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Given this problem:
There are several cards arranged in a row, and each card has an
associated number of points The points are given in the integer array
cardPoints.
In one step, you can take one card from the beginning or from the end
of the row. You have to take exactly k cards.
Your score is the sum of the points of the cards you have taken.
Given the integer array cardPoints and the integer k, return the
maximum score you can obtain.
Example 1:
Input: cardPoints = [1,2,3,4,5,6,1], k = 3
Output: 12
Explanation: After the first step, your score will always be 1. However,
choosing
the rightmost card first will maximize your total score. The optimal
strategy is to take the three cards on the right, giving a final score
of 1 + 6 + 5 = 12.
Constraints:
1 <= cardPoints.length <= 10^5
1 <= cardPoints[i] <= 10^4
1 <= k <= cardPoints.length
I believe I wrote a top-down dp solution with memoization, but after submitting the code I see a Time Limit Exceeded error. What is wrong with this solution?
class Solution {
Map<String, Integer> cache = new HashMap<>();
public int maxScore(int[] cardPoints, int k) {
return max(0, cardPoints.length - 1, cardPoints, k);
}
private int max(int start, int end, int[] cardPoints, int k) {
if (k == 1) return Math.max(cardPoints[start], cardPoints[end]);
String key = "" + start + end;
Integer value = cache.get(key);
if (value != null) {
return value;
}
value = Math.max(
cardPoints[start] + max(start + 1, end, cardPoints, k - 1),
cardPoints[end] + max(start, end - 1, cardPoints, k - 1)
);
cache.put(key, value);
return value;
}
}
Your caching-algorithm stores each and every intermediate step. And there's a lot of these. Take for example the simple case of picking four values, and all possible paths your algorithm takes to pick two on each side:
1 2 ... 4 3
1 3 ... 4 2
1 4 ... 3 2
3 4 ... 2 1
...
In total there's 6 different paths. And all of them lead to the same result. In total this simple example already generates 9 states in your cache. For the upper bound of 10^5, things look even worse. There's a total of
(10^5 + 1) * 10^5 / 2 = 5000050000
(yup, that's 5 billion) possible states. And each single one of them will be explored. So without the TLE you'd simply run out of memory.
Instead you could use the following considerations to build a more efficient algorithm:
the order in which values are picked from either side doesn't matter for the final result
any value that is not taken from the left side must be taken from the right side and vice versa. So if k values must be picked in total and l are taken from the left side of the array, then k - l values must be taken from the right side.
Given the number of rows and columns of a 2d matrix
Initially all elements of matrix are 0
Given the number of 1's that should be present in each row
Given the number of 1's that should be present in each column
Determine if it is possible to form such matrix.
Example:
Input: r=3 c=2 (no. of rows and columns)
2 1 0 (number of 1's that should be present in each row respectively)
1 2 (number of 1's that should be present in each column respectively)
Output: Possible
Explanation:
1 1
0 1
0 0
I tried solving this problem for like 12 hours by checking if summation of Ri = summation of Ci
But I wondered if wouldn't be possible for cases like
3 3
1 3 0
0 2 2
r and c can be upto 10^5
Any ideas how should I move further?
Edit: Constraints added and output should only be "possible" or "impossible". The possible matrix need not be displayed.
Can anyone help me now?
Hint: one possible solution utilizes Maximum Flow Problem by creating a special graph and running the standard maximum flow algorithm on it.
If you're not familiar with the above problem, you may start reading about it e.g. here https://en.wikipedia.org/wiki/Maximum_flow_problem
If you're interested in the full solution please comment and I'll update the answer. But it requires understading the above algorithm.
Solution as requested:
Create a graph of r+c+2 nodes.
Node 0 is the source, node r+c+1 is the sink. Nodes 1..r represent the rows, while r+1..r+c the columns.
Create following edges:
from source to nodes i=1..r of capacity r_i
from nodes i=r+1..r+c to sink of capacity c_i
between all the nodes i=1..r and j=r+1..r+c of capacity 1
Run maximum flow algorithm, the saturated edges between row nodes and column nodes define where you should put 1.
Or if it's not possible then the maximum flow value is less than number of expected ones in the matrix.
I will illustrate the algorithm with an example.
Assume we have m rows and n columns. Let rows[i] be the number of 1s in row i, for 0 <= i < m,
and cols[j] be the number of 1s in column j, for 0 <= j < n.
For example, for m = 3, and n = 4, we could have: rows = {4 2 3}, cols = {1 3 2 3}, and
the solution array would be:
1 3 2 3
+--------
4 | 1 1 1 1
2 | 0 1 0 1
3 | 0 1 1 1
Because we only want to know whether a solution exists, the values in rows and cols may be permuted in any order. The solution of each permutation is just a permutation of the rows and columns of the above solution.
So, given rows and cols, sort cols in decreasing order, and rows in increasing order. For our example, we have cols = {3 3 2 1} and rows = {2 3 4}, and the equivalent problem.
3 3 2 1
+--------
2 | 1 1 0 0
3 | 1 1 1 0
4 | 1 1 1 1
We transform cols into a form that is better suited for the algorithm. What cols tells us is that we have two series of 1s of length 3, one series of 1s of length 2, and one series of 1s of length 1, that are to be distributed among the rows of the array. We rewrite cols to capture just that, that is COLS = {2/3 1/2 1/1}, 2 series of length 3, 1 series of length 2, and 1 series of length 1.
Because we have 2 series of length 3, a solution exists only if we can put two 1s in the first row. This is possible because rows[0] = 2. We do not actually put any 1 in the first row, but record the fact that 1s have been placed there by decrementing the length of the series of length 3. So COLS becomes:
COLS = {2/2 1/2 1/1}
and we combine our two counts for series of length 2, yielding:
COLS = {3/2 1/1}
We now have the reduced problem:
3 | 1 1 1 0
4 | 1 1 1 1
Again we need to place 1s from our series of length 2 to have a solution. Fortunately, rows[1] = 3 and we can do this. We decrement the length of 3/2 and get:
COLS = {3/1 1/1} = {4/1}
We have the reduced problem:
4 | 1 1 1 1
Which is solved by 4 series of length 1, just what we have left. If at any step, the series in COLS cannot be used to satisfy a row count, then no solution is possible.
The general processing for each row may be stated as follows. For each row r, starting from the first element in COLS, decrement the lengths of as many elements count[k]/length[k] of COLS as needed, so that the sum of the count[k]'s equals rows[r]. Eliminate series of length 0 in COLS and combine series of same length.
Note that because elements of COLS are in decreasing order of lengths, the length of the last element decremented is always less than or equal to the next element in COLS (if there is a next element).
EXAMPLE 2 : Solution exists.
rows = {1 3 3}, cols = {2 2 2 1} => COLS = {3/2 1/1}
1 series of length 2 is decremented to satisfy rows[0] = 1, and the 2 other series of length 2 remains at length 2.
rows[0] = 1
COLS = {2/2 1/1 1/1} = {2/2 2/1}
The 2 series of length 2 are decremented, and 1 of the series of length 1.
The series whose length has become 0 is deleted, and the series of length 1 are combined.
rows[1] = 3
COLS = {2/1 1/0 1/1} = {2/1 1/1} = {3/1}
A solution exists for rows[2] can be satisfied.
rows[2] = 3
COLS = {3/0} = {}
EXAMPLE 3: Solution does not exists.
rows = {0 2 3}, cols = {3 2 0 0} => COLS = {1/3 1/2}
rows[0] = 0
COLS = {1/3 1/2}
rows[1] = 2
COLS = {1/2 1/1}
rows[2] = 3 => impossible to satisfy; no solution.
SPACE COMPLEXITY
It is easy to see that it is O(m + n).
TIME COMPLEXITY
We iterate over each row only once. For each row i, we need to iterate over at most
rows[i] <= n elements of COLS. Time complexity is O(m x n).
After finding this algorithm, I found the following theorem:
The Havel-Hakimi theorem (Havel 1955, Hakimi 1962) states that there exists a matrix Xn,m of 0’s and 1’s with row totals a0=(a1, a2,… , an) and column totals b0=(b1, b2,… , bm) such that bi ≥ bi+1 for every 0 < i < m if and only if another matrix Xn−1,m of 0’s and 1’s with row totals a1=(a2, a3,… , an) and column totals b1=(b1−1, b2−1,… ,ba1−1, ba1+1,… , bm) also exists.
from the post Finding if binary matrix exists given the row and column sums.
This is basically what my algorithm does, while trying to optimize the decrementing part, i.e., all the -1's in the above theorem. Now that I see the above theorem, I know my algorithm is correct. Nevertheless, I checked the correctness of my algorithm by comparing it with a brute-force algorithm for arrays of up to 50 cells.
Here is the C# implementation.
public class Pair
{
public int Count;
public int Length;
}
public class PairsList
{
public LinkedList<Pair> Pairs;
public int TotalCount;
}
class Program
{
static void Main(string[] args)
{
int[] rows = new int[] { 0, 0, 1, 1, 2, 2 };
int[] cols = new int[] { 2, 2, 0 };
bool success = Solve(cols, rows);
}
static bool Solve(int[] cols, int[] rows)
{
PairsList pairs = new PairsList() { Pairs = new LinkedList<Pair>(), TotalCount = 0 };
FillAllPairs(pairs, cols);
for (int r = 0; r < rows.Length; r++)
{
if (rows[r] > 0)
{
if (pairs.TotalCount < rows[r])
return false;
if (pairs.Pairs.First != null && pairs.Pairs.First.Value.Length > rows.Length - r)
return false;
DecrementPairs(pairs, rows[r]);
}
}
return pairs.Pairs.Count == 0 || pairs.Pairs.Count == 1 && pairs.Pairs.First.Value.Length == 0;
}
static void DecrementPairs(PairsList pairs, int count)
{
LinkedListNode<Pair> pair = pairs.Pairs.First;
while (count > 0 && pair != null)
{
LinkedListNode<Pair> next = pair.Next;
if (pair.Value.Count == count)
{
pair.Value.Length--;
if (pair.Value.Length == 0)
{
pairs.Pairs.Remove(pair);
pairs.TotalCount -= count;
}
else if (pair.Next != null && pair.Next.Value.Length == pair.Value.Length)
{
pair.Value.Count += pair.Next.Value.Count;
pairs.Pairs.Remove(pair.Next);
next = pair;
}
count = 0;
}
else if (pair.Value.Count < count)
{
count -= pair.Value.Count;
pair.Value.Length--;
if (pair.Value.Length == 0)
{
pairs.Pairs.Remove(pair);
pairs.TotalCount -= pair.Value.Count;
}
else if(pair.Next != null && pair.Next.Value.Length == pair.Value.Length)
{
pair.Value.Count += pair.Next.Value.Count;
pairs.Pairs.Remove(pair.Next);
next = pair;
}
}
else // pair.Value.Count > count
{
Pair p = new Pair() { Count = count, Length = pair.Value.Length - 1 };
pair.Value.Count -= count;
if (p.Length > 0)
{
if (pair.Next != null && pair.Next.Value.Length == p.Length)
pair.Next.Value.Count += p.Count;
else
pairs.Pairs.AddAfter(pair, p);
}
else
pairs.TotalCount -= count;
count = 0;
}
pair = next;
}
}
static int FillAllPairs(PairsList pairs, int[] cols)
{
List<Pair> newPairs = new List<Pair>();
int c = 0;
while (c < cols.Length && cols[c] > 0)
{
int k = c++;
if (cols[k] > 0)
pairs.TotalCount++;
while (c < cols.Length && cols[c] == cols[k])
{
if (cols[k] > 0) pairs.TotalCount++;
c++;
}
newPairs.Add(new Pair() { Count = c - k, Length = cols[k] });
}
LinkedListNode<Pair> pair = pairs.Pairs.First;
foreach (Pair p in newPairs)
{
while (pair != null && p.Length < pair.Value.Length)
pair = pair.Next;
if (pair == null)
{
pairs.Pairs.AddLast(p);
}
else if (p.Length == pair.Value.Length)
{
pair.Value.Count += p.Count;
pair = pair.Next;
}
else // p.Length > pair.Value.Length
{
pairs.Pairs.AddBefore(pair, p);
}
}
return c;
}
}
(Note: to avoid confusion between when I'm talking about the actual numbers in the problem vs. when I'm talking about the zeros in the ones in the matrix, I'm going to instead fill the matrix with spaces and X's. This obviously doesn't change the problem.)
Some observations:
If you're filling in a row, and there's (for example) one column needing 10 more X's and another column needing 5 more X's, then you're sometimes better off putting the X in the "10" column and saving the "5" column for later (because you might later run into 5 rows that each need 2 X's), but you're never better off putting the X in the "5" column and saving the "10" column for later (because even if you later run into 10 rows that all need an X, they won't mind if they don't all go in the same column). So we can use a somewhat "greedy" algorithm: always put an X in the column still needing the most X's. (Of course, we'll need to make sure that we don't greedily put an X in the same column multiple times for the same row!)
Since you don't need to actually output a possible matrix, the rows are all interchangeable and the columns are all interchangeable; all that matter is how many rows still need 1 X, how many still need 2 X's, etc., and likewise for columns.
With that in mind, here's one fairly simple approach:
(Optimization.) Add up the counts for all the rows, add up the counts for all the columns, and return "impossible" if the sums don't match.
Create an array of length r+1 and populate it with how many columns need 1 X, how many need 2 X's, etc. (You can ignore any columns needing 0 X's.)
(Optimization.) To help access the array efficiently, build a stack/linked-list/etc. of the indices of nonzero array elements, in decreasing order (e.g., starting at index r if it's nonzero, then index r−1 if it's nonzero, etc.), so that you can easily find the elements representing columns to put X's in.
(Optimization.) To help determine when there'll be a row can't be satisfied, also make note of the total number of columns needing any X's, and make note of the largest number of X's needed by any row. If the former is less than the latter, return "impossible".
(Optimization.) Sort the rows by the number of X's they need.
Iterate over the rows, starting with the one needing the fewest X's and ending with the one needing the most X's, and for each one:
Update the array accordingly. For example, if a row needs 12 X's, and the array looks like [..., 3, 8, 5], then you'll update the array to look like [..., 3+7 = 10, 8+5−7 = 6, 5−5 = 0]. If it's not possible to update the array because you run out of columns to put X's in, return "impossible". (Note: this part should never actually return "impossible", because we're keeping count of the number of columns left and the max number of columns we'll need, so we should have already returned "impossible" if this was going to happen. I mention this check only for clarity.)
Update the stack/linked-list of indices of nonzero array elements.
Update the total number of columns needing any X's. If it's now less than the greatest number of X's needed by any row, return "impossible".
(Optimization.) If the first nonzero array element has an index greater than the number of rows left, return "impossible".
If we complete our iteration without having returned "impossible", return "possible".
(Note: the reason I say to start with the row needing the fewest X's, and work your way to the row with the most X's, is that a row needing more X's may involve examining updating more elements of the array and of the stack, so the rows needing fewer X's are cheaper. This isn't just a matter of postponing the work: the rows needing fewer X's can help "consolidate" the array, so that there will be fewer distinct column-counts, making the later rows cheaper than they would otherwise be. In a very-bad-case scenario, such as the case of a square matrix where every single row needs a distinct positive number of X's and every single column needs a distinct positive number of X's, the fewest-to-most order means you can handle each row in O(1) time, for linear time overall, whereas the most-to-fewest order would mean that each row would take time proportional to the number of X's it needs, for quadratic time overall.)
Overall, this takes no worse than O(r+c+n) time (where n is the number of X's); I think that the optimizations I've listed are enough to ensure that it's closer to O(r+c) time, but it's hard to be 100% sure. I recommend trying it to see if it's fast enough for your purposes.
You can use brute force (iterating through all 2^(r * c) possibilities) to solve it, but that will take a long time. If r * c is under 64, you can accelerate it to a certain extent using bit-wise operations on 64-bit integers; however, even then, iterating through all 64-bit possibilities would take, at 1 try per ms, over 500M years.
A wiser choice is to add bits one by one, and only continue placing bits if no constraints are broken. This will eliminate the vast majority of possibilities, greatly speeding up the process. Look up backtracking for the general idea. It is not unlike solving sudokus through guesswork: once it becomes obvious that your guess was wrong, you erase it and try guessing a different digit.
As with sudokus, there are certain strategies that can be written into code and will result in speedups when they apply. For example, if the sum of 1s in rows is different from the sum of 1s in columns, then there are no solutions.
If over 50% of the bits will be on, you can instead work on the complementary problem (transform all ones to zeroes and vice-versa, while updating row and column counts). Both problems are equivalent, because any answer for one is also valid for the complementary.
This problem can be solved in O(n log n) using Gale-Ryser Theorem. (where n is the maximum of lengths of the two degree sequences).
First, make both sequences of equal length by adding 0's to the smaller sequence, and let this length be n.
Let the sequences be A and B. Sort A in non-decreasing order, and sort B in non-increasing order. Create another prefix sum array P for B such that ith element of P is equal to sum of first i elements of B.
Now, iterate over k's from 1 to n, and check for
The second sum can be calculated in O(log n) using binary search for index of last number in B smaller than k, and then using precalculated P.
Inspiring from the solution given by RobertBaron I have tried to build a new algorithm.
rows = [int(x)for x in input().split()]
cols = [int (ss) for ss in input().split()]
rows.sort()
cols.sort(reverse=True)
for i in range(len(rows)):
for j in range(len(cols)):
if(rows[i]!= 0 and cols[j]!=0):
rows[i] = rows[i] - 1;
cols[j] =cols[j]-1;
print("rows: ",rows)
print("cols: ",cols)
#if there is any non zero value, print NO else print yes
flag = True
for i in range(len(rows)):
if(rows[i]!=0):
flag = False
break
for j in range(len(cols)):
if(cols[j]!=0):
flag = False
if(flag):
print("YES")
else:
print("NO")
here, i have sorted the rows in ascending order and cols in descending order. later decrementing particular row and column if 1 need to be placed!
it is working for all the test cases posted here! rest GOD knows
Given a array with +ve and -ve integer , find the maximum sum such that you are not allowed to skip 2 contiguous elements ( i.e you have to select at least one of them to move forward).
eg :-
10 , 20 , 30, -10 , -50 , 40 , -50, -1, -3
Output : 10+20+30-10+40-1 = 89
This problem can be solved using Dynamic Programming approach.
Let arr be the given array and opt be the array to store the optimal solutions.
opt[i] is the maximum sum that can be obtained starting from element i, inclusive.
opt[i] = arr[i] + (some other elements after i)
Now to solve the problem we iterate the array arr backwards, each time storing the answer opt[i].
Since we cannot skip 2 contiguous elements, either element i+1 or element i+2 has to be included
in opt[i].
So for each i, opt[i] = arr[i] + max(opt[i+1], opt[i+2])
See this code to understand:
int arr[n]; // array of given numbers. array size = n.
nput(arr, n); // input the array elements (given numbers)
int opt[n+2]; // optimal solutions.
memset(opt, 0, sizeof(opt)); // Initially set all optimal solutions to 0.
for(int i = n-1; i >= 0; i--) {
opt[i] = arr[i] + max(opt[i+1], opt[i+2]);
}
ans = max(opt[0], opt[1]) // final answer.
Observe that opt array has n+2 elements. This is to avoid getting illegal memory access exception (memory out of bounds) when we try to access opt[i+1] and opt[i+2] for the last element (n-1).
See the working implementation of the algorithm given above
Use a recurrence that accounts for that:
dp[i] = max(dp[i - 1] + a[i], <- take two consecutives
dp[i - 2] + a[i], <- skip a[i - 1])
Base cases left as an exercise.
If you see a +ve integer add it to the sum. If you see a negative integer, then inspect the next integer pick which ever is maximum and add it to the sum.
10 , 20 , 30, -10 , -50 , 40 , -50, -1, -3
For this add 10, 20, 30, max(-10, -50), 40 max(-50, -1) and since there is no element next to -3 discard it.
The last element will go to sum if it was +ve.
Answer:
I think this algorithm will help.
1. Create a method which gives output the maximum sum of particular user input array say T[n], where n denotes the total no. of elements.
2. Now this method will keep on adding array elements till they are positive. As we want to maximize the sum and there is no point in dropping positive elements behind.
3. As soon as our method encounters a negative element, it will transfer all consecutive negative elements to another method which create a new array say N[i] such that this array will contain all the consecutive negative elements that we encountered in T[n] and returns N[i]'s max output.
In this way our main method is not affected and its keep on adding positive elements and whenever it encounters negative element, it instead of adding their real values adds the net max output of that consecutive array of negative elements.
for example: T[n] = 29,34,55,-6,-5,-4,6,43,-8,-9,-4,-3,2,78 //here n=14
Main Method Working:
29+34+55+(sends data & gets value from Secondary method of array [-6,-5,-4])+6+43+(sends data & gets value from Secondary method of array [-8,-9,-4,-3])+2+78
Process Terminates with max output.
Secondary Method Working:
{
N[i] = gets array from Main method or itself as and when required.
This is basically a recursive method.
say N[i] has elements like N1, N2, N3, N4, etc.
for i>=3:
Now choice goes like this.
1. If we take N1 then we can recurse the left off array i.e. N[i-1] which has all elements except N1 in same order. Such that the net max output will be
N1+(sends data & gets value from Secondary method of array N[i-1] recursively)
2. If we doesn't take N1, then we cannot skip N2. So, Now algorithm is like 1st choice but starting with N2. So max output in this case will be
N2+(sends data & gets value from Secondary method of array N[i-2] recursively).
Here N[i-2] is an array containing all N[i] elements except N1 & N2 in same order.
Termination: When we are left with the array of size one ( for N[i-2] ) then we have to choose that particular value as no option.
The recursions will finally yield the max outputs and we have to finally choose the output of that choice which is more.
and redirect the max output to wherever required.
for i=2:
we have to choose the value which is bigger
for i=1:
We can surely skip that value.
So max output in this case will be 0.
}
I think this answer will help to you.
Given array:
Given:- 10 20 30 -10 -50 40 -50 -1 -3
Array1:-10 30 60 50 10 90 40 89 86
Array2:-10 20 50 40 0 80 30 79 76
Take the max value of array1[n-1],array1[n],array2[n-1],array2[n] i.e 89(array1[n-1])
Algorithm:-
For the array1 value assign array1[0]=a[0],array1=a[0]+a[1] and array2[0]=a[0],array2[1]=a[1].
calculate the array1 value from 2 to n is max of sum of array1[i-1]+a[i] or array1[i-2]+a[i].
for loop from 2 to n{
array1[i]=max(array1[i-1]+a[i],array1[i-2]+a[i]);
}
similarly for array2 value from 2 to n is max of sum of array2[i-1]+a[i] or array2[i-2]+a[i].
for loop from 2 to n{
array2[i]=max(array2[i-1]+a[i],array2[i-2]+a[i]);
}
Finally find the max value of array1[n-1],array[n],array2[n-1],array2[n];
int max(int a,int b){
return a>b?a:b;
}
int main(){
int a[]={10,20,30,-10,-50,40,-50,-1,-3};
int i,n,max_sum;
n=sizeof(a)/sizeof(a[0]);
int array1[n],array2[n];
array1[0]=a[0];
array1[1]=a[0]+a[1];
array2[0]=a[0];
array2[1]=a[1];
for loop from 2 to n{
array1[i]=max(array1[i-1]+a[i],array1[i-2]+a[i]);
array2[i]=max(array2[i-1]+a[i],array2[i-2]+a[i]);
}
--i;
max_sum=max(array1[i],array1[i-1]);
max_sum=max(max_sum,array2[i-1]);
max_sum=max(max_sum,array2[i]);
printf("The max_sum is %d",max_sum);
return 0;
}
Ans: The max_sum is 89
public static void countSum(int[] a) {
int count = 0;
int skip = 0;
int newCount = 0;
if(a.length==1)
{
count = a[0];
}
else
{
for(int i:a)
{
newCount = count + i;
if(newCount>=skip)
{
count = newCount;
skip = newCount;
}
else
{
count = skip;
skip = newCount;
}
}
}
System.out.println(count);
}
}
Let the array be of size N, indexed as 1...N
Let f(n) be the function, that provides the answer for max sum of sub array (1...n), such that no two left over elements are consecutive.
f(n) = max (a[n-1] + f(n-2), a(n) + f(n-1))
In first option, which is - {a[n-1] + f(n-2)}, we are leaving the last element, and due to condition given in question selecting the second last element.
In the second option, which is - {a(n) + f(n-1)} we are selecting the last element of the subarray, so we have an option to select/deselect the second last element.
Now starting from the base case :
f(0) = 0 [Subarray (1..0) doesn't exist]
f(1) = (a[1] > 0 ? a[1] : 0); [Subarray (1..1)]
f(2) = max( a(2) + 0, a[1] + f(1)) [Choosing atleast one of them]
Moving forward we can calculate any f(n), where n = 1...N, and store them to calculate next results. And yes, obviously, the case f(N) will give us the answer.
Time complexity o(n)
Space complexity o(n)
n = arr.length().
Append a 0 at the end of the array to handle boundary case.
ans: int array of size n+1.
ans[i] will store the answer for array a[0...i] which includes a[i] in the answer sum.
Now,
ans[0] = a[0]
ans[1] = max(a[1], a[1] + ans[0])
for i in [2,n-1]:
ans[i] = max(ans[i-1] , ans[i-2]) + a[i]
Final answer would be a[n]
If you want to avoid using Dynamic Programming
To find the maximum sum, first, you've to add all the positive
numbers.
We'll be skipping only negative elements. Since we're not
allowed to skip 2 contiguous elements, we will put all contiguous
negative elements in a temp array, and can figure out the maximum sum
of alternate elements using sum_odd_even function as defined below.
Then we can add the maximum of all such temp arrays to our sum of all
positive numbers. And the final sum will give us the desired output.
Code:
def sum_odd_even(arr):
sum1 = sum2 = 0
for i in range(len(arr)):
if i%2 == 0:
sum1 += arr[i]
else:
sum2 += arr[i]
return max(sum1,sum2)
input = [10, 20, 30, -10, -50, 40, -50, -1, -3]
result = 0
temp = []
for i in range(len(input)):
if input[i] > 0:
result += input[i]
if input[i] < 0 and i != len(input)-1:
temp.append(input[i])
elif input[i] < 0:
temp.append(input[i])
result += sum_odd_even(temp)
temp = []
else:
result += sum_odd_even(temp)
temp = []
print result
Simple Solution: Skip with twist :). Just skip the smallest number in i & i+1 if consecutive -ve. Have if conditions to check that till n-2 elements and check for the last element in the end.
int getMaxSum(int[] a) {
int sum = 0;
for (int i = 0; i <= a.length-2; i++) {
if (a[i]>0){
sum +=a[i];
continue;
} else if (a[i+1] > 0){
i++;
continue;
} else {
sum += Math.max(a[i],a[i+1]);
i++;
}
}
if (a[a.length-1] > 0){
sum+=a[a.length-1];
}
return sum;
}
The correct recurrence is as follow:
dp[i] = max(dp[i - 1] + a[i], dp[i - 2] + a[i - 1])
The first case is the one we pick the i-th element. The second case is the one we skip the i-th element. In the second case, we must pick the (i-1)th element.
The problem of IVlad's answer is that it always pick i-th element, which can lead to incorrect answer.
This question can be solved using include,exclude approach.
For first element, include = arr[0], exclude = 0.
For rest of the elements:
nextInclude = arr[i]+max(include, exclude)
nextExclude = include
include = nextInclude
exclude = nextExclude
Finally, ans = Math.max(include,exclude).
Similar questions can be referred at (Not the same)=> https://www.youtube.com/watch?v=VT4bZV24QNo&t=675s&ab_channel=Pepcoding.
It's an interview question.
Given a number n, find out how many numbers have digit 2 in the range 0...n
For example ,
input = 13 output = 2 (2 and 12)
I gave the usual O(n^2) solution but is there a better approach.
is there any 'trick' formula that will help me to get the answer right away
Count the numbers that do not have the digit 2. Among the numbers less than 10k, there are exactly 9k of them. Then it remains to treat the numbers from 10k to n, where
10^k <= n < 10^(k+1)
which you can do by treating the first digits individually (the cases 2 and others have to be differentiated), and then the first 2 digits etc.
For example, for n = 2345, we find there are 9^3 = 729 numbers without the digit 2 below 1000. There are again 729 such numbers in the range from 1000 to 1999. Then in the range from 2000 to 2345, there are none, for a total of 1458, hence the numbers containing the digit 2 are
2345 - 1458 = 887
argument 'digit' is the one which we want to count and arg 'number' is till where we want to count. For eg: If we want to count occurrences of '1', from 0 to 12, call the function with digit=1, and number=12, and it will return the number of occurrences of '1'.
int countOccurrences(int digit, int number)
{
int counter = 0;
for(int i=1; i<number; i++)
{
int j = i;
while(j > 0)
{
if(j%10 == digit)
counter++;
j /= 10;
}
}
return counter;
}
Given the number with the digits ABCDEF you can count the number of '2's in the ranges [0,F], [0,E9], [0,D99], [0,C999], [0,B9999] and [0,A99999] and add them.
Then for the range [0, X9999...999], the top number T = X9999...999 can be written as (X+1) * 10<sup>nines</sup> -1.
The number of '2's in that range is:
((X >= 2 ? 1/(X + 1)) : 0) + nines/10 ) * (T + 1);
That is: if X >= 2, the fraction of numbers that have a '2' at the position nines+1 is 1/(X+1), In total there are (T+1)/(X+1) '2's at that position. If X < 2, then no number on [0..T] has a '2' at that position.
For the other digit positions, is easy to see that at every digit position, 1/10 of the numbers have a '2', so there are (T+1)/10 '2's at position 0, (T+1)/10 '2's at position 1, etc. In total, (T+1) * nines / 10.
The complexity of this solution is O(logN).
this is how I would go about coding my first draft (Python code)
def count2(n) :
return [p for p in range(n+1) if '2' in str(p)]
and that will return you a list with the containing number.
In terms of performance it is not that bad, for n=10,000,000 an average iteration takes about 5.5 seconds
Saw this question recently:
Given 2 arrays, the 2nd array containing some of the elements of the 1st array, return the minimum window in the 1st array which contains all the elements of the 2nd array.
Eg :
Given A={1,3,5,2,3,1} and B={1,3,2}
Output : 3 , 5 (where 3 and 5 are indices in the array A)
Even though the range 1 to 4 also contains the elements of A, the range 3 to 5 is returned Since it contains since its length is lesser than the previous range ( ( 5 - 3 ) < ( 4 - 1 ) )
I had devised a solution but I am not sure if it works correctly and also not efficient.
Give an Efficient Solution for the problem. Thanks in Advance
A simple solution of iterating through the list.
Have a left and right pointer, initially both at zero
Move the right pointer forwards until [L..R] contains all the elements (or quit if right reaches the end).
Move the left pointer forwards until [L..R] doesn't contain all the elements. See if [L-1..R] is shorter than the current best.
This is obviously linear time. You'll simply need to keep track of how many of each element of B is in the subarray for checking whether the subarray is a potential solution.
Pseudocode of this algorithm.
size = bestL = A.length;
needed = B.length-1;
found = 0; left=0; right=0;
counts = {}; //counts is a map of (number, count)
for(i in B) counts.put(i, 0);
//Increase right bound
while(right < size) {
if(!counts.contains(right)) continue;
amt = count.get(right);
count.set(right, amt+1);
if(amt == 0) found++;
if(found == needed) {
while(found == needed) {
//Increase left bound
if(counts.contains(left)) {
amt = count.get(left);
count.set(left, amt-1);
if(amt == 1) found--;
}
left++;
}
if(right - left + 2 >= bestL) continue;
bestL = right - left + 2;
bestRange = [left-1, right] //inclusive
}
}