2024 January 20 Problems
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daryl...@gmail.com
Jan 20, 2024, 12:39:16 PMJan 20
to leetcode-meetup
Feel free to work on any of the problems you want; we'll have people present their solutions at 11:30.
Please post your solutions to this thread so others can use this as a reference.
2144Minimum Cost of Buying Candies With Discount60.9%Easy
2178Maximum Split of Positive Even Integers59.3%Medium
2136Earliest Possible Day of Full Bloom72.0%Hard
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Flocela Maldonado
Jan 20, 2024, 1:22:08 PMJan 20
to leetcode-meetup
Time O(n), Memory O(1)
class Solution {
public:
int minimumCost(vector<int>& cost) {
sort(cost.begin(), cost.end());
int sum = 0;
int count = 0;
for (int ii=cost.size()-1; ii>=0; --ii)
{
if (count == 0 || count ==1)
{
sum += cost[ii];
}
count = (count==0 || count==1) ? (count+1) : 0;
}
return sum;
}
};
Andriy T
Jan 20, 2024, 1:49:03 PMJan 20
to leetcode-meetup
Time O(nlog(n)) Space O(1)
class Solution {
public int minimumCost(int[] cost) {
Arrays.sort(cost);
int sum = 0;
int cost1 = 0;
int cost2 = 0;
for (int i = cost.length - 1;i>=0;i--) {
if (cost1 == 0 && cost2 == 0) {
cost1 = cost[i];
}
else if (cost1 > 0 && cost2 == 0) {
cost2 = cost[i];
}
else {
if (cost[i] <= cost1 + cost2) {
sum += (cost1 + cost2);
cost1 = 0;
cost2 = 0;
}
}
}
sum += (cost1 + cost2);
return sum;
}
}
Flocela Maldonado
Jan 20, 2024, 2:06:13 PMJan 20
to leetcode-meetup
Time: O(n); Memory O(1)
class Solution {
public:
vector<long long> maximumEvenSplit(long long finalSum) {
vector<long long> ans{};
if (finalSum % 2 != 0)
{
return ans;
}
/*vector<long long> origVector{};
int ii=1;
while (ii * 2 <= finalSum)
{
origVector.push_back(ii*2);
++ii;
}*/
long long sum = 0;
int idx=1;
while (idx <= finalSum/2 + 1)
{
long long temp = sum;
long long origVal = idx*2;
temp += origVal;
if (temp == finalSum)
{
ans.push_back(origVal);
break;
}
else if (finalSum - temp > origVal)
{
ans.push_back(origVal);
sum += origVal;
}
++idx;
}
return ans;
}
};
Vivek H
Jan 21, 2024, 1:08:36 AMJan 21
to leetcod...@googlegroups.com
*****************************************************************************
*****************************************************************************
time = O(n)
space = O(1)
Took a greedy approach and started with 2,4 6,8,...
maintain two numbers :
1) currSum : which will be increased by 2 , in every iteration
2) FinalSum = which will be decrease by currSum every iteration
For eg finalSum = 28 , the two arrays will look like below
currSum 0 2 4 6 8
finalSum 28 26 22 16 8
we need to stop the look when currSum == finalSum ( ie when both values are 8). At this point, we have added 2, 4, 6, into our array
now the only number that is left to make finalSum = 0 is16 , which is currSum + finalSum ( 8+8), so we add 16.
this our array becomes 2,4,6,16
Code below
class Solution {
public:
vector<long long> maximumEvenSplit(long long finalSum) {
long long currSum = 0;
long long i = 2;
vector<long long> result;
//if number is odd return empty vector
if((finalSum & 1) == 1)
return result;
while(currSum < finalSum) {
currSum = currSum+2;
finalSum = finalSum-currSum;
if(currSum < finalSum)
result.push_back(currSum);
else
break;
} //while ends
result.push_back(finalSum+currSum);
return result;
}
};
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Bhupathi Kakarlapudi
Jan 23, 2024, 10:53:36 PMJan 23
to leetcod...@googlegroups.com
class Solution:
def minimumCost(self, cost: List[int]) -> int:
cost.sort(reverse=True)
result:int = 0
idx:int = 0
while idx < len(cost):
result = result + sum(cost[idx:idx+2])
idx += 3
return result
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Vivek H
Jan 23, 2024, 10:53:36 PMJan 23
to leetcod...@googlegroups.com
*********************************************************************************
*********************************************************************************
time = O(n)
space = O(1)
class Solution {
public:
int minimumCost(vector<int>& cost) {
if(cost.size() == 0)
return 0;
sort(cost.begin(), cost.end());
int totalCost = 0;
while(cost.size() > 0) {
int cost1=0; int cost2=0;
cost1 = cost.back();
cost.pop_back();
if(cost.size() > 0) {
cost2 = cost.back();
cost.pop_back();
}
//cout << "cost1:" << cost1 << "cost2:" << cost2 << endl;
totalCost += cost1+cost2;
if(cost.size() > 0)
cost.pop_back();
}
return totalCost;
}
};
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