Latest DATA STRUCTURE Aptitude Question SOLUTION: Problem Statement: Problem statement is very easy . On a positive integer, you can perform any one of the following 3 steps. 1.) Subtract 1 from it. ( n = n - 1 ) 2.) If it

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Previous Question: EXAMPLE: Input=['1','0','1','2','3','0','3','4','0'] Output: ['1', '1', '2', '3', '3', '4', '0', '0', '0'] // Function to move all zeros present in the array to the end void reorder(int A[], int n) { // k stores index of next available position int k = 0; // do for each element for (int i = 0; i < n; i++) { // if current element is non-zero, put the element at // next free position in the array if (A[i] != 0) A[k++] = A[i]; } // move all 0's to the end of the array (remaining indices) for (int i = k; i < n; i++) A[i] = 0; } // Move all zeros present in the array to the end int main(void) { int A[] = { ..... }; int n = sizeof(A) / sizeof(A[0]); reorder(A, n); for (int i = 0; i < n; i++) printf("%d ", A[i]); return 0; } Next Question: Which of the following data structure is better for storing the sorted data on which often insert and deletion operations are performed? A. Linked list B. Queue C. Array D. Doubly linked-list

(#M40167070) DATA STRUCTURE Programming question Given a positive integer n and you task is find the minimum number of steps that takes n to one . Keep an EYE

Problem Statement:

Problem statement is very easy . On a positive integer, you can perform any one of the following 3 steps.

1.) Subtract 1 from it. ( n = n - 1 )

2.) If its divisible by 2, divide by 2. ( if n % 2 == 0 , then n = n / 2 )

3.) If its divisible by 3, divide by 3. ( if n % 3 == 0 , then n = n / 3 )

Given a positive integer n and you task is find the minimum number of steps that takes n to one .: Asked In DATA STRUCTURE SUDHEER BABU PALLANTI (3 years ago); Unsolved; Is this Puzzle helpful? (2) (1) Submit Your Solution

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Previous Question: EXAMPLE: Input=['1','0','1','2','3','0','3','4','0'] Output: ['1', '1', '2', '3', '3', '4', '0', '0', '0'] // Function to move all zeros present in the array to the end void reorder(int A[], int n) { // k stores index of next available position int k = 0; // do for each element for (int i = 0; i < n; i++) { // if current element is non-zero, put the element at // next free position in the array if (A[i] != 0) A[k++] = A[i]; } // move all 0's to the end of the array (remaining indices) for (int i = k; i < n; i++) A[i] = 0; } // Move all zeros present in the array to the end int main(void) { int A[] = { ..... }; int n = sizeof(A) / sizeof(A[0]); reorder(A, n); for (int i = 0; i < n; i++) printf("%d ", A[i]); return 0; } Next Question: Which of the following data structure is better for storing the sorted data on which often insert and deletion operations are performed? A. Linked list B. Queue C. Array D. Doubly linked-list

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