I maintain an array of integers. It is important that at all times the integers in this array are in sequence from 0. For example, if there are 5 integers in the array, their values must be 0, 1, 2, 3, 4 (though in any order).
I would like to design a simple, efficient method that checks this. It will return true if the array contains all positive integers in sequence from 0 to array.count - 1.
I would love to hear some different ideas for handling this!
Or, given the integers [0..N-1] if you raise 2 to the power of each in turn the sum will be -1+2^N. This is not a property that any other set of N integers has.
I offer this as an alternative, making no claim about suitability, performance or efficiency, and I recognise that there will be problems as N gets large.
Basically what you want to test is if your array is a permutation of [0...n-1]. There are easy algorithms for this that are O(n) in both time and memory. See for example this PDF file.
Sometimes I use a very simple check that is O(n) in time and O(1) in memory. It can in theory return false positives, but it is a good way to find most mistakes. It is based on the following facts:
0 + 1 + 2 + ... + n-1 == n * (n-1) / 2
0 + 1² + 2² + ... + (n-1)² == n * (n-1) * (2 * n - 1) / 6
I don't know objective-c, but the code would look like this in C#:
bool IsPermutation(int[] array)
{
long length = array.Length;
long total1 = 0;
long total2 = 0;
for (int i = 0; i<length; i++)
{
total1 += array[i];
total2 += (long)array[i] * array[i];
}
return
2 * total1 == length * (length - 1) &&
6 * total2 == length * (length - 1) * (2 * length - 1);
}
This isn't too different from your itemsSequencedCorrectlyInSet: method, but it uses a mutable index set which will be faster than doing -[NSSet containsObject:]. Probably not an issue until you've got thousands of table rows. Anyway, the key insight here is that the Pigeonhole Principle says if you've got N integers less than N and none is duplicated, then you have each of 0...N-1 exactly once.
-(BOOL)listIsValid:(NSArray*)list
{
NSMutableIndexSet* seen = [NSMutableIndexSet indexSet];
for ( NSNumber* number in list )
{
NSUInteger n = [number unsignedIntegerValue];
if ( n >= [array count] || [seen containsIndex:n] )
return NO;
[seen addIndex:n];
}
return YES;
}
Here's my current implementation (testSet is a set of NSNumbers) -
- (BOOL)itemsSequencedCorrectlyInSet:(NSSet *)testSet{
for (NSInteger i = 0; i < testSet.count; i++) {
if (![testSet containsObject:[NSNumber numberWithInteger:i]]) {
return NO;
}
}
return YES;
}
Related
I am trying to find the tightest upper bound for the following upper bound. However, I am not able to get the correct answer. The algorithm is as follows:
public staticintrecursiveloopy(int n){
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
System.out.println("Hello.");
}
} if(n <= 2) {
return 1;
} else if(n % 2 == 0) {
return(staticintrecursiveloopy(n+1));
} else{
return(staticintrecursiveloopy(n-2));
}
}
I tried to draw out the recursion tree for this. I know that for each run of the algorithm the time complexity will be O(n2) plus the time taken for each of the recursive calls. Also, the recursion tree will have n levels. I then calculated the total time taken for each level:
For the first level, the time taken will be n2. For the second level, since there are two recursive calls, the time taken will be 2n2. For the third level, the time taken will be 4n 2 and so on until n becomes <= 2.
Thus, the time complexity should be n2 * (1 + 2 + 4 + .... + 2n). 1 + 2 + 4 + .... + 2n is a geometric sequence and its sum is equal to 2n - 1.Thus, the total time complexity should be O(2nn2). However, the answer says O(n3). What am I doing wrong?
Consider the below fragment
for(int i = 0; i < n; i++) {
for(int j = 0; j < n; j++) {
System.out.println("Hello.");
}
}
This doesn't need any introduction and is O(n2)
Now consider the below fragment
if(n <= 2) {
return 1;
} else if(n % 2 == 0) {
return(staticintrecursiveloopy(n+1));
} else {
return(staticintrecursiveloopy(n-2));
}
How many times do you think this fragment will be executed?
If n%2 == 0 then the method staticintrecursiveloopy will be executed 1 extra time. Otherwise it goes about decresing it by 2, thus it'll be executed n/2 times (or (n+1)/2 if you include the other condition).
Thus the total number of times the method staticintrecursiveloopy will be executed is roughly n/2 which when expressed in terms of complexity becomes O(n).
And the method staticintrecursiveloopy calls a part with complexity O(n2) in each iteration, thus the total time complexity becomes
O(n) * O(n2) = O(n3).
I'm referring to the leetcode question: Kth Smallest Element in a Sorted Matrix
There are two well-known solutions to the problem. One using Heap/PriorityQueue and other is using Binary Search. The Binary Search solution goes like this (top post):
public class Solution {
public int kthSmallest(int[][] matrix, int k) {
int lo = matrix[0][0], hi = matrix[matrix.length - 1][matrix[0].length - 1] + 1;//[lo, hi)
while(lo < hi) {
int mid = lo + (hi - lo) / 2;
int count = 0, j = matrix[0].length - 1;
for(int i = 0; i < matrix.length; i++) {
while(j >= 0 && matrix[i][j] > mid) j--;
count += (j + 1);
}
if(count < k) lo = mid + 1;
else hi = mid;
}
return lo;
}
}
While I understand how this works, I have trouble figuring out one issue.
How can we be sure that the returned lo is always in the matrix?
Since the search space is min and max value of the array, the mid need NOT be a value that is in the array. However, the returned lo always is.
Why is this happening?
For the sake of argument, we can move the calculation of count to a separate function like the following:
bool valid(int mid, int[][] matrix, int k) {
int count = 0, m = matrix.length;
for (int i = 0; i < m; i++) {
int j = 0;
while (j < m && matrix[i][j] <= mid) j++;
count += j;
}
return (count < k);
}
This predicate will do exactly same as your specified operation. Here, the loop invariant is that, the range [lo, hi] always contains the kth smallest number of the 2D array.
In other words, lo <= solution <= hi
Now, when the loop terminates, it is evident that lo >= hi
Merging those two properties, we get, lo = solution = hi, since solution is a member of array, it can be said that, lo is always in the array after loop termination and will rightly point to the kth smallest element.
Because We are finding the lower_bound using binary search and there cannot be any number smaller than the number(lo) in the array which could be the kth smallest element.
I have a method that finds 3 numbers in an array that add up to a desired number.
code:
public static void threeSum(int[] arr, int sum) {
quicksort(arr, 0, arr.length - 1);
for (int i = 0; i < arr.length - 2; i++) {
for (int j = 1; j < arr.length - 1; j++) {
for (int k = arr.length - 1; k > j; k--) {
if ((arr[i] + arr[j] + arr[k]) == sum) {
System.out.println(Integer.toString(i) + "+" + Integer.toString(j) + "+" + Integer.toString(k) + "=" + sum);
}
}
}
}
}
I'm not sure about the big O of this method. I have a hard time wrapping my head around this right now. My guess is O(n^2) or O(n^2logn). But these are complete guesses. I can't prove this. Could someone help me wrap my head around this?
You have three runs over the array (the i, j and k loops), in sizes that depend primarily on n, the size of the array. Hence, this is an O(n3) operation.
Even though your quicksort is O(nlogn), it is overshadowed by the fact that you have 3 nested for loops. So the time complexity w.r.t number of elements (n) is O(n^3)
It's an O(n^3) complexity because there are three nested forloops. The inner forloop only runs if k>j so you can think of n^2*(n/2) but in big O notation you can ignore this.
Methodically speaking, the order of growth complexity can be accurately inferred like the following:
I've been looking for some simple coding challenges recently, and discovered about Pascal's triangle (here), and I've tried to generate one myself in C/Objective-C. For those that don't know what it is, that link explains it pretty well.
I'm starting to get oddness after the fourth row, and I just can't figure out why.
My output for 5 iterations currently looks like this:
1
1 1
1 2 1
1 3 3 1
4 6 3 1
It should look like this:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
Here is my code so far. The first loop is just a reset loop (setting all the values to 0). The actual logic happens mostly in the second loop. The third loop is where the values are concatenated and formatted in a string.
I've commented this code much more than I would for myself just to aid readability.
int iterations, i, b, mid, chars, temp;
NSLog(#"Please enter the number of itereations");
scanf("%i",&iterations); // take users input and store it in iterations
// calculate where the first 1 should go.
if (iterations % 2 == 0) mid = (iterations)/2;
else mid = (iterations+1)/2;
chars = iterations*2;
int solutions[iterations][chars];
// reset loop
for (i = 0; i<iterations; i++) {
for (b = 0; b<chars; b++) {
solutions[i][b] = 0;
}
}
solutions[0][mid] = 1; // place the initial 1 in first row
for (int row = 1; row<iterations; row++) {
for (int chi = 0; chi<chars; chi++) {
temp = 0;
if (chi > 0) {
temp += solutions[row-1][chi-1]; // add the one diagonally left
}
if (chi < iterations) {
temp += solutions[row-1][chi+1]; // add the one diagonally right
}
solutions[row][chi] = temp; // set the value
}
}
// printing below...
NSMutableString *result = [[NSMutableString alloc] initWithString:#"\n"];
NSMutableString *rowtmp;
for (i = 0; i<iterations; i++) {
rowtmp = [NSMutableString stringWithString:#""];
for (b = 0; b<chars; b++) {
if (solutions[i][b] != 0) [rowtmp appendFormat:#"%i",solutions[i][b]];
else [rowtmp appendString:#" "]; // replace any 0s with spaces.
}
[result appendFormat:#"%#\n",rowtmp];
}
NSLog(#"%#",result);
[result release];
I have a feeling the problem may be to do with the offset, but I have no idea how to fix it. If anyone can spot where my code is going wrong, that would be great.
It appears (from a brief look) that the original midpoint calculation is incorrect. I think it should simply be:
mid = iterations - 1;
In the example of 5 iterations, the midpoint needs to be at array position 4. Each iteration "moves" one more position to the left. The 2nd iteration (2nd row) would then place a 1 at positions 3 and 5. The 3rd iteration at 2 and 6. The 4th at 1 and 7. And the 5th and last iteration would fill in the 1s at 0 and 8.
Also, the second if statement for the temp addition should be as follows otherwise it reads past the end of the array bounds:
if (chi < iterations - 1) {
How many possible combinations of the variables a,b,c,d,e are possible if I know that:
a+b+c+d+e = 500
and that they are all integers and >= 0, so I know they are finite.
#Torlack, #Jason Cohen: Recursion is a bad idea here, because there are "overlapping subproblems." I.e., If you choose a as 1 and b as 2, then you have 3 variables left that should add up to 497; you arrive at the same subproblem by choosing a as 2 and b as 1. (The number of such coincidences explodes as the numbers grow.)
The traditional way to attack such a problem is dynamic programming: build a table bottom-up of the solutions to the sub-problems (starting with "how many combinations of 1 variable add up to 0?") then building up through iteration (the solution to "how many combinations of n variables add up to k?" is the sum of the solutions to "how many combinations of n-1 variables add up to j?" with 0 <= j <= k).
public static long getCombos( int n, int sum ) {
// tab[i][j] is how many combinations of (i+1) vars add up to j
long[][] tab = new long[n][sum+1];
// # of combos of 1 var for any sum is 1
for( int j=0; j < tab[0].length; ++j ) {
tab[0][j] = 1;
}
for( int i=1; i < tab.length; ++i ) {
for( int j=0; j < tab[i].length; ++j ) {
// # combos of (i+1) vars adding up to j is the sum of the #
// of combos of i vars adding up to k, for all 0 <= k <= j
// (choosing i vars forces the choice of the (i+1)st).
tab[i][j] = 0;
for( int k=0; k <= j; ++k ) {
tab[i][j] += tab[i-1][k];
}
}
}
return tab[n-1][sum];
}
$ time java Combos
2656615626
real 0m0.151s
user 0m0.120s
sys 0m0.012s
The answer to your question is 2656615626.
Here's the code that generates the answer:
public static long getNumCombinations( int summands, int sum )
{
if ( summands <= 1 )
return 1;
long combos = 0;
for ( int a = 0 ; a <= sum ; a++ )
combos += getNumCombinations( summands-1, sum-a );
return combos;
}
In your case, summands is 5 and sum is 500.
Note that this code is slow. If you need speed, cache the results from summand,sum pairs.
I'm assuming you want numbers >=0. If you want >0, replace the loop initialization with a = 1 and the loop condition with a < sum. I'm also assuming you want permutations (e.g. 1+2+3+4+5 plus 2+1+3+4+5 etc). You could change the for-loop if you wanted a >= b >= c >= d >= e.
I solved this problem for my dad a couple months ago...extend for your use. These tend to be one time problems so I didn't go for the most reusable...
a+b+c+d = sum
i = number of combinations
for (a=0;a<=sum;a++)
{
for (b = 0; b <= (sum - a); b++)
{
for (c = 0; c <= (sum - a - b); c++)
{
//d = sum - a - b - c;
i++
}
}
}
This would actually be a good question to ask on an interview as it is simple enough that you could write up on a white board, but complex enough that it might trip someone up if they don't think carefully enough about it. Also, you can also for two different answers which cause the implementation to be quite different.
Order Matters
If the order matters then any solution needs to allow for zero to appear for any of the variables; thus, the most straight forward solution would be as follows:
public class Combos {
public static void main() {
long counter = 0;
for (int a = 0; a <= 500; a++) {
for (int b = 0; b <= (500 - a); b++) {
for (int c = 0; c <= (500 - a - b); c++) {
for (int d = 0; d <= (500 - a - b - c); d++) {
counter++;
}
}
}
}
System.out.println(counter);
}
}
Which returns 2656615626.
Order Does Not Matter
If the order does not matter then the solution is not that much harder as you just need to make sure that zero isn't possible unless sum has already been found.
public class Combos {
public static void main() {
long counter = 0;
for (int a = 1; a <= 500; a++) {
for (int b = (a != 500) ? 1 : 0; b <= (500 - a); b++) {
for (int c = (a + b != 500) ? 1 : 0; c <= (500 - a - b); c++) {
for (int d = (a + b + c != 500) ? 1 : 0; d <= (500 - a - b - c); d++) {
counter++;
}
}
}
}
System.out.println(counter);
}
}
Which returns 2573155876.
One way of looking at the problem is as follows:
First, a can be any value from 0 to 500. Then if follows that b+c+d+e = 500-a. This reduces the problem by one variable. Recurse until done.
For example, if a is 500, then b+c+d+e=0 which means that for the case of a = 500, there is only one combination of values for b,c,d and e.
If a is 300, then b+c+d+e=200, which is in fact the same problem as the original problem, just reduced by one variable.
Note: As Chris points out, this is a horrible way of actually trying to solve the problem.
link text
If they are a real numbers then infinite ... otherwise it is a bit trickier.
(OK, for any computer representation of a real number there would be a finite count ... but it would be big!)
It has general formulae, if
a + b + c + d = N
Then number of non-negative integral solution will be C(N + number_of_variable - 1, N)
#Chris Conway answer is correct. I have tested with a simple code that is suitable for smaller sums.
long counter = 0;
int sum=25;
for (int a = 0; a <= sum; a++) {
for (int b = 0; b <= sum ; b++) {
for (int c = 0; c <= sum; c++) {
for (int d = 0; d <= sum; d++) {
for (int e = 0; e <= sum; e++) {
if ((a+b+c+d+e)==sum) counter=counter+1L;
}
}
}
}
}
System.out.println("counter e "+counter);
The answer in math is 504!/(500! * 4!).
Formally, for x1+x2+...xk=n, the number of combination of nonnegative number x1,...xk is the binomial coefficient: (k-1)-combination out of a set containing (n+k-1) elements.
The intuition is to choose (k-1) points from (n+k-1) points and use the number of points between two chosen points to represent a number in x1,..xk.
Sorry about the poor math edition for my fist time answering Stack Overflow.
Just a test for code block
Just a test for code block
Just a test for code block
Including negatives? Infinite.
Including only positives? In this case they wouldn't be called "integers", but "naturals", instead. In this case... I can't really solve this, I wish I could, but my math is too rusty. There is probably some crazy integral way to solve this. I can give some pointers for the math skilled around.
being x the end result,
the range of a would be from 0 to x,
the range of b would be from 0 to (x - a),
the range of c would be from 0 to (x - a - b),
and so forth until the e.
The answer is the sum of all those possibilities.
I am trying to find some more direct formula on Google, but I am really low on my Google-Fu today...