Solving Leetcode Interviews in Seconds with AI: Find the Number of Winning Players
Introduction
In this blog post, we will explore how to solve the LeetCode problem "3238" 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
You are given an integer n representing the number of players in a game and a 2D array pick where pick[i] = [xi, yi] represents that the player xi picked a ball of color yi. Player i wins the game if they pick strictly more than i balls of the same color. In other words, Player 0 wins if they pick any ball. Player 1 wins if they pick at least two balls of the same color. ... Player i wins if they pick at leasti + 1 balls of the same color. Return the number of players who win the game. Note that multiple players can win the game. Example 1: Input: n = 4, pick = [[0,0],[1,0],[1,0],[2,1],[2,1],[2,0]] Output: 2 Explanation: Player 0 and player 1 win the game, while players 2 and 3 do not win. Example 2: Input: n = 5, pick = [[1,1],[1,2],[1,3],[1,4]] Output: 0 Explanation: No player wins the game. Example 3: Input: n = 5, pick = [[1,1],[2,4],[2,4],[2,4]] Output: 1 Explanation: Player 2 wins the game by picking 3 balls with color 4. Constraints: 2 <= n <= 10 1 <= pick.length <= 100 pick[i].length == 2 0 <= xi <= n - 1 0 <= yi <= 10
Explanation
- Count Color Frequencies per Player: Iterate through the
pickarray and store the frequency of each color picked by each player using a dictionary (or a similar data structure).- Check Winning Condition: For each player, iterate through the color frequencies and check if any color's frequency is greater than the player's index (i). If yes, the player wins.
- Count Winners: Keep track of the number of players who win and return the final count.
- Time Complexity: O(n*m), where n is the number of players and m is the number of picks. Space Complexity: O(n*k), where k is the number of distinct colors (at most the number of picks).
Code
def solve():
n = int(input())
pick = []
for _ in range(int(input())):
pick.append(list(map(int, input().split())))
wins = 0
for player in range(n):
color_counts = {}
for p in pick:
if p[0] == player:
color = p[1]
if color not in color_counts:
color_counts[color] = 0
color_counts[color] += 1
won = False
for color in color_counts:
if color_counts[color] > player:
won = True
break
if won:
wins += 1
print(wins)
# Taking input for n and pick array in one line (like in the examples) is not supported.
# The code below simulates the problem described with input using multiple lines.
def main():
n = int(input())
m = int(input())
picks = []
for _ in range(m):
picks.append(list(map(int, input().split())))
wins = 0
for player in range(n):
color_counts = {}
for p in picks:
if p[0] == player:
color = p[1]
if color not in color_counts:
color_counts[color] = 0
color_counts[color] += 1
won = False
for color in color_counts:
if color_counts[color] > player:
won = True
break
if won:
wins += 1
print(wins)
if __name__ == "__main__":
main()