# CSPT19 Lecture 2

## [Richest Customer Wealth](https://leetcode.com/problems/richest-customer-wealth/)

class Solution:
    """
    Understand
    
    [[1,2,3],
     [3,2,1]]
    output: 6
    
    [[1,2,3]]
    output: 6
    
    [[1,2,3],
     [3,2,100]]
    output: 105
    
    Plan
    Go through each customer and find the max wealth seen so far
    """
    def maximumWealth(self, accounts: List[List[int]]) -> int:
        maxWealth = 0
        for customer in accounts:
            customerWealth = sum(customer)
            if customerWealth > maxWealth:
                maxWealth = customerWealth
        return maxWealth             

## [Number of Good Pairs](https://leetcode.com/problems/number-of-good-pairs/)

class Solution:
    """
    Understand
    nums = [1,2,3,1,1,3]
    output: 4
    
    nums = [1,2,3]
    output: 0
    
    nums = [0,0,0,0]
    output: 6
    
    nums = []
    output: 0
    
    Plan
    Keep track of numGoodPairs
    We can iterate through all possible pairs (using two for-loops).
    If a pair is a valid "good pair" then increment numGoodPairs
    return numGoodPairs
    """
    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

## [N-Repeated Element in Size 2N Array](https://leetcode.com/problems/n-repeated-element-in-size-2n-array/)

class Solution:
    """
    Understand
    
    [1,2,3,3] length = 4, N = 2
    output = 3
    
    [6,1,6,5,6,1] length = 6, N = 3
    output = 6
    
    Plan
    Go through entire list and keep track of how many times each
    element occurs using a dictionary (element -> numTimesOccurence)
    
    We iterate through the dictionary again and return the number that occurred
    N times
    """
    def repeatedNTimes(self, A: List[int]) -> int:
        counts = {}
        for num in A:
            if num not in counts:
                counts[num] = 1
            else:
                counts[num] += 1
        target = len(A) / 2
        for (num, timesAppeared) in counts.items():
            if timesAppeared == target:
                return num