // Returns 2.  The stacks are now:     4
                                                       1  3  5
                                                       ﹈ ﹈ ﹈
D.push(20);        // The stacks are now: 20  4
                                           1  3  5
                                           ﹈ ﹈ ﹈
D.push(21);        // The stacks are now: 20  4 21
                                           1  3  5
                                           ﹈ ﹈ ﹈
D.popAtStack(0);   // Returns 20.  The stacks are now:     4 21
                                                        1  3  5
                                                        ﹈ ﹈ ﹈
D.popAtStack(2);   // Returns 21.  The stacks are now:     4
                                                        1  3  5
                                                        ﹈ ﹈ ﹈
D.pop()            // Returns 5.  The stacks are now:      4
                                                        1  3
                                                        ﹈ ﹈
D.pop()            // Returns 4.  The stacks are now:   1  3
                                                        ﹈ ﹈
D.pop()            // Returns 3.  The stacks are now:   1
                                                        ﹈
D.pop()            // Returns 1.  There are no stacks.
D.pop()            // Returns -1.  There are still no stacks.

Constraints:

• 1 <= capacity <= 20000
• 1 <= val <= 20000
• 0 <= index <= 100000
• At most 200000 calls will be made to push, pop, and popAtStack.

Documentation

Index

• type DinnerPlates
  • func Constructor(capacity int) DinnerPlates
  • func (d *DinnerPlates) Pop() int
  • func (d *DinnerPlates) PopAtStack(i int) (res int)
  • func (d *DinnerPlates) Push(val int)

Types

type DinnerPlates

type DinnerPlates struct {
	// contains filtered or unexported fields
}

DinnerPlates is ...

func Constructor

func Constructor(capacity int) DinnerPlates

Constructor is ...

func (*DinnerPlates) Pop

func (d *DinnerPlates) Pop() int

Pop is a special condition of PopAtStack

func (*DinnerPlates) PopAtStack

func (d *DinnerPlates) PopAtStack(i int) (res int)

PopAtStack is ...

func (*DinnerPlates) Push

func (d *DinnerPlates) Push(val int)

Push is ...