Weekly Contest 121
984. String Without AAA or BBB
Given two integers A
and B
, return any string S
such that:
S
has lengthA + B
and contains exactlyA
'a'
letters, and exactlyB
'b'
letters;- The substring
'aaa'
does not occur inS
; - The substring
'bbb'
does not occur inS
.
Example 1:
Input: A = 1, B = 2
Output: "abb"
Explanation: "abb", "bab" and "bba" are all correct answers.
Example 2:
Input: A = 4, B = 1
Output: "aabaa"
Note:
0 <= A <= 100
0 <= B <= 100
- It is guaranteed such an
S
exists for the givenA
andB
.
Approach #1:
class Solution {
public:
string strWithout3a3b(int A, int B) {
string ans = "";
char a = 'a';
char b = 'b';
if (B > A) {
swap(A, B);
swap(a, b);
}
while (A != 0 || B != 0) {
if (A > 0) ans += a, A--;
if (A > B) ans += a, A--;
if (B > 0) ans += b, B--;
if (B > A) ans += b, B--;
} return ans;
}
};
981. Time Based Key-Value Store
Create a timebased key-value store class TimeMap
, that supports two operations.
1. set(string key, string value, int timestamp)
- Stores the
key
andvalue
, along with the giventimestamp
.
2. get(string key, int timestamp)
- Returns a value such that
set(key, value, timestamp_prev)
was called previously, withtimestamp_prev <= timestamp
. - If there are multiple such values, it returns the one with the largest
timestamp_prev
. - If there are no values, it returns the empty string (
""
).
Example 1:
Input: inputs = ["TimeMap","set","get","get","set","get","get"], inputs = [[],["foo","bar",1],["foo",1],["foo",3],["foo","bar2",4],["foo",4],["foo",5]]
Output: [null,null,"bar","bar",null,"bar2","bar2"]
Explanation:
TimeMap kv;
kv.set("foo", "bar", 1); // store the key "foo" and value "bar" along with timestamp = 1
kv.get("foo", 1); // output "bar"
kv.get("foo", 3); // output "bar" since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 ie "bar"
kv.set("foo", "bar2", 4);
kv.get("foo", 4); // output "bar2"
kv.get("foo", 5); //output "bar2"
Example 2:
Input: inputs = ["TimeMap","set","set","get","get","get","get","get"], inputs = [[],["love","high",10],["love","low",20],["love",5],["love",10],["love",15],["love",20],["love",25]]
Output: [null,null,null,"","high","high","low","low"]
Note:
- All key/value strings are lowercase.
- All key/value strings have length in the range
[1, 100]
- The
timestamps
for allTimeMap.set
operations are strictly increasing. 1 <= timestamp <= 10^7
TimeMap.set
andTimeMap.get
functions will be called a total of120000
times (combined) per test case.
Approach #1:
class TimeMap {
public:
/** Initialize your data structure here. */
vector<string> ans;
map<string, vector<pair<int, string>>> mp;
TimeMap() { } void set(string key, string value, int timestamp) {
mp[key].push_back({timestamp, value});
} string get(string key, int timestamp) {
if (mp.count(key)) {
for (int i = mp[key].size()-1; i >= 0 ; --i) {
if (mp[key][i].first <= timestamp) {
return mp[key][i].second;
}
}
}
return "";
} }; /**
* Your TimeMap object will be instantiated and called as such:
* TimeMap* obj = new TimeMap();
* obj->set(key,value,timestamp);
* string param_2 = obj->get(key,timestamp);
*/
982. Triples with Bitwise AND Equal To Zero
Given an array of integers A
, find the number of triples of indices (i, j, k) such that:
0 <= i < A.length
0 <= j < A.length
0 <= k < A.length
A[i] & A[j] & A[k] == 0
, where&
represents the bitwise-AND operator.
Example 1:
Input: [2,1,3]
Output: 12
Explanation: We could choose the following i, j, k triples:
(i=0, j=0, k=1) : 2 & 2 & 1
(i=0, j=1, k=0) : 2 & 1 & 2
(i=0, j=1, k=1) : 2 & 1 & 1
(i=0, j=1, k=2) : 2 & 1 & 3
(i=0, j=2, k=1) : 2 & 3 & 1
(i=1, j=0, k=0) : 1 & 2 & 2
(i=1, j=0, k=1) : 1 & 2 & 1
(i=1, j=0, k=2) : 1 & 2 & 3
(i=1, j=1, k=0) : 1 & 1 & 2
(i=1, j=2, k=0) : 1 & 3 & 2
(i=2, j=0, k=1) : 3 & 2 & 1
(i=2, j=1, k=0) : 3 & 1 & 2
Note:
1 <= A.length <= 1000
0 <= A[i] < 2^16
Approach #1:
class Solution {
public:
int countTriplets(vector<int>& A) {
int size = A.size();
int ans = 0;
unordered_map<int, int> mp;
for (int i = 0; i < size; ++i) {
for (int j = 0; j < size; ++j) {
++mp[A[i] & A[j]];
}
}
for (int i = 0; i < size; ++i) {
for (auto m : mp) {
if ((A[i] & m.first) == 0)
ans += m.second;
}
}
return ans;
}
};
983. Minimum Cost For Tickets
In a country popular for train travel, you have planned some train travelling one year in advance. The days of the year that you will travel is given as an array days
. Each day is an integer from 1
to 365
.
Train tickets are sold in 3 different ways:
- a 1-day pass is sold for
costs[0]
dollars; - a 7-day pass is sold for
costs[1]
dollars; - a 30-day pass is sold for
costs[2]
dollars.
The passes allow that many days of consecutive travel. For example, if we get a 7-day pass on day 2, then we can travel for 7 days: day 2, 3, 4, 5, 6, 7, and 8.
Return the minimum number of dollars you need to travel every day in the given list of days
.
Example 1:
Input: days = [1,4,6,7,8,20], costs = [2,7,15]
Output: 11
Explanation:
For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 1-day pass for costs[0] = $2, which covered day 1.
On day 3, you bought a 7-day pass for costs[1] = $7, which covered days 3, 4, ..., 9.
On day 20, you bought a 1-day pass for costs[0] = $2, which covered day 20.
In total you spent $11 and covered all the days of your travel.
Example 2:
Input: days = [1,2,3,4,5,6,7,8,9,10,30,31], costs = [2,7,15]
Output: 17
Explanation:
For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 30-day pass for costs[2] = $15 which covered days 1, 2, ..., 30.
On day 31, you bought a 1-day pass for costs[0] = $2 which covered day 31.
In total you spent $17 and covered all the days of your travel.
Note:
1 <= days.length <= 365
1 <= days[i] <= 365
days
is in strictly increasing order.costs.length == 3
1 <= costs[i] <= 1000
Approach #1:
class Solution {
public:
int mincostTickets(vector<int>& days, vector<int>& costs) {
vector<int> dp(366, 0);
vector<bool> isday(366, false);
for (int day: days) {
isday[day] = true;
}
for (int i = 1; i <= 365; ++i) {
if (!isday[i]) {
dp[i] = dp[i-1];
continue;
}
dp[i] = costs[0] + dp[i-1];
if (i >= 7) {
dp[i] = min(dp[i], costs[1]+dp[i-7]);
} else {
dp[i] = min(dp[i], costs[1]);
}
if (i >= 30) {
dp[i] = min(dp[i], costs[2]+dp[i-30]);
} else {
dp[i] = min(dp[i], costs[2]);
}
}
return dp[365];
}
};
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