I am coding brute force approach for one coding problem - I need to count the maximum score path in the array with maximum step k.
Input: nums = [1,-1,-2,4,-7,3], k = 2
Output: 7
Explanation: You can choose your jumps forming the subsequence [1,-1,4,3] (underlined above). The sum is 7.
And I encountered a problem with calculating complexity. My thought was that on each elemnt we may call function k times, so time and space are O(k^n), where n is length of the array. My second guess: for first element we call function at most 1 time, for second 2 times (that is if k > i) and so on. So we have sum 1 + 2 + ... + k + k + ... + k = ((1 + k) / 2)k + ((k + k) / 2) / (n-k) = O(k^2). I think the first one is correct, but I can't tell for sure why :/
Here's my Java code:
public int maxResult(int[] nums, int k) {
return maxResult(nums, k, nums.length - 1);
}
private int maxResult(int[] nums, int k, int index) {
if (index == 0)
return nums[0];
int max = Integer.MIN_VALUE;
int start = index - k < 0 ? 0 : index - k;
for ( int i = start; i < index; i++ ) {
int res = maxResult(nums, k, i);
System.out.println(i);
max = Math.max(res, max);
}
return max + nums[index];
}
The recurrence relation for your code for a particular k is
C(n) = sum(C(n-i) for i = 1...k) for n>k
C(n) = C(1) + C(2) + ... + C(n-1) for n <= k
C(1) = 1
These are the recurrence relations for the higher-order Fibonacci numbers, shifted by k-1 places. That is, C(n) = kFib(k, n+k-1). The k-Fibonacci numbers grow as Theta(alpha^n) where alpha is some constant based on k -- for k=2, alpha is the golden ratio, and as k increases, alpha gets closer and closer to 2. (Specifically, alpha is is the positive root of (x^k - x^(k-1) - ... - x - 1)).
Therefore C(n) = kFib(k, n+k-1) = Theta(alpha^(n+k)).
Because alpha is always less than 2, O(2^(n+k)) is a simple correct bound, although not a tight one.
Related
Title says it all.
I need to split n as sum of k parts where each part ki should be in the range of
1 <= ki <= ri for given array r.
for example -
n = 4, k = 3 and r = [2, 2, 1]
ans = 2
#[2, 1, 1], [1, 2, 1]
Order matters. (2, 1, 1) and (1, 2, 1) are different.
I taught of solving it using stars and bars method, but be because of upper bound ri i dont know to to approach it.
i implemented a direct recursion function and it works fine for small values only.
Constraints of original problem are
1 <= n <= 107
1 <= k <= 105
1 <= ri <= 51
All calculations will be done under prime Modulo.
i found a similar problem here but i don't know how to implement in program. HERE
My brute-force recursive function -
#define MAX 1000
const int md = 1e9 + 7;
vector <int> k;
vector <map<int, int>> mapper;
vector <int> hold;
int solve(int sum, int cur){
if(cur == (k.size() - 1) && sum >= 1 && sum <= k[cur]) return 1;
if(cur == (k.size() - 1) && (sum < 1 || sum > k[cur])) return 0;
if(mapper[cur].find(sum) != mapper[cur].end())
return mapper[cur][sum];
int ans = 0;
int start = 1;
for(int i=start; i<=k[cur]; ++i){
int remain = sum - i;
int seg = (k.size() - cur) - 1;
if(remain < seg) break;
int res = solve(sum - i, cur + 1);
ans = (1LL * ans + res) % md;
}
mapper[cur][sum] = ans;
return ans;
}
int main(){
for(int i=0; i<MAX; ++i) k.push_back(51); // restriction for each part default 51
mapper.resize(MAX);
cout << solve(MAX + MAX, 0) << endl;
}
Instead of using a map for storing result of computation i used a two dimensional array and it gave very good performance boost but i cannot use it because of large n and k values.
How could i improve my recursive function or what are other ways of solving this problem.
That's interesting problem.
First lets say r_i = r_i - 1, n = n - k, numbers in [0, r_i] just for convenience. Now it's possible to add some fictitious numbers to make m the power of 2 without changing answer.
Now let's represent each interval of [0, r_i] as polynomial 1 * x ^ 0 + 1 * x ^ 1 + ... + 1 * x & r_i. Now if we multiply all these polynomials, coefficient at x ^ n will be answer.
Here is structure called Number Theoretic Transform (NTT) which allows to multiply two polynomials modulo p in O(size * log(size)).
If you will just multiply it using NTT, code will work in something like O(n * k * log (k * max(r))). It's very slow.
But now our fictive numbers help. Let's use divide and conquer technics. We'll make O(log m) steps, on each step multiply 2 * i-th and 2 * i + 1-th polynomials. In the next step we'll multiply resulting polynomials of this step.
Each step works in O(k * log(k)) and there is O(log(k)) steps, so algorhitm works in O(k * log^2 (k)). It's fast asymptotically, but I'm not sure if it fits TL for this problem. I think it will work about 20 seconds on max test.
how to find the maximum sum of the numbers starting from 1 to n but at any time the current sum should never be equal to a given number k.
This is what I have done but its not correct all the time.
static int maxMoney(int n, long k) {
int currentSum = 0;
for(int i = 1; i <= n; i++){
currentSum += i;
if(currentSum == k){
currentSum -= 1;
continue;
}
}
return currentSum;
}
constraints
1<= n <= 2 * 10^9
1<= k <= 4*10^15
So if n = 3 and k = 3;
and if we add 1 + 2 we get 3, we reject 1 and start at 2 and do 2+3 to get 5.
We can also do 1+3 and we get the value 4 in this case but since we want maximum we choose 5 over 4 and hence 5 is the answer in this case.
Assume first you can just sum all numbers: sum = 1+2+...+n, which can be calculated as sum = n(n+1)/2.
Then check if there is an i between 1 and n, such that 1+2+...+i = k. If not, then the previously calculated sum is the solution.
If there is such i, then reject 1, and return sum - 1.
Determine whether such i exists by solving the following equation:
k = i(i+1)/2
This quadratic equation has this positive solution:
i = (sqrt(1+8k) - 1) / 2
If this i is an integer number, and it is within the interval 1...n, then we have indeed found such an i. Otherwise, we can conclude that k will never be equal to 1+2+...+i for any i in the interval 1...n.
So here is a little JavaScript function that implements this:
function getSum(n, k) {
var sum, i;
sum = n*(n+1)/2;
i = (Math.sqrt(1+8*k) - 1) / 2;
if (i === Math.floor(i) && i <= n)
sum--;
return sum;
}
// Sample call
console.log(getSum(3, 3)); // = 5
Given N numbers I need to count subsets whose sum is S.
Note : Numbers in array need not to be distinct.
My current code is :
int countSubsets(vector<int> numbers,int sum)
{
vector<int> DP(sum+1);
DP[0]=1;
int currentSum=0;
for(int i=0;i<numbers.size();i++)
{
currentSum+=numbers[i];
for (int j=min(sum,currentSum);j>=numbers[i];j--)
DP[j]+=DP[j - numbers[i]];
}
return DP[sum];
}
Can their be any efficient way than this ?
Constraints are :
1 ≤ N ≤ 14
1 ≤ S ≤ 100000
1 ≤ A[i] ≤ 10000
Also their are 100 test cases in a single file. So please help if their exist better solution than this one
N is small (2^20 - is about 1 milion - 2^14 is really small value) - just iterate over all subsets, below I wrote pretty fast way to do that (bithacking). Treat integers as sets (that's enumerating subsets in Lexicographical order)
int length = array.Length;
int subsetCount = 0;
for (int i=0; i<(1<<length); ++i)
{
int currentSet = i;
int tempIndex = length-1;
int currentSum = 0;
while (currentSet > 0) // iterate over bits "from the right side"
{
if (currentSet & 1 == 1) // if current bit is "1"
currentSum += array[tempIndex];
currentSet >>= 1;
tempIndex--;
}
subsetCount += (currentSum == targetSum) ? 1 : 0;
}
You can use the fact that N is small: it is possible to generate all possible subsets of the given array and check if its sum is S for each of them. The time complexity is O(N * 2 ** N) or O(2 ** N)(it depends on the way of the generation). This solution should be fast enough for the given constraints.
Here is a pseudo code of an O(2 ** N) solution:
result = 0
void generate(int curPos, int curSum):
if curPos == N:
if curSum == S:
result++
return
// Do not take the current element.
generate(curPos + 1, curSum)
// Take it.
generate(curPos + 1, curSum + numbers[curPos])
generate(0, 0)
A faster solution based on the meet in the middle technique:
Let's generate all subsets for the first half of the array using the algorithm described above and put their sums into a map(which maps a sum to the number of subsets that have it. It can be either a hash table or just an array because S is relatively small). This step takes O(2 ** (N / 2)) time.
Now let's generate all subsets for the second half and for each of them add the number of subset that sum up to S - currentSum e in the first half(using the map constructed in 1.), where the currentSum is the sum of all elements in the current subseta. Again, we have O(2 ** (N / 2)) subsets and each of them is processed in O(1).
The total time complexity is O(2 ** (N / 2)).
A pseudo code for this solution:
Map<int, int> count = new HashMap<int, int>() // or an array of size S + 1.
result = 0
void generate1(int[] numbers, int pos, int currentSum):
if pos == numbers.length:
count[currentSum]++
return
generate1(numbers, pos + 1, currentSum)
generate1(numbers, pos + 1, currentSum + numbers[pos])
void generate2(int[] numbers, int pos, int currentSum):
if pos == numbers.length:
result += count[S - currentSum]
return
generate2(numbers, pos + 1, currentSum)
generate2(numbers, pos + 1, currentSum + numbers[pos])
generate1(the first half of numbers, 0, 0)
generate2(the second half of numbers, 0, 0)
If N is odd, the middle element can go to either the first half or to the second one. It doesn't matter where it goes as long as it goes to exactly one of them.
I need to calculate combinations for a number.
What is the fastest way to calculate nCp where n>>p?
I need a fast way to generate binomial coefficients for an polynomial equation and I need to get the coefficient of all the terms and store it in an array.
(a+b)^n = a^n + nC1 a^(n-1) * b + nC2 a^(n-2) * ............
+nC(n-1) a * b^(n-1) + b^n
What is the most efficient way to calculate nCp ??
You cau use dynamic programming in order to generate binomial coefficients
You can create an array and than use O(N^2) loop to fill it
C[n, k] = C[n-1, k-1] + C[n-1, k];
where
C[1, 1] = C[n, n] = 1
After that in your program you can get the C(n, k) value just looking at your 2D array at [n, k] indices
UPDATE smth like that
for (int k = 1; k <= K; k++) C[0][k] = 0;
for (int n = 0; n <= N; n++) C[n][0] = 1;
for (int n = 1; n <= N; n++)
for (int k = 1; k <= K; k++)
C[n][k] = C[n-1][k-1] + C[n-1][k];
where the N, K - maximum values of your n, k
If you need to compute them for all n, Ribtoks's answer is probably the best.
For a single n, you're better off doing like this:
C[0] = 1
for (int k = 0; k < n; ++ k)
C[k+1] = (C[k] * (n-k)) / (k+1)
The division is exact, if done after the multiplication.
And beware of overflowing with C[k] * (n-k) : use large enough integers.
If you want complete expansions for large values of n, FFT convolution might be the fastest way. In the case of a binomial expansion with equal coefficients (e.g. a series of fair coin tosses) and an even order (e.g. number of tosses) you can exploit symmetries thus:
Theory
Represent the results of two coin tosses (e.g. half the difference between the total number of heads and tails) with the expression A + A*cos(Pi*n/N). N is the number of samples in your buffer - a binomial expansion of even order O will have O+1 coefficients and require a buffer of N >= O/2 + 1 samples - n is the sample number being generated, and A is a scale factor that will usually be either 2 (for generating binomial coefficients) or 0.5 (for generating a binomial probability distribution).
Notice that, in frequency, this expression resembles the binomial distribution of those two coin tosses - there are three symmetrical spikes at positions corresponding to the number (heads-tails)/2. Since modelling the overall probability distribution of independent events requires convolving their distributions, we want to convolve our expression in the frequency domain, which is equivalent to multiplication in the time domain.
In other words, by raising our cosine expression for the result of two tosses to a power (e.g. to simulate 500 tosses, raise it to the power of 250 since it already represents a pair), we can arrange for the binomial distribution for a large number to appear in the frequency domain. Since this is all real and even, we can substitute the DCT-I for the DFT to improve efficiency.
Algorithm
decide on a buffer size, N, that is at least O/2 + 1 and can be conveniently DCTed
initialise it with the expression pow(A + A*cos(Pi*n/N),O/2)
apply the forward DCT-I
read out the coefficients from the buffer - the first number is the central peak where heads=tails, and subsequent entries correspond to symmetrical pairs successively further from the centre
Accuracy
There's a limit to how high O can be before accumulated floating-point rounding errors rob you of accurate integer values for the coefficients, but I'd guess the number is pretty high. Double-precision floating-point can represent 53-bit integers with complete accuracy, and I'm going to ignore the rounding loss involved in the use of pow() because the generating expression will take place in FP registers, giving us an extra 11 bits of mantissa to absorb the rounding error on Intel platforms. So assuming we use a 1024-point DCT-I implemented via the FFT, that means losing 10 bits' accuracy to rounding error during the transform and not much else, leaving us with ~43 bits of clean representation. I don't know what order of binomial expansion generates coefficients of that size, but I dare say it's big enough for your needs.
Asymmetrical expansions
If you want the asymmetrical expansions for unequal coefficients of a and b, you'll need to use a two-sided (complex) DFT and a complex pow() function. Generate the expression A*A*e^(-Pi*i*n/N) + A*B + B*B*e^(+Pi*i*n/N) [using the complex pow() function to raise it to the power of half the expansion order] and DFT it. What you have in the buffer is, again, the central point (but not the maximum if A and B are very different) at offset zero, and it is followed by the upper half of the distribution. The upper half of the buffer will contain the lower half of the distribution, corresponding to heads-minus-tails values that are negative.
Notice that the source data is Hermitian symmetrical (the second half of the input buffer is the complex conjugate of the first), so this algorithm is not optimal and can be performed using a complex-to-complex FFT of half the required size for optimum efficiency.
Needless to say, all the complex exponentiation will chew more CPU time and hurt accuracy compared to the purely real algorithm for symmetrical distributions above.
This is my version:
def binomial(n, k):
if k == 0:
return 1
elif 2*k > n:
return binomial(n,n-k)
else:
e = n-k+1
for i in range(2,k+1):
e *= (n-k+i)
e /= i
return e
I recently wrote a piece of code that needed to call for a binary coefficient about 10 million times. So I did a combination lookup-table/calculation approach that's still not too wasteful of memory. You might find it useful (and my code is in the public domain). The code is at
http://www.etceterology.com/fast-binomial-coefficients
It's been suggested that I inline the code here. A big honking lookup table seems like a waste, so here's the final function, and a Python script that generates the table:
extern long long bctable[]; /* See below */
long long binomial(int n, int k) {
int i;
long long b;
assert(n >= 0 && k >= 0);
if (0 == k || n == k) return 1LL;
if (k > n) return 0LL;
if (k > (n - k)) k = n - k;
if (1 == k) return (long long)n;
if (n <= 54 && k <= 54) {
return bctable[(((n - 3) * (n - 3)) >> 2) + (k - 2)];
}
/* Last resort: actually calculate */
b = 1LL;
for (i = 1; i <= k; ++i) {
b *= (n - (k - i));
if (b < 0) return -1LL; /* Overflow */
b /= i;
}
return b;
}
#!/usr/bin/env python3
import sys
class App(object):
def __init__(self, max):
self.table = [[0 for k in range(max + 1)] for n in range(max + 1)]
self.max = max
def build(self):
for n in range(self.max + 1):
for k in range(self.max + 1):
if k == 0: b = 1
elif k > n: b = 0
elif k == n: b = 1
elif k == 1: b = n
elif k > n-k: b = self.table[n][n-k]
else:
b = self.table[n-1][k] + self.table[n-1][k-1]
self.table[n][k] = b
def output(self, val):
if val > 2**63: val = -1
text = " {0}LL,".format(val)
if self.column + len(text) > 76:
print("\n ", end = "")
self.column = 3
print(text, end = "")
self.column += len(text)
def dump(self):
count = 0
print("long long bctable[] = {", end="");
self.column = 999
for n in range(self.max + 1):
for k in range(self.max + 1):
if n < 4 or k < 2 or k > n-k:
continue
self.output(self.table[n][k])
count += 1
print("\n}}; /* {0} Entries */".format(count));
def run(self):
self.build()
self.dump()
return 0
def main(args):
return App(54).run()
if __name__ == "__main__":
sys.exit(main(sys.argv))
If you really only need the case where n is much larger than p, one way to go would be to use the Stirling's formula for the factorials. (if n>>1 and p is order one, Stirling approximate n! and (n-p)!, keep p! as it is etc.)
The fastest reasonable approximation in my own benchmarking is the approximation used by the Apache Commons Maths library: http://commons.apache.org/proper/commons-math/apidocs/org/apache/commons/math3/special/Gamma.html#logGamma(double)
My colleagues and I tried to see if we could beat it, while using exact calculations rather than approximates. All approaches failed miserably (many orders slower) except one, which was 2-3 times slower. The best performing approach uses https://math.stackexchange.com/a/202559/123948, here is the code (in Scala):
var i: Int = 0
var binCoeff: Double = 1
while (i < k) {
binCoeff *= (n - i) / (k - i).toDouble
i += 1
}
binCoeff
The really bad approaches where various attempts at implementing Pascal's Triangle using tail recursion.
nCp = n! / ( p! (n-p)! ) =
( n * (n-1) * (n-2) * ... * (n - p) * (n - p - 1) * ... * 1 ) /
( p * (p-1) * ... * 1 * (n - p) * (n - p - 1) * ... * 1 )
If we prune the same terms of the numerator and the denominator, we are left with minimal multiplication required. We can write a function in C to perform 2p multiplications and 1 division to get nCp:
int binom ( int p, int n ) {
if ( p == 0 ) return 1;
int num = n;
int den = p;
while ( p > 1 ) {
p--;
num *= n - p;
den *= p;
}
return num / den;
}
I was looking for the same thing and couldn't find it, so wrote one myself that seems optimal for any Binomial Coeffcient for which the endresult fits into a Long.
// Calculate Binomial Coefficient
// Jeroen B.P. Vuurens
public static long binomialCoefficient(int n, int k) {
// take the lowest possible k to reduce computing using: n over k = n over (n-k)
k = java.lang.Math.min( k, n - k );
// holds the high number: fi. (1000 over 990) holds 991..1000
long highnumber[] = new long[k];
for (int i = 0; i < k; i++)
highnumber[i] = n - i; // the high number first order is important
// holds the dividers: fi. (1000 over 990) holds 2..10
int dividers[] = new int[k - 1];
for (int i = 0; i < k - 1; i++)
dividers[i] = k - i;
// for every dividers there is always exists a highnumber that can be divided by
// this, the number of highnumbers being a sequence that equals the number of
// dividers. Thus, the only trick needed is to divide in reverse order, so
// divide the highest divider first trying it on the highest highnumber first.
// That way you do not need to do any tricks with primes.
for (int divider: dividers) {
boolean eliminated = false;
for (int i = 0; i < k; i++) {
if (highnumber[i] % divider == 0) {
highnumber[i] /= divider;
eliminated = true;
break;
}
}
if(!eliminated) throw new Error(n+","+k+" divider="+divider);
}
// multiply remainder of highnumbers
long result = 1;
for (long high : highnumber)
result *= high;
return result;
}
If I understand the notation in the question, you don't just want nCp, you actually want all of nC1, nC2, ... nC(n-1). If this is correct, we can leverage the following relationship to make this fairly trivial:
for all k>0: nCk = prod_{from i=1..k}( (n-i+1)/i )
i.e. for all k>0: nCk = nC(k-1) * (n-k+1) / k
Here's a python snippet implementing this approach:
def binomial_coef_seq(n, k):
"""Returns a list of all binomial terms from choose(n,0) up to choose(n,k)"""
b = [1]
for i in range(1,k+1):
b.append(b[-1] * (n-i+1)/i)
return b
If you need all coefficients up to some k > ceiling(n/2), you can use symmetry to reduce the number of operations you need to perform by stopping at the coefficient for ceiling(n/2) and then just backfilling as far as you need.
import numpy as np
def binomial_coef_seq2(n, k):
"""Returns a list of all binomial terms from choose(n,0) up to choose(n,k)"""
k2 = int(np.ceiling(n/2))
use_symmetry = k > k2
if use_symmetry:
k = k2
b = [1]
for i in range(1, k+1):
b.append(b[-1] * (n-i+1)/i)
if use_symmetry:
v = k2 - (n-k)
b2 = b[-v:]
b.extend(b2)
return b
Time Complexity : O(denominator)
Space Complexity : O(1)
public class binomialCoeff {
static double binomialcoeff(int numerator, int denominator)
{
double res = 1;
//invalid numbers
if (denominator>numerator || denominator<0 || numerator<0) {
res = -1;
return res;}
//default values
if(denominator==numerator || denominator==0 || numerator==0)
return res;
// Since C(n, k) = C(n, n-k)
if ( denominator > (numerator - denominator) )
denominator = numerator - denominator;
// Calculate value of [n * (n-1) *---* (n-k+1)] / [k * (k-1) *----* 1]
while (denominator>=1)
{
res *= numerator;
res = res / denominator;
denominator--;
numerator--;
}
return res;
}
/* Driver program to test above function*/
public static void main(String[] args)
{
int numerator = 120;
int denominator = 20;
System.out.println("Value of C("+ numerator + ", " + denominator+ ") "
+ "is" + " "+ binomialcoeff(numerator, denominator));
}
}
Someone asked me a brainteaser, and I don't know; my knowledge slows down after amortized analysis, and in this case, this is O(n).
public int findMax(array) {
int count = 0;
int max = array[0];
for (int i=0; i<array.length; i++) {
if (array[i] > max) {
count++;
max = array[i];
}
}
return count;
}
What's the expected value of count for an array of size n?
Numbers are randomly picked from a uniform distribution.
Let f(n) be the average number of assignments.
Then if the last element is not the largest, f(n) = f(n-1).
If the last element is the largest, then f(n) = f(n-1) + 1.
Since the last number is largest with probability 1/n, and not the largest with probability (n-1)/n, we have:
f(n) = (n-1)/n*f(n-1) + 1/n*(f(n-1) + 1)
Expand and collect terms to get:
f(n) = f(n-1) + 1/n
And f(1) = 0. So:
f(1) = 0
f(2) = 0 + 1/2
f(3) = 0 + 1/2 + 1/3
f(4) = 0 + 1/2 + 1/3 + 1/4
That is, f(n) is the n_th "Harmonic number", which you can get in closed form only approximately. (Well, one less than the n_th Harmonic number. The problem would be prettier if you initialized max to INT_MIN and just let the loop run, so that f(1) = 1.)
The above is not a rigorous proof, since I was sloppy about expected values versus actual values. But I believe the answer is right anyway :-).
I would like to comment on Nemo's answer, but I don't have the reputation to comment. His correct answer can be simplified:
The chance that the second number is larger than the first is 1/2. Regardless of that, the chance that the 3rd number is larger than two before, is 1/3. These are all independent chances and the total expectation is therefore
1/2 + 1/3 + 1/4 + .. + 1/n
You can actually take this analysis a step further when the value of each item comes from a finite set. Let E(N, M) be the expected number of assignments when finding the max of N elements that come uniformly from an alphabet of size M. Then we can say...
E(0, M) = E(N, 0) = 0
E(N, M) = 1 + SUM[SUM[E(j, i) * (N - 1 Choose j) * ((M - i) / M)^(N-j-1) * (i / M) ^ j : j from 0 to N - 1] : i from 0 to M - 1]
This is a bit hard to come up with a closed form for but we can be sure that E(N, M) is in O(log(min(N, M))). This is because E(N, INF) is in THETA(log(N)) as the harmonic series sum grows proportional to the log function and E(N, M) < E(N, M + 1). Likewise when M < N we have E(N, M) < E(M, INF) as there is at M unique values.
And here's some code to compute E(N, M) yourself. I wonder if anyone can get this to a closed form?
#define N 100
#define M 100
double NCR[N + 1][M + 1];
double E[N + 1][M + 1];
int main() {
NCR[0][0] = 1;
for(int i = 1; i <= N; i++) {
NCR[i][0] = NCR[i][i] = 1;
for(int j = 1; j < i; j++) {
NCR[i][j] = NCR[i - 1][j - 1] + NCR[i - 1][j];
}
}
for(int n = 1; n <= N; n++) {
for(int m = 1; m <= M; m++) {
E[n][m] = 1;
for(int i = 1; i < m; i++) {
for(int j = 1; j < n; j++) {
E[n][m] += NCR[n - 1][j] *
pow(1.0 * (m - i) / m, n - j - 1) *
pow(1.0 * i / m, j) * E[j][i] / m;
}
}
}
}
cout << E[N][M] << endl;
}
I am assuming all elements are distinct and counting the initial assignment to max outside the for loop.
If the array is sorted in increasing order, the variable max gets assigned to exactly n times (each time it gets a greater value).
If the array is sorted in decreasing order, the variable max gets assigned to exactly once (it gets assigned the first time and all subsequent values are smaller).
Edit:
My formulation for a randomly permuted array was actually wrong, as pointed out in the comments. I think #Nemo posts the correct answer to this.
I think just counting the number of assignments is not really a true measure of the cost of this function. whether or not we actually update the value of max, we are actually comparing it exactly n times. So, fewer assignments does not really imply less work done.
Also observe that there are actually no swaps being done. Only assignments and comparisons.