LeetCode 1172. Dinner Plate Stacks Solution in Java, C++, Python & More | Explanation + Code

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1172. Dinner Plate Stacks

Description

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 capacity.
  • void push(int val) Pushes the given integer val into the leftmost stack with a 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 the stacks are empty.
  • int popAtStack(int index) Returns the value at the top of the stack with the given index index and removes it from that stack or returns -1 if the stack with that given index is empty.

 

Example 1:

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 <= 2 * 104
  • 1 <= val <= 2 * 104
  • 0 <= index <= 105
  • At most 2 * 105 calls will be made to push, pop, and popAtStack.

Solutions

Solution 1: Stack Array + Ordered Set

We define the following data structures or variables:

  • capacity: The capacity of each stack;
  • stacks: Stack array, used to store all stacks, each with a maximum capacity of capacity;
  • not_full: Ordered set, used to store the indices of all non-full stacks in the stack array.

For the push(val) operation:

  • We first check if not_full is empty. If it is, it means there are no non-full stacks, so we need to create a new stack and push val into it. At this point, we check if the capacity capacity is greater than 1. If it is, we add the index of this stack to not_full.
  • If not_full is not empty, it means there are non-full stacks. We take out the smallest index index from not_full, and push val into stacks[index]. At this point, if the capacity of stacks[index] equals capacity, we remove index from not_full.

For the popAtStack(index) operation:

  • We first check if index is within the index range of stacks. If it is not, we directly return -1. If stacks[index] is empty, we also directly return -1.
  • If stacks[index] is not empty, we pop the top element val from stacks[index]. If index equals the length of stacks minus 1, it means stacks[index] is the last stack. If it is empty, we loop to remove the index of the last stack from not_full, and remove the last stack from the stack array stacks, until the last stack is not empty, or the stack array stacks is empty. Otherwise, if stacks[index] is not the last stack, we add index to not_full.
  • Finally, return val.

For the pop() operation:

  • We directly call popAtStack(stacks.length - 1).

The time complexity is (n × log n), and the space complexity is O(n). Here, n is the number of operations.

PythonJavaC++GoTypeScript
class DinnerPlates: def __init__(self, capacity: int): self.capacity = capacity self.stacks = [] self.not_full = SortedSet() def push(self, val: int) -> None: if not self.not_full: self.stacks.append([val]) if self.capacity > 1: self.not_full.add(len(self.stacks) - 1) else: index = self.not_full[0] self.stacks[index].append(val) if len(self.stacks[index]) == self.capacity: self.not_full.discard(index) def pop(self) -> int: return self.popAtStack(len(self.stacks) - 1) def popAtStack(self, index: int) -> int: if index < 0 or index >= len(self.stacks) or not self.stacks[index]: return -1 val = self.stacks[index].pop() if index == len(self.stacks) - 1 and not self.stacks[-1]: while self.stacks and not self.stacks[-1]: self.not_full.discard(len(self.stacks) - 1) self.stacks.pop() else: self.not_full.add(index) return val # Your DinnerPlates object will be instantiated and called as such: # obj = DinnerPlates(capacity) # obj.push(val) # param_2 = obj.pop() # param_3 = obj.popAtStack(index)(code-box)

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