Make one pass through the sequence to find the largest value, the second largest value, and the smallest value. Swap the largest to one end, the second largest to the other end, and the smallest to the middle. Voila: largest items are on the ends and the smallest is in the middle. Calling this a … Read more
How would you solve the same problem if you had to only report the largest number? The pseudocode would be something like the following consider the first number to be largest for each of the rest of the number if it is larger then the current largest assign to largest How would you do it … Read more
Using ^ below a letter to indicate it’s a pivot, and * to indicate it’s already been pivoted. If a partition has odd length, I pick the “rounded down middle” (partition length 4 would have the 2nd item as pivot). Note how in the 4th line, there are two R‘s, with one of them being … Read more
The only way to do this in O(1) would be an array of all prime numbers. So, you’d only be able to support a certain number depending on your computer’s memory. Edit: There might be some way to compute this involving a bunch of calculus, but that’s beyond me 🙂
it is actually just O(n).. (not O(n+n), not O(2n), not O(n^n), just O(n) ) because since you increment j up to n in the inner loop as soon as you finish the inner loop of n elements, the outer condition will be false.. so it will exit, after n iterations
Let fabs be a function that returns the absolute value of a float. Then the formula could look like this: function calcRank(float v1, float v2) { float k = fabs(v2-30.0); if (k < 30.0) { // Rule 1 and 2 rank = k/60.0 + (60.0-k) * v1 / 60.0; } else { // Rule 3 … Read more
Denote A[i] – the i’th digit of A. We first need to understand that to get N+M=10^X-1, we need N[i]+M[i]=9 for all i. Since M[i]=N[X-i], it means we need N[i] + N[X-i] = 9. This means, once N[i] is set, also N[X-i]. We can now derive a recursive formula: F(X) = 10*F(X-2) The idea is … Read more
|O(f(n)) – O(f(n))| = O(f(n)). That is basically the only thing you can say about it. This is because you’re subtracting, from a function growing at most like f(n), another function growing at most like f(n). The result (pre the absolute value) can be even negative (e.g., when subtracting 2 f(n) = O(f(n)), but it … Read more
A possible approach could be to start from an empty string and add characters one by one to it using a recursive function and printing it. Here is my code: #include<stdio.h> #include<stdlib.h> #include<string.h> void print_string(char str[],char new_str[],int current_len,int n,int len) { /* str=orignal set, new_str=empty char array, current_len=0(Intially) n=no of elements to be used len=the … Read more
Correct order should be : 1–>2–>4–>3 i.e. 1’st=O(2)=O(4)=O(3)