#1172 Leetcode Dinner Plate Stacks Solution in C, C++, Java, JavaScript, Python, C# Leetcode

Algorithm

Problem Name: 1172. Dinner Plate Stacks

Problem Link: https://leetcode.com/problems/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 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.

Code Examples

#1 Code Example with Javascript Programming

Code - Javascript Programming


const DinnerPlates = function (capacity) {
  this.pushIndex = 0
  this.popIndex = 0
  this.capacity = capacity
  this.stacks = [[]]
}

DinnerPlates.prototype.push = function (val) {
  while (
    this.pushIndex < this.stacks.length &&
    this.stacks[this.pushIndex].length === this.capacity
  ) {
    this.pushIndex++
  }
  if (this.stacks.length === this.pushIndex) {
    this.stacks[this.pushIndex] = [val]
  } else {
    this.stacks[this.pushIndex].push(val)
  }
  if (this.popIndex < this.pushIndex) {
    this.popIndex = this.pushIndex
  }
}

DinnerPlates.prototype.pop = function () {
  while (this.stacks[this.popIndex].length === 0) {
    if (this.popIndex > 0) {
      this.popIndex--
    } else {
      return -1
    }
  }
  const valueAtIndex = this.stacks[this.popIndex].pop()
  if (this.pushIndex > this.popIndex) {
    this.pushIndex = this.popIndex
  }
  return valueAtIndex
}

DinnerPlates.prototype.popAtStack = function (index) {
  if (index >= this.stacks.length) return -1
  if (index < this.pushIndex) this.pushIndex = index
  return this.stacks[index].length > 0 ? this.stacks[index].pop() : -1
}

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Input

–

cmd

["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

–

cmd

[null, null, null, null, null, null, 2, null, null, 20, 21, 5, 4, 3, 1, -1]

#2 Code Example with Python Programming

Code - Python Programming


class DinnerPlates:

    def __init__(self, capacity: int):
        self.c = capacity
        self.stacks = []
        self.l = [] # pushable
        self.r = [] # poppable
        
    def push(self, val: int) -> None:
        if not self.l:
            # if the rightmost is empty, clear it from self.r
            while self.r and not self.stacks[-self.r[0]]:
                heapq.heappop(self.r)
            i = 0 if not self.r else -self.r[0] + 1
            self.stacks.append([val])
            heapq.heappush(self.r, -i)
            if len(self.stacks[i]) < self.c:
                heapq.heappush(self.l, i)
        else:
            i = self.l[0]
            # new poppable
            if not self.stacks[i]:
                heapq.heappush(self.r, -i)
            self.stacks[i].append(val)
            if len(self.stacks[i]) == self.c:
                heapq.heappop(self.l)
                
    def pop(self) -> int:
        while self.r and not self.stacks[-self.r[0]]:
            heapq.heappop(self.r)
        return self.popAtStack(-self.r[0]) if self.r else -1

    def popAtStack(self, index: int) -> int:
        if index < len(self.stacks) and self.stacks[index]:
            ret = self.stacks[index].pop()
            if len(self.stacks[index]) == self.c - 1:
                heapq.heappush(self.l, index)
            return ret
        return -1

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Input

–

cmd

Output

–

cmd

[null, null, null, null, null, null, 2, null, null, 20, 21, 5, 4, 3, 1, -1]

Demonstration

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