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What is the index number of the last element of an array with 9 elements?
(array.length - 1) will find the index of the last element in an array (or -1 if the array is empty).
How will you Write a c program to find the kth smallest element in an array in c?
//This is for kth largest element. (So this is for n-k smallest element) //Sudipta Kundu [Wipro Technologies] #include <stdio.h> //Input: array with index range [first, last) //Output: new index of the pivot. An element in the middle is chosen to be a pivot. Then the array's elements are //placed in such way that all elements <= pivot are to the left and all elements >= pivot are to the right. int positionPivot(int* array, int first, int last); //Input: array with index range [first, last) and integer K (first <= K < last) //Output: array whose Kth element (i.e. array[K]) has the "correct" position. More precisely, //array[first ... K - 1] <= array[K] <= array[K + 1 ... last - 1] void positionKthElement(int* array, int first, int last, int k); int main() { int array[] = {7,1,8,3,1,9,4,8}; int i; for (i = 0; i < 8; i++) { positionKthElement(array, 0, sizeof(array) / sizeof(array[0]),i); printf("%d is at position %d\n", array[i], i); } return 0; } int positionPivot(int* array, int first, int last) { if (first last) return first; int tmp = (first + last) / 2; int pivot = array[tmp]; int movingUp = first + 1; int movingDown = last - 1; array[tmp] = array[first]; array[first] = pivot; while (movingUp <= movingDown) { while (movingUp <= movingDown && array[movingUp] < pivot) ++movingUp; while (pivot < array[movingDown]) --movingDown; if (movingUp <= movingDown) { tmp = array[movingUp]; array[movingUp] = array[movingDown]; array[movingDown] = tmp; ++movingUp; --movingDown; } } array[first] = array[movingDown]; array[movingDown] = pivot; return movingDown; } void positionKthElement(int* array, int first, int last, int k) { int index; while ((index = positionPivot(array, first, last)) != k) { if (k < index) last = index; else first = index + 1; } }
What is typically the last subscript in an array?
In most programming languages, the last subscript (or index) in an array is typically one less than the total number of elements in the array. This is because array indexing usually starts at zero. For example, in an array with 10 elements, the last subscript would be 9.
How can you fix an ArrayIndexOutOfBoundsException in java?
This is very common exception and name of the exception class tells exactly what kind of problem you have. If you would look into stack trace, which should have been generated too you would be able to find exact place where it happened. The problem is that "Array Index Out Of Bounds". This means that you used index on array which is invalid. That could negative number, because all arrays starts from 0. If you array has N items and you will try to get item with index N or higher you will get this exception too. Only available indexes are from 0 to N - 1, where 0 points to the first item and N - 1 to the last one.
The maximum number of records you might want to store in an array is defined as MAX When using 0 based indexing what index will reference the last allocated position of an array of size MAX?
If all elements of the array are in use then the last record is referred to as MAX-1. If you are using a count variable to remember how far into the array you are using then this variable will keep track of the last allocated value in the array.
Why in c plus plus index always start with 0?
C++ array indices are zero-based because the first element in any array is offset 0 elements from the start address. The second element is offset by 1 element and the third by 2 elements, and so on. To put it another way, the index refers to the number of elements that come before the desired element. The first element has zero elements before it, so it is index 0. For an array of n elements, the last element is at index n-1.
How To find a maximum number in a 2D array?
// Pseudocode int findMax( int[][] data ) { // Return if data is empty if( data.length 0 ) { return 0; } int max = data[0][0]; // Iterate through each element in the array for( int r = 0; r < data.length; ++r ) { for( int c = 0; c < data[0].length; ++c ) { // If we find a value greater than the current max, update max if( data[r][c] > max ) { max = data[r][c]; } } } return max; }
What are the address calculations for accessing arrays?
Arrays use a Base Pointer and an Index. They are calculated using the notation BP+IDX*Size, where IDX 0 is the first index location, and Size is the size of each element (at a hardware level, the size is determined by the register the data is loaded into). Languages that start array indexes at 0 use this hardware notation. Other languages start at 1 as a means of being "human friendly", since the zeroth location is the primary cause for developers going "out of bounds" on an array (they forget that the size of the array is one larger than the last index of the array).
Algorithm of linear search in c?
int linearSearch(int a[], int first, int last, int key) { // function: // Searches a[first]..a[last] for key. // returns: index of the matching element if it finds key, // otherwise -1. // parameters: // a in array of (possibly unsorted) values. // first, last in lower and upper subscript bounds // key in value to search for. // returns: // index of key, or -1 if key is not in the array. for (int i=first; i<=last; i++) { if (key == a[i]) { return i; } } return -1; // failed to find key }
If an array is declared in user defined function and pointer to this array is returned to main function is it possible to access that array?
Either put some array terminator as the last element (like '\0') or return the length of the array as an output from user defined function.
How do you set rules for the last two digits of a number?
I assume you are using C.I also assume that you stored the input in an int.All you need to do is store that int into an array.You can do this using itoa.Just search about it.When you have done this, use the index of the last two digits of the array to check their values.If the number of digits is not definite, create a loop like this to count the number of digits:array digits[MAX];int c = 0;while(array[c]!=null){c++;}// c will then be the number of digits.example:inputted digit: 12345// the value of c is now 4//Take note that the index of arrays is 0//index: [0][1][2][3][4]//values: [1][2][3][4][5]To access the last two digits:digits[4] // this is 5digit[3] // this is 4
Is an array not suitable?
That depends on what you want to use it for. An array offers constant-time random access to any element in the array, but is restricted by the fact it must be re-allocated in order to cater for new elements or to remove redundancy. Linked lists overcome this limitation at the expense of random access, which is linear-time other than for the first and/or last nodes which remain constant-time.
