A emoticon consists of an arbitrary positive number of underscores between two semicolons. Hence, the shortest possible emoticon is ;_;. The strings ;__; and ;_____________; are also valid emoticons.
given a String containing only(;,_).The problem is to divide string into one or more emoticons and count how many division are possible. Each emoticon must be a subsequence of the message, and each character of the message must belong to exactly one emoticon. Note that the subsequences are not required to be contiguous. subsequence definition.
The approach I thought of is to write a recursive method as follows:
countDivision(string s){
//base cases
if(s.empty()) return 1;
if(s.length()<=3){
if(s.length()!=3) return 0;
return s[0]==';' && s[1]=='_' && s[2]==';';
}
result=0;
//subproblems
genrate all valid emocticon and remove it from s let it be w
result+=countDivision(w);
return result;
}
The solution above will easily timeout when n is large such as 100. What kind of approach should I use to convert this brute force solution to a dynamic programming solution?
Few examples
1. ";_;;_____;" ans is 2
2. ";;;___;;;" ans is 36
Example 1.
";_;;_____;" Returns: 2
There are two ways to divide this string into two emoticons.
One looks as follows: ;_;|;_____; and the other looks like
this(rembember we can pick subsequence it need not be contigous): ;_ ;|; _____;
I'll describe an O(n^4)-time and -space dynamic programming solution (that can easily be improved to use just O(n^3) space) that should work for up to n=100 or so.
Call a subsequence "fresh" if consists of a single ;.
Call a subsequence "finished" if it corresponds to an emoticon.
Call a subsequence "partial" if it has nonzero length and is a proper prefix of an emoticon. (So for example, ;, ;_, and ;___ are all partial subsequences, while the empty string, _, ;; and ;___;; are not.)
Finally, call a subsequence "admissible" if it is fresh, finished or partial.
Let f(i, j, k, m) be the number of ways of partitioning the first i characters of the string into exactly j+k+m admissible subsequences, of which exactly j are fresh, k are partial and m are finished. Notice that any prefix of a valid partition into emoticons determines i, j, k and m uniquely -- this means that no prefix of a valid partition will be counted by more than one tuple (i, j, k, m), so if we can guarantee that, for each tuple (i, j, k, m), the partition prefixes within that tuple are all counted once and only once, then we can add together the counts for tuples to get a valid total. Specifically, the answer to the question will then be the sum over all 1 <= j <= n of f(n, 0, j, 0).
If s[i] = "_":
f(i, j, k, m) =
(j+1) * f(i-1, j+1, k, m-1) // Convert any of the j+1 fresh subsequences to partial
+ m * f(i-1, j, k, m) // Add _ to any of the m partial subsequences
Else if s[i] = ";":
f(i, j, k, m) =
f(i-1, j-1, k, m) // Start a fresh subsequence
+ (m+1) * f(i-1, j, k-1, m+1) // Finish any of the m+1 partial subsequences
We also need the base cases
f(0, 0, 0, 0) = 1
f(0, _, _, _) = 0
f(i, j, k, m) = 0 if any of i, j, k or m are negative
My own C++ implementation gives the correct answer of 36 for ;;;___;;; in a few milliseconds, and e.g. for ;;;___;;;_;_; it gives an answer of 540 (also in a few milliseconds). For a string consisting of 66 ;s followed by 66 _s followed by 66 ;s, it takes just under 2s and reports an answer of 0 (probably due to overflow of the long long).
Here's a fairly straightforward memoized recursion that returns an answer immediately for a string of 66 ;s followed by 66 _s followed by 66 ;s. The function has three parameters: i = index in the string, j = number of accumulating emoticons with only a left semi-colon, and k = number of accumulating emoticons with a left semi-colon and one or more underscores.
An array is also constructed for how many underscores and semi-colons are available to the right of each index, to help decide on the next possibilities.
Complexity is O(n^3) and the problem constrains the search space, where j is at most n/2 and k at most n/4.
Commented JavaScript code:
var s = ';_;;__;_;;';
// record the number of semi-colons and
// underscores to the right of each index
var cs = new Array(s.length);
cs.push(0);
var us = new Array(s.length);
us.push(0);
for (var i=s.length-1; i>=0; i--){
if (s[i] == ';'){
cs[i] = cs[i+1] + 1;
us[i] = us[i+1];
} else {
us[i] = us[i+1] + 1;
cs[i] = cs[i+1];
}
}
// memoize
var h = {};
function f(i,j,k){
// memoization
var key = [i,j,k].join(',');
if (h[key] !== undefined){
return h[key];
}
// base case
if (i == s.length){
return 1;
}
var a = 0,
b = 0;
if (s[i] == ';'){
// if there are still enough colons to start an emoticon
if (cs[i] > j + k){
// start a new emoticon
a = f(i+1,j+1,k);
}
// close any of k partial emoticons
if (k > 0){
b = k * f(i+1,j,k-1);
}
}
if (s[i] == '_'){
// if there are still extra underscores
if (j < us[i] && k > 0){
// apply them to partial emoticons
a = k * f(i+1,j,k);
}
// convert started emoticons to partial
if (j > 0){
b = j * f(i+1,j-1,k+1);
}
}
return h[key] = a + b;
}
console.log(f(0,0,0)); // 52
Given a matrix of characters and a string, find whether the string can be obtained from the matrix. From each character in the matrix, we can move up/down/right/left. For example, if the matrix[3][4] is:
o f a s
l l q w
z o w k
and the string is follow, then the function should return true.
The only approach I can think of is a backtracking algorithm that searches whether the word is possible or not. Is there any other faster algorithm to approach this problem?
And suppose I have a lot of queries (on finding whether a word exists or not). Then can there be some preprocessing done to answer the queries faster?
You can solve this using DFS. Let's define a graph for the problem. The vertices of the graph will comprise of the cell of a combination of cell of the matrix and a length of prefix of the string we are searching for. When we are at a given vertex this will mean that all the characters of the specified prefix were matched so far and that we currently are at the given cell.
We define edges as connecting cells adjacent by a side and doing a "valid" transaction. That is the cell we are going to should be the next in the string we are searching for.
To solve the problem we do a DFS from all cells that contain the first letter of the string and prefix length 1(meaning we've matched this first letter). From there on we continue the search and on each step we compute which are the edges going out of the current position(cell/string prefix length combination). We terminate the first time we reach a prefix of length L - the length of the string.
Note that DFS may be considered backtracking but what is more important is to keep track of the nodes in the graph we've already visited. Thus the overall complexity is bound by N * M * L where N and M are the dimensions of the matrix and L - the length of the string.
You could of course find all possible strings (start with a charater and go as far as you can). This can be done with a recursive function.
grid:
abc
def
ghi
strings:
abcfedghi
abcfehgd
abcfehi
abedghif
abefc
abefighd
abehgd
abehifc
ad...
...
Then sort these strings and when looking for a word use a binary search on the list. (When looking for an n letter word you would of course only consider the first n letters of the strings in the list.) A lot of preparation and much memory needed, but searching will be fast. So if you use the same grid again and again, the preparation may finally pay :-)
Below is the pseudo code for finding if the given string is present in a given matrix. Here visited keeps track of the location of the string in the matrix and it uses backtracking for keeping track of that. I hope this is helpful.
bool isSafe(matrix[n][m], int visited[n][m], int i, int j, int n, int m){
if(i<m && j<n && i>=0 && j>=0 && visited[i][j] == 0)
return true;
return false;
}
bool dfs(char matrix[n][m], int i, int j, int visited[n][m], char str[], int index){
if(index == strlen(str))
return true;
// row moves
int x[] = {-1, 0, 1, -1};
// col moves
int y[] = {0, -1, 1, 0};
if(str[index] == matrix[i][j]){
visited[i][j] = 1;
// for all the neighbours
for(int k = 0; k<4; k++){
// mark given position visited
next_x = i + x[k];
next_y = j + y[k];
if(isSafe(matrix, visited, next_x, next_y, n, m)){
if(dfs(matrix, next_x, next_y, visited, str, index+1) == true)
return true;
}
}
// backtrack
visited[i][j] = 0;
}
return false;
}
bool isPresent(char matrix[n][m], char str[]){
// visited initialized to 0
int visited[n][m] = {0};
for(int i=0;i<n;i++)
for(int j=0;j<n;j++){
if(dfs(matrix, i, j, n, m ,visited, str, 0) == true)
return true;
}
return false;
}
This problem is trying to find the lexicographical max suffix of a given list.
Suppose we have an array/list [e1;e2;e3;e4;e5].
Then all suffixes of [e1;e2;e3;e4;e5] are:
[e1;e2;e3;e4;e5]
[e2;e3;e4;e5]
[e3;e4;e5]
[e4;e5]
[e5]
Then our goal is to find the lexicographical max one among the above 5 lists.
for example, all suffixes of [1;2;3;1;0] are
[1;2;3;1;0]
[2;3;1;0]
[3;1;0]
[1;0]
[0].
The lexicographical max suffix is [3;1;0] from above example.
The straightforward algorithm is just to compare all suffixes one by one and always record the max. The time complexity is O(n^2) as comparing two lists need O(n).
However, the desired time complexity is O(n) and no suffix tree (no suffix array either) should be used.
please note that elements in the list may not be distinct
int max_suffix(const vector<int> &a)
{
int n = a.size(),
i = 0,
j = 1,
k;
while (j < n)
{
for (k = 0; j + k < n && a[i + k] == a[j + k]; ++k);
if (j + k == n) break;
(a[i + k] < a[j + k] ? i : j) += k + 1;
if (i == j)
++j;
else if (i > j)
swap(i, j);
}
return i;
}
My solution is a little modification of the solution to the problem Minimum Rotations.
In the above code, each time it step into the loop, it's keeped that i < j, and all a[p...n] (0<=p<j && p!=i) are not the max suffix. Then in order to decide which of a[i...n] and a[j...n] is less lexicographical, use the for-loop to find the least k that make a[i+k]!=a[j+k], then update i and j according to k.
We can skip k elements for i or j, and still keep it true that all a[p...n] (0<=p<j && p!=i) are not the max suffix. For example, if a[i+k]<a[j+k], then a[i+p...n](0<=p<=k) is not max suffix, since a[j+p...n] is lexicographically greater than it.
Imagine in a two player game, two opponents A and B work against each other, on finding the max suffix of a given string s. Whoever first finds the max suffix will win the game. In the first round, A picks suffix s[i..], and B picks suffix s[j..].
i: _____X
j: _____Y
Matched length = k
A judge compares two suffixes and finds there is mismatch after k comparisons, as shown in the fig above.
Without the loss of generality, we assume X > Y, then B is lost in this round. So he has to pick a different suffix in order to (possibly) beat A in next round. If B is smart, he will not pick any suffix starting at position j, j + 1, ..., j + k, because s[j..] is already beaten by s[i..] and he knows s[j+1..] will be beaten by s[i+1..], and s[j+2..] will be beaten by s[i+2..] and so on. So B should pick suffix S[j + k + 1..] for next round. One extra observation is that B should not pick the same suffix as A either because the first person who finds the max suffix wins the game. If j + k + 1 happens to be equal to i, B should skip to the next position.
Finally, after many rounds, either A or B will run out choices and lose the game, because the number of choices are limited for both A and B, and some choices will be eliminated after each round.
When this happens, the current suffix that the winner holds is the max suffix (Remember the loser runs out all choices. A choice is given up because either it cannot possibly be max suffix, or it is currently held by the other person. So the only reason that the loser gives up the actual max suffix in some round is that his opponent is holding it. Once a player holds max suffix, he will never lose and give it up).
The program below in C++ is almost literal translation of this game.
int maxSuffix(const std::string& s) {
std::size_t i = 0, j = 1, k;
while (i < s.size() && j < s.size()) {
for (k = 0; i + k < s.size() && j + k < s.size() && s[i + k] == s[j +k]; ++k) { } //judge
if (j + k >= s.size()) return i; //B is finally lost
if (i + k >= s.size()) return j; //A is finally lost
if (s[i + k] > s[j + k]) { //B is lost in this round so he needs a new choice
j = j + k + 1;
if (j == i) ++j;
} else { //A is lost in this round so he needs a new choice
i = i + k + 1;
if (i == j) ++i;
}
}
return j >= s.size() ? i : j;
}
Running time analysis: Initially each player has n choices. After each round, the judge makes k comparisons, and at least k possible choices are eliminated from either A or B. So the total number of comparisons are bounded by 2n when the game is over.
The discussion above is in the context of string, but it should work with minor modification on any container that supports sequential access only.
An interesting interview question that a colleague of mine uses:
Suppose that you are given a very long, unsorted list of unsigned 64-bit integers. How would you find the smallest non-negative integer that does not occur in the list?
FOLLOW-UP: Now that the obvious solution by sorting has been proposed, can you do it faster than O(n log n)?
FOLLOW-UP: Your algorithm has to run on a computer with, say, 1GB of memory
CLARIFICATION: The list is in RAM, though it might consume a large amount of it. You are given the size of the list, say N, in advance.
If the datastructure can be mutated in place and supports random access then you can do it in O(N) time and O(1) additional space. Just go through the array sequentially and for every index write the value at the index to the index specified by value, recursively placing any value at that location to its place and throwing away values > N. Then go again through the array looking for the spot where value doesn't match the index - that's the smallest value not in the array. This results in at most 3N comparisons and only uses a few values worth of temporary space.
# Pass 1, move every value to the position of its value
for cursor in range(N):
target = array[cursor]
while target < N and target != array[target]:
new_target = array[target]
array[target] = target
target = new_target
# Pass 2, find first location where the index doesn't match the value
for cursor in range(N):
if array[cursor] != cursor:
return cursor
return N
Here's a simple O(N) solution that uses O(N) space. I'm assuming that we are restricting the input list to non-negative numbers and that we want to find the first non-negative number that is not in the list.
Find the length of the list; lets say it is N.
Allocate an array of N booleans, initialized to all false.
For each number X in the list, if X is less than N, set the X'th element of the array to true.
Scan the array starting from index 0, looking for the first element that is false. If you find the first false at index I, then I is the answer. Otherwise (i.e. when all elements are true) the answer is N.
In practice, the "array of N booleans" would probably be encoded as a "bitmap" or "bitset" represented as a byte or int array. This typically uses less space (depending on the programming language) and allows the scan for the first false to be done more quickly.
This is how / why the algorithm works.
Suppose that the N numbers in the list are not distinct, or that one or more of them is greater than N. This means that there must be at least one number in the range 0 .. N - 1 that is not in the list. So the problem of find the smallest missing number must therefore reduce to the problem of finding the smallest missing number less than N. This means that we don't need to keep track of numbers that are greater or equal to N ... because they won't be the answer.
The alternative to the previous paragraph is that the list is a permutation of the numbers from 0 .. N - 1. In this case, step 3 sets all elements of the array to true, and step 4 tells us that the first "missing" number is N.
The computational complexity of the algorithm is O(N) with a relatively small constant of proportionality. It makes two linear passes through the list, or just one pass if the list length is known to start with. There is no need to represent the hold the entire list in memory, so the algorithm's asymptotic memory usage is just what is needed to represent the array of booleans; i.e. O(N) bits.
(By contrast, algorithms that rely on in-memory sorting or partitioning assume that you can represent the entire list in memory. In the form the question was asked, this would require O(N) 64-bit words.)
#Jorn comments that steps 1 through 3 are a variation on counting sort. In a sense he is right, but the differences are significant:
A counting sort requires an array of (at least) Xmax - Xmin counters where Xmax is the largest number in the list and Xmin is the smallest number in the list. Each counter has to be able to represent N states; i.e. assuming a binary representation it has to have an integer type (at least) ceiling(log2(N)) bits.
To determine the array size, a counting sort needs to make an initial pass through the list to determine Xmax and Xmin.
The minimum worst-case space requirement is therefore ceiling(log2(N)) * (Xmax - Xmin) bits.
By contrast, the algorithm presented above simply requires N bits in the worst and best cases.
However, this analysis leads to the intuition that if the algorithm made an initial pass through the list looking for a zero (and counting the list elements if required), it would give a quicker answer using no space at all if it found the zero. It is definitely worth doing this if there is a high probability of finding at least one zero in the list. And this extra pass doesn't change the overall complexity.
EDIT: I've changed the description of the algorithm to use "array of booleans" since people apparently found my original description using bits and bitmaps to be confusing.
Since the OP has now specified that the original list is held in RAM and that the computer has only, say, 1GB of memory, I'm going to go out on a limb and predict that the answer is zero.
1GB of RAM means the list can have at most 134,217,728 numbers in it. But there are 264 = 18,446,744,073,709,551,616 possible numbers. So the probability that zero is in the list is 1 in 137,438,953,472.
In contrast, my odds of being struck by lightning this year are 1 in 700,000. And my odds of getting hit by a meteorite are about 1 in 10 trillion. So I'm about ten times more likely to be written up in a scientific journal due to my untimely death by a celestial object than the answer not being zero.
As pointed out in other answers you can do a sort, and then simply scan up until you find a gap.
You can improve the algorithmic complexity to O(N) and keep O(N) space by using a modified QuickSort where you eliminate partitions which are not potential candidates for containing the gap.
On the first partition phase, remove duplicates.
Once the partitioning is complete look at the number of items in the lower partition
Is this value equal to the value used for creating the partition?
If so then it implies that the gap is in the higher partition.
Continue with the quicksort, ignoring the lower partition
Otherwise the gap is in the lower partition
Continue with the quicksort, ignoring the higher partition
This saves a large number of computations.
To illustrate one of the pitfalls of O(N) thinking, here is an O(N) algorithm that uses O(1) space.
for i in [0..2^64):
if i not in list: return i
print "no 64-bit integers are missing"
Since the numbers are all 64 bits long, we can use radix sort on them, which is O(n). Sort 'em, then scan 'em until you find what you're looking for.
if the smallest number is zero, scan forward until you find a gap. If the smallest number is not zero, the answer is zero.
For a space efficient method and all values are distinct you can do it in space O( k ) and time O( k*log(N)*N ). It's space efficient and there's no data moving and all operations are elementary (adding subtracting).
set U = N; L=0
First partition the number space in k regions. Like this:
0->(1/k)*(U-L) + L, 0->(2/k)*(U-L) + L, 0->(3/k)*(U-L) + L ... 0->(U-L) + L
Find how many numbers (count{i}) are in each region. (N*k steps)
Find the first region (h) that isn't full. That means count{h} < upper_limit{h}. (k steps)
if h - count{h-1} = 1 you've got your answer
set U = count{h}; L = count{h-1}
goto 2
this can be improved using hashing (thanks for Nic this idea).
same
First partition the number space in k regions. Like this:
L + (i/k)->L + (i+1/k)*(U-L)
inc count{j} using j = (number - L)/k (if L < number < U)
find first region (h) that doesn't have k elements in it
if count{h} = 1 h is your answer
set U = maximum value in region h L = minimum value in region h
This will run in O(log(N)*N).
I'd just sort them then run through the sequence until I find a gap (including the gap at the start between zero and the first number).
In terms of an algorithm, something like this would do it:
def smallest_not_in_list(list):
sort(list)
if list[0] != 0:
return 0
for i = 1 to list.last:
if list[i] != list[i-1] + 1:
return list[i-1] + 1
if list[list.last] == 2^64 - 1:
assert ("No gaps")
return list[list.last] + 1
Of course, if you have a lot more memory than CPU grunt, you could create a bitmask of all possible 64-bit values and just set the bits for every number in the list. Then look for the first 0-bit in that bitmask. That turns it into an O(n) operation in terms of time but pretty damned expensive in terms of memory requirements :-)
I doubt you could improve on O(n) since I can't see a way of doing it that doesn't involve looking at each number at least once.
The algorithm for that one would be along the lines of:
def smallest_not_in_list(list):
bitmask = mask_make(2^64) // might take a while :-)
mask_clear_all (bitmask)
for i = 1 to list.last:
mask_set (bitmask, list[i])
for i = 0 to 2^64 - 1:
if mask_is_clear (bitmask, i):
return i
assert ("No gaps")
Sort the list, look at the first and second elements, and start going up until there is a gap.
We could use a hash table to hold the numbers. Once all numbers are done, run a counter from 0 till we find the lowest. A reasonably good hash will hash and store in constant time, and retrieves in constant time.
for every i in X // One scan Θ(1)
hashtable.put(i, i); // O(1)
low = 0;
while (hashtable.get(i) <> null) // at most n+1 times
low++;
print low;
The worst case if there are n elements in the array, and are {0, 1, ... n-1}, in which case, the answer will be obtained at n, still keeping it O(n).
You can do it in O(n) time and O(1) additional space, although the hidden factor is quite large. This isn't a practical way to solve the problem, but it might be interesting nonetheless.
For every unsigned 64-bit integer (in ascending order) iterate over the list until you find the target integer or you reach the end of the list. If you reach the end of the list, the target integer is the smallest integer not in the list. If you reach the end of the 64-bit integers, every 64-bit integer is in the list.
Here it is as a Python function:
def smallest_missing_uint64(source_list):
the_answer = None
target = 0L
while target < 2L**64:
target_found = False
for item in source_list:
if item == target:
target_found = True
if not target_found and the_answer is None:
the_answer = target
target += 1L
return the_answer
This function is deliberately inefficient to keep it O(n). Note especially that the function keeps checking target integers even after the answer has been found. If the function returned as soon as the answer was found, the number of times the outer loop ran would be bound by the size of the answer, which is bound by n. That change would make the run time O(n^2), even though it would be a lot faster.
Thanks to egon, swilden, and Stephen C for my inspiration. First, we know the bounds of the goal value because it cannot be greater than the size of the list. Also, a 1GB list could contain at most 134217728 (128 * 2^20) 64-bit integers.
Hashing part
I propose using hashing to dramatically reduce our search space. First, square root the size of the list. For a 1GB list, that's N=11,586. Set up an integer array of size N. Iterate through the list, and take the square root* of each number you find as your hash. In your hash table, increment the counter for that hash. Next, iterate through your hash table. The first bucket you find that is not equal to it's max size defines your new search space.
Bitmap part
Now set up a regular bit map equal to the size of your new search space, and again iterate through the source list, filling out the bitmap as you find each number in your search space. When you're done, the first unset bit in your bitmap will give you your answer.
This will be completed in O(n) time and O(sqrt(n)) space.
(*You could use use something like bit shifting to do this a lot more efficiently, and just vary the number and size of buckets accordingly.)
Well if there is only one missing number in a list of numbers, the easiest way to find the missing number is to sum the series and subtract each value in the list. The final value is the missing number.
int i = 0;
while ( i < Array.Length)
{
if (Array[i] == i + 1)
{
i++;
}
if (i < Array.Length)
{
if (Array[i] <= Array.Length)
{//SWap
int temp = Array[i];
int AnoTemp = Array[temp - 1];
Array[temp - 1] = temp;
Array[i] = AnoTemp;
}
else
i++;
}
}
for (int j = 0; j < Array.Length; j++)
{
if (Array[j] > Array.Length)
{
Console.WriteLine(j + 1);
j = Array.Length;
}
else
if (j == Array.Length - 1)
Console.WriteLine("Not Found !!");
}
}
Here's my answer written in Java:
Basic Idea:
1- Loop through the array throwing away duplicate positive, zeros, and negative numbers while summing up the rest, getting the maximum positive number as well, and keep the unique positive numbers in a Map.
2- Compute the sum as max * (max+1)/2.
3- Find the difference between the sums calculated at steps 1 & 2
4- Loop again from 1 to the minimum of [sums difference, max] and return the first number that is not in the map populated in step 1.
public static int solution(int[] A) {
if (A == null || A.length == 0) {
throw new IllegalArgumentException();
}
int sum = 0;
Map<Integer, Boolean> uniqueNumbers = new HashMap<Integer, Boolean>();
int max = A[0];
for (int i = 0; i < A.length; i++) {
if(A[i] < 0) {
continue;
}
if(uniqueNumbers.get(A[i]) != null) {
continue;
}
if (A[i] > max) {
max = A[i];
}
uniqueNumbers.put(A[i], true);
sum += A[i];
}
int completeSum = (max * (max + 1)) / 2;
for(int j = 1; j <= Math.min((completeSum - sum), max); j++) {
if(uniqueNumbers.get(j) == null) { //O(1)
return j;
}
}
//All negative case
if(uniqueNumbers.isEmpty()) {
return 1;
}
return 0;
}
As Stephen C smartly pointed out, the answer must be a number smaller than the length of the array. I would then find the answer by binary search. This optimizes the worst case (so the interviewer can't catch you in a 'what if' pathological scenario). In an interview, do point out you are doing this to optimize for the worst case.
The way to use binary search is to subtract the number you are looking for from each element of the array, and check for negative results.
I like the "guess zero" apprach. If the numbers were random, zero is highly probable. If the "examiner" set a non-random list, then add one and guess again:
LowNum=0
i=0
do forever {
if i == N then leave /* Processed entire array */
if array[i] == LowNum {
LowNum++
i=0
}
else {
i++
}
}
display LowNum
The worst case is n*N with n=N, but in practice n is highly likely to be a small number (eg. 1)
I am not sure if I got the question. But if for list 1,2,3,5,6 and the missing number is 4, then the missing number can be found in O(n) by:
(n+2)(n+1)/2-(n+1)n/2
EDIT: sorry, I guess I was thinking too fast last night. Anyway, The second part should actually be replaced by sum(list), which is where O(n) comes. The formula reveals the idea behind it: for n sequential integers, the sum should be (n+1)*n/2. If there is a missing number, the sum would be equal to the sum of (n+1) sequential integers minus the missing number.
Thanks for pointing out the fact that I was putting some middle pieces in my mind.
Well done Ants Aasma! I thought about the answer for about 15 minutes and independently came up with an answer in a similar vein of thinking to yours:
#define SWAP(x,y) { numerictype_t tmp = x; x = y; y = tmp; }
int minNonNegativeNotInArr (numerictype_t * a, size_t n) {
int m = n;
for (int i = 0; i < m;) {
if (a[i] >= m || a[i] < i || a[i] == a[a[i]]) {
m--;
SWAP (a[i], a[m]);
continue;
}
if (a[i] > i) {
SWAP (a[i], a[a[i]]);
continue;
}
i++;
}
return m;
}
m represents "the current maximum possible output given what I know about the first i inputs and assuming nothing else about the values until the entry at m-1".
This value of m will be returned only if (a[i], ..., a[m-1]) is a permutation of the values (i, ..., m-1). Thus if a[i] >= m or if a[i] < i or if a[i] == a[a[i]] we know that m is the wrong output and must be at least one element lower. So decrementing m and swapping a[i] with the a[m] we can recurse.
If this is not true but a[i] > i then knowing that a[i] != a[a[i]] we know that swapping a[i] with a[a[i]] will increase the number of elements in their own place.
Otherwise a[i] must be equal to i in which case we can increment i knowing that all the values of up to and including this index are equal to their index.
The proof that this cannot enter an infinite loop is left as an exercise to the reader. :)
The Dafny fragment from Ants' answer shows why the in-place algorithm may fail. The requires pre-condition describes that the values of each item must not go beyond the bounds of the array.
method AntsAasma(A: array<int>) returns (M: int)
requires A != null && forall N :: 0 <= N < A.Length ==> 0 <= A[N] < A.Length;
modifies A;
{
// Pass 1, move every value to the position of its value
var N := A.Length;
var cursor := 0;
while (cursor < N)
{
var target := A[cursor];
while (0 <= target < N && target != A[target])
{
var new_target := A[target];
A[target] := target;
target := new_target;
}
cursor := cursor + 1;
}
// Pass 2, find first location where the index doesn't match the value
cursor := 0;
while (cursor < N)
{
if (A[cursor] != cursor)
{
return cursor;
}
cursor := cursor + 1;
}
return N;
}
Paste the code into the validator with and without the forall ... clause to see the verification error. The second error is a result of the verifier not being able to establish a termination condition for the Pass 1 loop. Proving this is left to someone who understands the tool better.
Here's an answer in Java that does not modify the input and uses O(N) time and N bits plus a small constant overhead of memory (where N is the size of the list):
int smallestMissingValue(List<Integer> values) {
BitSet bitset = new BitSet(values.size() + 1);
for (int i : values) {
if (i >= 0 && i <= values.size()) {
bitset.set(i);
}
}
return bitset.nextClearBit(0);
}
def solution(A):
index = 0
target = []
A = [x for x in A if x >=0]
if len(A) ==0:
return 1
maxi = max(A)
if maxi <= len(A):
maxi = len(A)
target = ['X' for x in range(maxi+1)]
for number in A:
target[number]= number
count = 1
while count < maxi+1:
if target[count] == 'X':
return count
count +=1
return target[count-1] + 1
Got 100% for the above solution.
1)Filter negative and Zero
2)Sort/distinct
3)Visit array
Complexity: O(N) or O(N * log(N))
using Java8
public int solution(int[] A) {
int result = 1;
boolean found = false;
A = Arrays.stream(A).filter(x -> x > 0).sorted().distinct().toArray();
//System.out.println(Arrays.toString(A));
for (int i = 0; i < A.length; i++) {
result = i + 1;
if (result != A[i]) {
found = true;
break;
}
}
if (!found && result == A.length) {
//result is larger than max element in array
result++;
}
return result;
}
An unordered_set can be used to store all the positive numbers, and then we can iterate from 1 to length of unordered_set, and see the first number that does not occur.
int firstMissingPositive(vector<int>& nums) {
unordered_set<int> fre;
// storing each positive number in a hash.
for(int i = 0; i < nums.size(); i +=1)
{
if(nums[i] > 0)
fre.insert(nums[i]);
}
int i = 1;
// Iterating from 1 to size of the set and checking
// for the occurrence of 'i'
for(auto it = fre.begin(); it != fre.end(); ++it)
{
if(fre.find(i) == fre.end())
return i;
i +=1;
}
return i;
}
Solution through basic javascript
var a = [1, 3, 6, 4, 1, 2];
function findSmallest(a) {
var m = 0;
for(i=1;i<=a.length;i++) {
j=0;m=1;
while(j < a.length) {
if(i === a[j]) {
m++;
}
j++;
}
if(m === 1) {
return i;
}
}
}
console.log(findSmallest(a))
Hope this helps for someone.
With python it is not the most efficient, but correct
#!/usr/bin/env python3
# -*- coding: UTF-8 -*-
import datetime
# write your code in Python 3.6
def solution(A):
MIN = 0
MAX = 1000000
possible_results = range(MIN, MAX)
for i in possible_results:
next_value = (i + 1)
if next_value not in A:
return next_value
return 1
test_case_0 = [2, 2, 2]
test_case_1 = [1, 3, 44, 55, 6, 0, 3, 8]
test_case_2 = [-1, -22]
test_case_3 = [x for x in range(-10000, 10000)]
test_case_4 = [x for x in range(0, 100)] + [x for x in range(102, 200)]
test_case_5 = [4, 5, 6]
print("---")
a = datetime.datetime.now()
print(solution(test_case_0))
print(solution(test_case_1))
print(solution(test_case_2))
print(solution(test_case_3))
print(solution(test_case_4))
print(solution(test_case_5))
def solution(A):
A.sort()
j = 1
for i, elem in enumerate(A):
if j < elem:
break
elif j == elem:
j += 1
continue
else:
continue
return j
this can help:
0- A is [5, 3, 2, 7];
1- Define B With Length = A.Length; (O(1))
2- initialize B Cells With 1; (O(n))
3- For Each Item In A:
if (B.Length <= item) then B[Item] = -1 (O(n))
4- The answer is smallest index in B such that B[index] != -1 (O(n))