A. Find Closest Number to Zero (opens new window)

Iterate over all numbers, find the one with the smallest absolute value.

DETAILS

class Solution {
    public int findClosestNumber(int[] nums) {
        int ans = -1000000, dist = 1000000;
        for (int num : nums) {
            if (Math.abs(num) < dist) {
                ans = num;
                dist = Math.abs(num);
            } else if (Math.abs(num) == dist) {
                ans = Math.max(num, ans);
            }
        }
        return ans;
    }
}

B. Number of Ways to Buy Pens and Pencils (opens new window)

Brute Force

DETAILS

def waysToBuyPensPencils(self, total: int, cost1: int, cost2: int) -> int:
        ans = 0
        while T >= 0:
            ans += (T // c2) + 1
            T -= c1
        return ans2240. Number of Ways to Buy Pens and Pencils

C. Design an ATM Machine (opens new window)

DETAILS

class ATM:
    def __init__(self):
        self.cnt = [0] * 5
        self.d1 = {0:20, 1:50, 2:100, 3:200, 4:500}
        
    def deposit(self, A: List[int]) -> None:
        for i, a in enumerate(A):
            self.cnt[i] += a

    def withdraw(self, amount: int) -> List[int]:
        i, ans = 4, [0] * 5
        while i >= 0:
            c = min(self.cnt[i], amount // self.d1[i])
            amount -= c * self.d1[i]
            ans[i] = c
            i -= 1
        if amount == 0:
            for i, a in enumerate(ans):
                self.cnt[i] -= a
            return ans
        else:
            return [-1]

D. Maximum Score of a Node Sequence (opens new window)

Since the chain's length is limited to 4, meaning there will ONLY be 3 edges in any valid chain. Thus we just focus on the middle edge consists of node a, b, then we can just find the longest edge that from a and b, For example: c-a-b-d, where c, d are the nodes that have the longest edge from a, b. Notice that some edge cases happen that a, b be such furthest neighbor of each other, thus we will save the longest 3 neighbors instead.

DETAILS

class Solution:
    def maximumScore(self, A: List[int], E: List[List[int]]) -> int:
        n = len(A)

        top3 = collections.defaultdict(list)
        
        def func(a, b, e):
            bisect.insort_left(top3[a], [e, b])
            if len(top3[a]) > 3:
                top3[a].pop(0)

        for a, b in E:
            func(a, b, A[b])
            func(b, a, A[a])
        
        ans = -1
        for a, b in E:
            if len(top3[a]) < 2 or len(top3[b]) < 2:
                continue
            for c in top3[a]:
                for d in top3[b]:
                    if c[1] not in [a, b] and d[1] not in [a, b] and c[1] != d[1]:
                        ans = max(ans, A[a] + A[b] + c[0] + d[0])
        return ans