# CSPT29 Lecture 2
## Number of Good Pairs

class Solution:
    """
    nums = [1,2,3,1,1,3]
    output = 4
    
    nums = [1,1,1]
    output = 3
    
    nums = [1,2,3]
    output = 0
    
    Plan
    Brute-force - generate all possible pairs and count which ones
    are good
    
    nums = [1,2,3,1,1,3]
    numGoodPairs = 4
    """
    def numIdenticalPairs(self, nums: List[int]) -> int:
        numGoodPairs = 0
        for i in range(len(nums)):
            for j in range(i + 1, len(nums)):
                if nums[i] == nums[j]:
                    numGoodPairs += 1
        return numGoodPairs

## Max Wealth

class Solution:
    """
    accounts = [[1,2,3],
                [3,2,1]]
    output = 6
    
    accounts = [[1,2,3],
                [3,2,10]]
    output = 15
    
    accounts = [[1,2,3]]
    output = 6
    
    Plan
    Look at every single customer
    While doing that, keep track of max wealth found from a customer
    Return highest wealth found
    """
    def maximumWealth(self, accounts: List[List[int]]) -> int:
        maxWealth = 0
        for customer in accounts:
            currCustomerWealth = 0
            for bankAccount in customer:
                currCustomerWealth += bankAccount
            if currCustomerWealth > maxWealth:
                maxWealth = currCustomerWealth
        return maxWealth

## N-repeated Element

class Solution:
    """
    nums.length = 2 * n
    n = nums.length / 2
    
    nums = [1,2,3,3]
    output = 3
    
    nums = [2,1,2,5,3,2]
    output = 2
    
    Plan
    You need to go through all elements at least once
    Use a dictionary that maps from element --> num times occuered
    n = nums.length / 2
    Go through the dictionary one more time and return
    the key that occurred n times
    """
    def repeatedNTimes(self, nums: List[int]) -> int:
        counts = {}
        for num in nums:
            if num not in counts:
                counts[num] = 1
            else:
                counts[num] += 1
        n = len(nums) / 2
        for num, numTimesOccurred in counts.items():
            if numTimesOccurred == n:
                return num