Solution Explanation: Design a Stack with Increment Operation

This problem requires designing a stack data structure with the ability to perform increment operations on its elements. The solution uses an array-based approach to efficiently handle these operations. Let's break down the solution's logic and complexity analysis.

Approach: Array Simulation with Incremental Updates

The core idea is to simulate a stack using an array (stk) to store the elements and another array (add) to track cumulative increments for each position in the stack. Instead of directly modifying the stack elements during increment operations, we store the increment values in the add array. This allows us to efficiently calculate the actual value of an element when it's popped.

Data Structures:

• stk: An array to store the stack's elements.
• add: An array to store the cumulative increment values for each position. add[i] represents the total increment applied to elements at index i or below.
• i: An integer representing the current top of the stack (index of the next available position).

Operations:

• push(x): Adds x to the top of the stack if there's space (i < maxSize).
• pop(): If the stack is not empty, it retrieves the element at the top (stk[i-1] + add[i-1]), updates the add array (propagating increments downward), and returns the value.
• increment(k, val): Increments the bottom k elements by val. It adds val to add[min(i, k) - 1]. This effectively adds val to all elements at index min(i, k) - 1 and below.

Time and Space Complexity Analysis

• Time Complexity: All three operations (push, pop, increment) have a time complexity of O(1) in the average case. This is because they involve a constant number of array accesses and arithmetic operations regardless of the stack size.

• Space Complexity: The space complexity is O(maxSize), where maxSize is the maximum size of the stack. This is due to the use of two arrays, stk and add, both of size maxSize.

Code Examples (Python3 and Java)

The Python3 and Java code implementations efficiently implement the described approach. The other languages (C++, Go, TypeScript) follow a very similar structure.

Python3:

class CustomStack:
    def __init__(self, maxSize: int):
        self.stk = [0] * maxSize
        self.add = [0] * maxSize
        self.i = 0
 
    def push(self, x: int) -> None:
        if self.i < len(self.stk):
            self.stk[self.i] = x
            self.i += 1
 
    def pop(self) -> int:
        if self.i <= 0:
            return -1
        self.i -= 1
        ans = self.stk[self.i] + self.add[self.i]
        if self.i > 0:
            self.add[self.i - 1] += self.add[self.i]
        self.add[self.i] = 0
        return ans
 
    def increment(self, k: int, val: int) -> None:
        if self.i > 0:
            self.add[min(k, self.i) - 1] += val
 

Java:

class CustomStack {
    private int[] stk;
    private int[] add;
    private int i;
 
    public CustomStack(int maxSize) {
        stk = new int[maxSize];
        add = new int[maxSize];
    }
 
    public void push(int x) {
        if (i < stk.length) {
            stk[i++] = x;
        }
    }
 
    public int pop() {
        if (i <= 0) {
            return -1;
        }
        i--;
        int ans = stk[i] + add[i];
        if (i > 0) {
            add[i - 1] += add[i];
        }
        add[i] = 0;
        return ans;
    }
 
    public void increment(int k, int val) {
        if (i > 0) {
            add[Math.min(i, k) - 1] += val;
        }
    }
}

This detailed explanation, along with the code examples, provides a comprehensive understanding of the solution to the "Design a Stack with Increment Operation" problem. The efficient use of arrays and the clever handling of incremental updates contribute to the optimal time and space complexity of the solution.