Introduction

In this blog post, we will explore how to solve the LeetCode problem "895" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Design a stack-like data structure to push elements to the stack and pop the most frequent element from the stack. Implement the FreqStack class: FreqStack() constructs an empty frequency stack. void push(int val) pushes an integer val onto the top of the stack. int pop() removes and returns the most frequent element in the stack. If there is a tie for the most frequent element, the element closest to the stack's top is removed and returned. Example 1: Input ["FreqStack", "push", "push", "push", "push", "push", "push", "pop", "pop", "pop", "pop"] [[], [5], [7], [5], [7], [4], [5], [], [], [], []] Output [null, null, null, null, null, null, null, 5, 7, 5, 4] Explanation FreqStack freqStack = new FreqStack(); freqStack.push(5); // The stack is [5] freqStack.push(7); // The stack is [5,7] freqStack.push(5); // The stack is [5,7,5] freqStack.push(7); // The stack is [5,7,5,7] freqStack.push(4); // The stack is [5,7,5,7,4] freqStack.push(5); // The stack is [5,7,5,7,4,5] freqStack.pop(); // return 5, as 5 is the most frequent. The stack becomes [5,7,5,7,4]. freqStack.pop(); // return 7, as 5 and 7 is the most frequent, but 7 is closest to the top. The stack becomes [5,7,5,4]. freqStack.pop(); // return 5, as 5 is the most frequent. The stack becomes [5,7,4]. freqStack.pop(); // return 4, as 4, 5 and 7 is the most frequent, but 4 is closest to the top. The stack becomes [5,7]. Constraints: 0 <= val <= 109 At most 2 * 104 calls will be made to push and pop. It is guaranteed that there will be at least one element in the stack before calling pop.

Explanation

Here's a breakdown of the approach, complexities, and the Python code for the FreqStack problem:

• Key Idea: Maintain a frequency map to track element counts and a stack of stacks to organize elements by frequency levels. The stack at each frequency level stores elements in the order they were pushed.
• Pop Operation: Pop from the highest frequency stack. Update the frequency map.

• Push Operation: Increment the frequency of element. Create a new stack if element's frequency is higher than the maximum frequency observed so far.

• Complexity:

  • Runtime: O(1) for both push and pop operations on average.
  • Storage: O(N) where N is the number of push operations.

Code

    class FreqStack:
    def __init__(self):
        self.freq = {}  # Frequency of each element
        self.group = {} # group[freq] = stack of elements with frequency freq
        self.maxfreq = 0 # Maximum frequency observed so far

    def push(self, val):
        self.freq[val] = self.freq.get(val, 0) + 1
        f = self.freq[val]

        if f > self.maxfreq:
            self.maxfreq = f
            self.group[f] = []

        self.group[f].append(val)

    def pop(self):
        x = self.group[self.maxfreq].pop()
        self.freq[x] -= 1
        if not self.group[self.maxfreq]:
            self.maxfreq -= 1

        return x