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Complexity analysis.txt
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Complexity analysis.txt
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/*
Session : Time Complexity analysis
By : Muhammad Magdi
On : 20/08/2017
*/
* What's time Complexity?
How does the running time change as the size of input change.
* Running time depends on:
1- Processor.
2- Read/Write speed to memory.
3- 32 bit VS 64 bit.
4- INPUT.
5- ...
* What are the cases your code may face?
1- Best case.
2- Worst case.
3- Average case.
* The notations to represent your code's Complexity:
1- Big O Notation ----> the upper bound of the time.
2- Omega Notation ----> the lower bound of the time.
3- Theta Notation ----> the bound itself.
* Rules:
1- Running time is the sum of running times of all consecutive blocks.
2- Nested loops are multiplied.
In general -> Nested repetitive Blocks are multiplied.
3- In Conditional statements pick the "Worst case" one.
4- Drop Constants (addition, subtraction, multiplication or division).
5- Drop all lower order terms.
* Some useful Observations:
Big O Name Max n
-------------------------------------------------------------------------------------------
O(1) ----> Constant ----> 1e18 ----> Math, Observation
O(Log(n)) ----> Logarithmic ----> 1e18 ----> Binary Search (lower -upper- bound)
O(n) ----> Linear ----> 1e8 ----> one loop
O(n*Log(n)) ----> LogLinear ----> 4e5 ----> Sorting, loop + binary search
O(n^2) ----> Quadratic ----> 1e4 ----> nested loop
O(2^n) ----> Exponential ----> 25 ----> Bitmasks, finding all possible answers
O(n!) ----> factorial ----> 11 ----> finding all permutations
int calcSum(int a, int b){ //O(1)
int sum = a+b;
return sum;
}
double calcAverage(int a, int b){ //O(1)
double avg = (a+b)/2.0;
return avg;
}
bool isAlphabit(char x){ //O(1)
return (x>='A' && x<='Z' || x>='a' && x<='z');
}
double sumHarmonicSeries(int n){ //O(n)
double sum = 0;
for(int i = 1 ; i <= n ; ++i){
sum += (1.0/i);
}
return sum;
}
long long calcSumSegment(int a, int b){ //O(b)
long long sum = 0;
for(int i = a ; i<=b ; ++i)
sum += i;
return sum;
}
int stepper(int n, int s){ //O(n/s)
int ret = 0;
for(int i = 1 ; i <= n ; i += s){
ret += i;
}
return ret;
}
void merge(int* A, int szA, int* B, int szB){ //O(sz)
int idxA = 0, idxB = 0, idxC = 0;
while(idxA < szA && idxB < szB){
if(A[idxA] < B[idxB]) C[idxC++] = A[idxA++];
else C[idxC++] = B[idxB++];
}
while(idxA < szA) C[idxC++] = A[idxA++];
while(idxB < szB) C[idxC++] = B[idxB++];
}
int fact(int n){ //O(n)
if(!n || n==1) return 1;
return n*fact(n-1);
}
int power1(int base, int power){ //O(power)
if(!power) return 1;
return base*power1(base, power-1);
}
int calcLog(int n){ //O(log(n))
int ret = 0;
while(n > 1){
++ret;
n /= 2;
}
return ret;
}
bool binarySearch(int val, int n){ //O(log(n))
int lo = 0, hi = n, mid;
while(hi-lo > 0){
mid = ((lo+hi)>>1);
if(A[mid] == val) return 1;
if(A[mid] < val)
lo = mid+1;
else
hi = mid-1;
}
return 0;
}
void printPowersOfTwoTill(int n){ //O(log(n))
for(int p = 1 ; p <= n ; p *= 2)
printf("%d\n", p);
}
int power2(int base, int power){ //O(log(n))
if(!power) return 1;
int sub = power2(base, power>>1);
return (power&1? sub*sub*base : sub*sub);
}
for(int i = 0 ; i < (1<<n) ; ++i){ //O(2^n)
//some O(1) operations
}
/*
8 4 2 1
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
1<<0 = 2^0 = 1
1<<1 = 2^1 = 2
1<<2 = 2^2 = 4
1<<n = 2^n
*/
int fib(int n){ //O(2^n)
if(!n || n==1) return n;
return fib(n-1)+fib(n-2);
}
for(int i = 0 ; i < (1<<n) ; ++i){ //O(n * (2^n))
for(int i = 0 ; i < n ; ++i){
//some constant order statements go here
}
}
void searchArray(){ //O(n*log(n))
for(int i = 0 ; i < n ; ++i){
if(binarySearch(B[i]))
puts("Found");
else
puts("Not Found");
}
}
void something(int n){ //O(n*log(n))
for(int i = 1 ; i <= n ; ++i)
for(int j = i ; j <= n ; j+=i)
//Something
}
void mergeSort(int st = 0, int en = n-1){ //O(n*log(n))
if(st == en) return;
int mid = (st+en)>>1;
mergeSort(st, mid);
mergeSort(mid+1, en);
merge(A, mid-st+1, A+mid+1, en-mid);
}
void printPermutations(string s){ //O(n!)
sort(s.begin(), s.end());
do {
cout << s << endl;
}while(next_permutation(s.begin(), s.end()));
}