http://www.geeksforgeeks.org/k-th-prime-factor-given-number/
http://www.geeksforgeeks.org/probability-getting-least-k-heads-n-tosses-coins/
Method 1 (Naive)
A Naive approach is to store the value of factorial in dp[] array and call it directly whenever it is required. But the problem of this approach is that we can only able to store it up to certain value, after that it will lead to overflow.
Method 2 (Dynamic Programming and Log)
Another way is to use Dynamic programming and logarithm. log() is indeed useful to store the factorial of any number without worrying about overflow. Let’s see how we use it:
http://www.geeksforgeeks.org/sum-of-all-subarrays/
If we take a close look then we observe a pattern. Let take an example
http://www.geeksforgeeks.org/subarraysubstring-vs-subsequence-and-programs-to-generate-them/
http://www.geeksforgeeks.org/maximum-sum-of-smallest-and-second-smallest-in-an-array/
Given an array, find maximum sum of smallest and second smallest elements chosen from all possible subarrays. More formally, if we write all (nC2) subarrays of array of size >=2 and find the sum of smallest and second smallest, then our answer will be maximum sum among them.
http://www.geeksforgeeks.org/find-coordinates-triangle-given-midpoint-side/
Given three coordinates (x, y), which are the midpoint of the sides of the triangle. The task is to find the coordinates of the triangle.
Let’s separately solve for X-coordinates and Y-coordinates. For X coordinate of vertices, let them be x1, x2, x3. Then, X-coordinate of middel points will be (x1 + x2)/2, (x2 + x3)/2, (x3 + x1)/2. Observe, sum of these 3 expressions is equal to sum of X-coordinates. Now, we have sum of 3 varibles and 3 expressions for sum of every pair of them, find out the values of coordinates by solving equations.
Given two numbers n and k, print k-th prime factor among all prime factors of n. For example, if the input number is 15 and k is 2, then output should be “5”. And if the k is 3, then output should be “-1” (there are less than k prime factors).
An Efficient Solution is to use Sieve of Eratosthenes. Note that this solution is efficient when we need k-th prime factor for multiple test cases. For a single case, previous approach is better.
The idea is to do preprocessing and store least prime factor of all numbers in given range. Once we have least prime factors stored in an array, we can find k-th prime factor by repeatedly dividing n with least prime factor while it is divisible, then repeating the process for reduced n.
The idea is to do preprocessing and store least prime factor of all numbers in given range. Once we have least prime factors stored in an array, we can find k-th prime factor by repeatedly dividing n with least prime factor while it is divisible, then repeating the process for reduced n.
// Using SieveOfEratosthenes to find smallest prime// factor of all the numbers.// For example, if MAX is 10,// s[2] = s[4] = s[6] = s[10] = 2// s[3] = s[9] = 3// s[5] = 5// s[7] = 7void sieveOfEratosthenes(int s[]){ // Create a boolean array "prime[0..MAX]" and // initialize all entries in it as false. vector <bool> prime(MAX+1, false); // Initializing smallest factor equal to 2 // for all the even numbers for (int i=2; i<=MAX; i+=2) s[i] = 2; // For odd numbers less then equal to n for (int i=3; i<=MAX; i+=2) { if (prime[i] == false) { // s(i) for a prime is the number itself s[i] = i; // For all multiples of current prime number for (int j=i; j*i<=MAX; j+=2) { if (prime[i*j] == false) { prime[i*j] = true; // i is the smallest prime factor for // number "i*j". s[i*j] = i; } } } }}// Function to generate prime factors and return its// k-th prime factor. s[i] stores least prime factor// of i.int kPrimeFactor(int n, int k, int s[]){ // Keep dividing n by least prime factor while // either n is not 1 or count of prime factors // is not k. while (n > 1) { if (k == 1) return s[n]; // To keep tract of count of prime factors k--; // Divide n to find next prime factor n /= s[n]; } return -1;}
A Simple Solution is to first find prime factors of n. While finding prime factors, keep track of count. If count becomes k, we return current prime factor.
int kPrimeFactor(int n, int k){ // Find the number of 2's that divide k while (n%2 == 0) { k--; n = n/2; if (k == 0) return 2; } // n must be odd at this point. So we can skip // one element (Note i = i +2) for (int i = 3; i <= sqrt(n); i = i+2) { // While i divides n, store i and divide n while (n%i == 0) { if (k == 1) return i; k--; n = n/i; } } // This condition is to handle the case where // n is a prime number greater than 2 if (n > 2 && k == 1) return n; return -1;}
Given N number of coins, the task is to find probability of getting at least K number of heads after tossing all the N coins simultaneously.
The probability of exactly k success in n trials with probability p of success in any trial is given by:
So Probability ( getting at least 4 heads )=
So Probability ( getting at least 4 heads )=
A Naive approach is to store the value of factorial in dp[] array and call it directly whenever it is required. But the problem of this approach is that we can only able to store it up to certain value, after that it will lead to overflow.
double fact[MAX];// Returns probability of getting at least k// heads in n tosses.double probability(int k, int n){ double ans = 0; for (int i=k; i <= n; ++i) // Probability of getting exactly i // heads out of n heads ans += fact[n]/(fact[i] * fact[n-i]); // Note: 1 << n = pow(2, n) ans = ans/(1LL << n); return ans;}void precompute(){ // Preprocess all factorial only upto 19, // as after that it will overflow fact[0] = fact[1] = 1; for (int i=2; i < 20; ++i) fact[i] = fact[i-1] * i;}Another way is to use Dynamic programming and logarithm. log() is indeed useful to store the factorial of any number without worrying about overflow. Let’s see how we use it:
At first let see how n! can be written.
n! = n * (n-1) * (n-2) * (n-3) * ... * 3 * 2 * 1
Now take log on base 2 both the sides as:
=> log(n!) = log(n) + log(n-1) + log(n-2) + ... + log(3)
+ log(2) + log(1)
Now whenever we need to find the factorial of any number, we can use
this precomputed value. For example:
Suppose if we want to find the value of nCi which can be written as:
=> nCi = n! / (i! * (n-i)! )
Taking log2() both sides as:
=> log2 (nCi) = log2 ( n! / (i! * (n-i)! ) )
=> log2 (nCi) = log2 ( n! ) - log2(i!) - log2( (n-i)! ) `
Putting dp[num] = log2 (num!), we get:
=> log2 (nCi) = dp[n] - dp[i] - dp[n-i]
But as we see in above relation there is an extra factor of 2n which
tells the probability of getting i heads, so
=> log2 (2n) = n.
We will subtract this n from above result to get the final answer:
=> Pi (log2 (nCi)) = dp[n] - dp[i] - dp[n-i] - n
Now: Pi (nCi) = 2 dp[n] - dp[i] - dp[n-i] - n
Tada! Now the questions boils down the summation of Pi for all i in
[k, n] will yield the answer which can be calculated easily without
overflow.
// dp[i] is going to store Log ( i !) in base 2double dp[MAX];double probability(int k, int n){ double ans = 0; // Initialize result // Iterate from k heads to n heads for (int i=k; i <= n; ++i) { int res = dp[n] - dp[i] - dp[n-i] - n; ans += pow(2.0, res); } return ans;}void precompute(){ // Preprocess all the logarithm value on base 2 for (int i=2; i < MAX; ++i) dp[i] = log2(i) + dp[i-1];}
Given an integer array ‘arr[]’ of size n, find sum of all sub-arrays of given array.
long int SubArraySum(int arr[], int n){ long int result = 0; // Pick starting point for (int i=0; i <n; i++) { // Pick ending point for (int j=i; j<n; j++) { // sum subarray between current // starting and ending points for (int k=i; k<=j; k++) result += arr[k] ; } } return result ;}arr[] = [1, 2, 3], n = 3
All subarrays : [1], [1, 2], [1, 2, 3],
[2], [2, 3], [3]
here first element 'arr[0]' appears 3 times
second element 'arr[1]' appears 4 times
third element 'arr[2]' appears 3 times
Every element arr[i] appears in two types of subsets:
i) In sybarrays beginning with arr[i]. There are
(n-i) such subsets. For example [2] appears
in [2] and [2, 3].
ii) In (n-i)*i subarrays where this element is not
first element. For example [2] appears in
[1, 2] and [1, 2, 3].
Total of above (i) and (ii) = (n-i) + (n-i)*i
= (n-i)(i+1)
For arr[] = {1, 2, 3}, sum of subarrays is:
arr[0] * ( 0 + 1 ) * ( 3 - 0 ) +
arr[1] * ( 1 + 1 ) * ( 3 - 1 ) +
arr[2] * ( 2 + 1 ) * ( 3 - 2 )
= 1*3 + 2*4 + 3*3
= 20
long int SubArraySum( int arr[] , int n ){ long int result = 0; // computing sum of subarray using formula for (int i=0; i<n; i++) result += (arr[i] * (i+1) * (n-i)); // return all subarray sum return result ;}
We can run two nested loops, the outer loop picks starting element and inner loop considers all elements on right of the picked elements as ending element of subarray.
void subArray(int arr[], int n){ // Pick starting point for (int i=0; i <n; i++) { // Pick ending point for (int j=i; j<n; j++) { // Print subarray between current starting // and ending points for (int k=i; k<=j; k++) cout << arr[k] << " "; cout << endl; } }}Given an array, find maximum sum of smallest and second smallest elements chosen from all possible subarrays. More formally, if we write all (nC2) subarrays of array of size >=2 and find the sum of smallest and second smallest, then our answer will be maximum sum among them.
A simple solution is to generate all subarrays, find sum of smallest and second smallest of every subarray. Finally return maximum of all sums.
An efficient solution is based on the observation that this problem reduces to finding a maximum sum of two consecutive elements in array.
int pairWithMaxSum(int arr[], int N){ if (N < 2) return -1; // Find two consecutive elements with maximum // sum. int res = arr[0] + arr[1]; for (int i=1; i<N-1; i++) res = max(res, arr[i] + arr[i+1]); return res;}Given three coordinates (x, y), which are the midpoint of the sides of the triangle. The task is to find the coordinates of the triangle.
Let’s separately solve for X-coordinates and Y-coordinates. For X coordinate of vertices, let them be x1, x2, x3. Then, X-coordinate of middel points will be (x1 + x2)/2, (x2 + x3)/2, (x3 + x1)/2. Observe, sum of these 3 expressions is equal to sum of X-coordinates. Now, we have sum of 3 varibles and 3 expressions for sum of every pair of them, find out the values of coordinates by solving equations.
vector<int> solve(int v[]){ vector<int> res; // Finding sum of all three coordinate. int all3 = v[0] + v[1] + v[2]; // Solving the equation. res.push_back(all3 - v[1]*2); res.push_back(all3 - v[2]*2); res.push_back(all3 - v[0]*2); return res;}// Finds vertices of a triangles from given// middle vertices.void findVertex(int xmid[], int ymid[]){ // Find X coordinates of verices. vector<int> V1 = solve(xmid); // Find Y coordinates of verices. vector<int> V2 = solve(ymid); // Output the solution. for (int i = 0; i < 3; i++) cout << V1[i] << " " << V2[i] <<endl;}