The idea is to recursively sort the array from index 1 to n-1 and then insert the nth element at apt place. Algorithm: insertion_sort(Arr, n): if(n <= 1) return insertion_sort(Arr, n-1) temp = Arr[n] for (i = n; i > 0; i–): if(Arr[i] > temp): Arr[i+1] = Arr[i] else: break Arr[i+1] = temp
The while loop never stops, that’s probably the reason your code freezes. Your code only changes the address when the if statement is true. Which the array in your main() will stuck on the second (a[1]) element. So if we change change the address when the if statement is false… #include<stdio.h> #include<stdlib.h> void swap(int *p, … Read more
Point 1 should be easy: loop through the whole array { if (array[THIS INDEX] > array[THIS INDEX + 1]) //the element before is greater than the next one ARRAY IS NOT SORTED } //IF LOOP IS COMPLETED AND CODE REACHES THIS POINT ARRAY IS SORTED
include <time.h> include <stdio.h> int main(){ clock_t start = clock(); // Execuatable code clock_t stop = clock(); double elapsed = (double)(stop – start) * 1000.0 / CLOCKS_PER_SEC; printf(“Time elapsed in ms: %f”, elapsed); }
Create a post value and sort by it: postValue = likes + (isVetted ? 10 : 0) This makes “vetting” worth 10 likes. In other words, if 10 people like something not vetted, it will be ranked equivalently to something vetted with 0 likes. If you want to sort by date instead, then just sort … Read more
This is quite basic program called selection sort. The wiki article is: Selection sort. Basically in this program, i is first pointing to the first element in the array. Then, j points to the next element. If, the element j is smaller than element i, they get swapped. After swapping, the inner for loop still … Read more
Use String.localeCompare() with the numeric option to sort using possibility, but use – (minus) to get descending results. If the comparison returns 0 (equals), use localeCompare again to compare the names: const array = [{“name”:”fever”,”possibility”:”20%”},{“name”:”hiv”,”possibility”:”25%”},{“name”:”heart-attack”,”possibility”:”20%”},{“name”:”covid”,”possibility”:”40%”}] const result = array.sort((a, b) => -a.possibility.localeCompare(b.possibility, undefined, { numeric: true }) || a.name.localeCompare(b.name) ) console.log(result)
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
You can use pattern matching to determine which type of Array you’re dealing with. def arrTest[A](a :Array[A]) :Boolean = a match { case sa :Array[String] => sa.length < 2 || sa.sliding(2).forall(x => x(0) < x(1)) case ia :Array[Int] => ia.length > 2 && ia(0) == 1 && ia(1) == 1 && ia.sliding(3).forall(x => x(0) + … Read more
After sorting the array, you can simple iterate through each of the array elements and print out the index for each of them. for(int i = 0; i < arr.length; i++) { System.out.println(arr[i] + ” is at the position ” + (i + 1)); } Note that you need to use i + 1 since … Read more