Only by addition and subtraction… Simply add numbers from 1 and check whether the sum is the given number. In pseudocode. Given 5050. number := 5050 next, sum := 0 while sum <= number next := next + 1 sum := sum + next return next 3 solved How to get the n if I … Read more
This is almost the same as BLUEPIXY‘s answer since I believe it’s the straight forward solution, but since you are confused by the function pointer, I removed that. #include <stdio.h> #include <stdlib.h> void printValue() { static unsigned int y; printf(“%d\n”, y); y += 1; } void recursiveFunction(int counter) { printValue(); if (–counter == 0) return; … Read more
Can we solve the longest increasing subsequence prob by designing the binary search tree and traverse its longest right? solved Can we solve the longest increasing subsequence prob by designing the binary search tree and traverse its longest right?
On RosettaCode you have implemented generating combinations in many languages, choose yourself. solved How to choose all possible combinations?
Looks like a homework… Here’s a straight forward approach: lst = [‘apple’,’cherry’,’coffee’,’apple’,’coffee’,’coffee’] res = {} for obj in lst: if obj not in res: res[obj] = 0 res[obj] += 1 print res solved python algorithm, find duplicates in list [closed]
printf(“%s\n”,str2[k]); str2[k] is a char, but you tell printf it is a char* But this program still will not work properly – the first call to gets() will just read the carriage-return that is left in the input queue after reading the initial int value. And you never null-terminate str2. solved Program to remove all … Read more
if array length is odd take element # (length+1)/2 else take average of element # length/2 and # length/2 – 1 solved What is Median of Sorted Arrays in O(log n) complexity? [closed]
I think it’s a very bad idea to calculate two very large numbers and hope that the quotient comes out as something sensible. Start by taking the natural log: ln(n!/(q!)^r) = ln(n!) – r*ln(q!) You can use gammaln() for the two function values, simplify, then take exp() to get the result you want: value = … 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
I suggest A* Pathfinding. It’s a really neat algorithm that should do what you want. 2 solved Shortest path on a grid with obstacles [closed]
Yes, the BFS algorithm works for every graph with unweighted edges. Thus it can be used for a graph with directed edges as well as long as it does not contain any weights. solved Can BFS be used to find single source shortest path in an unweighted directed graph? [closed]
your code problem is int i = 0 start with 0 and next line if (N%i==0) so 10/0 is not possible throw a error something like java.lang.ArithmeticException: / by zero is not possible and you loop through result.length, you need to loop through i your parent loop and put condition inside if (N%i==0) and you … Read more
There are many possible solutions to this problem. Here’s a reusable solution that simply decrements from 4 to 1 and adds the expected number of 1’s and 0’s. Loops used : 1 def sequence(n): string = “” for i in range(n): string+=’1’*(n-i) string+=’0’*(n-i) return string print sequence(4) There’s another single-line elegant and more pythonic way … Read more
Just make simple algebraic transformations start * c^n = stop c^n = stop / start c = (stop / start)^(1/n) (math.pow or power function in some programming languages) solved Fast algorithm needed [closed]
How about this solution #include<stdio.h> int main() { int i; int arr[]={ 2, 1 ,4 ,3 ,6 ,5 ,8 ,7 ,9}; int arr_size = sizeof(arr)/sizeof(arr[0]); int sorted[arr_size]; int sp = 0; for(i=0;i<arr_size;i++){ if(arr[i]&1){ sorted[sp++]=arr[i]; } } for(i=0;i<arr_size;i++){ if(!(arr[i]&1)){ sorted[sp++]=arr[i]; } } for(i=0;i< arr_size ;i++) printf(“%d “, sorted[i]); return 0; } the output is 1 3 … Read more