Implement these methods:
public static int smallest(int[] arr) {
int small = arr[0];
for(int i = 1; i < arr.size(); i++)
if(arr[i] < small)
small = arr[i];
return small;
}
public static int largest(int[] arr) {
int large = arr[0];
for(int i = 1; i < arr.size(); i++)
if(arr[i] > large)
large = arr[i];
return large;
}
To compute the largest value in an array, assume that the first element is the largest and store the value. Then traverse the remainder of the array. Each time a larger value is encountered, update the stored value. Once all values are traversed, return the stored value. In pseudocode, this algorithm would be implemented as follows: Algorithm: largest Input: array A of length N Output: largest value in A let largest = A[0] // store first value for index = 1 to N-1 // traverse remaining elements if A[index] > largest then largest = A[index] // update stored value if current value is larger next index return largest To determine the position of the largest value, we alter the algorithm as follows: Algorithm: largest_by_index Input: array A of length N Output: index of the largest value in A let largest = 0; // store index 0 for index = 1 to N-1 // traverse remaining elements if A[index] > A[largest] then largest = index // update stored index next index return largest We can do the same to find the position of the smallest element: Algorithm: smallest_by_index Input: array A of length N Output: index of the smallest value in A let smallest = 0; // store index 0 for index = 1 to N-1 // traverse remaining elements if A[index] < A[smallest] then smallest = index // update stored index next index return smallest To perform both algorithms simultaneously, we need to return two values. To achieve this we can use a simple data structure known as a pair: struct pair { int smallest; int largest; }; Algorithm: range_by_index Input: array A of length N Output: a pair indicating the position of the smallest and largest values in A pair result = {0, 0} // initialise the pair for index = 1 to N-1 // traverse remaining elements if A[index] < A[result.smallest] then result.smallest = index // update stored index else if A[index] > A[result.largest] then result.largest = index // update stored index next index return result
Arrange the any one of the order and store the memory in order vice (ie.Ascending for Descending) Then print the second data of that array it is simple way
Your best bet would probably be to iterate through the array using a for loop and compare each value to the current low and high values (which you would store in a local variable) for example: for each element in array { if current is less than lowest_value lowest_value = current else if current is greater than highest_value highest_value = current }
You data has to be stored in the array arr[] of size 10.double min(const arr[], int arrSize){double minimum = arr[0];for (int j = 0; j < arrSize; j++){if (minimum > arr[j]){minimum = arr[j];}}return minimum;}
If you have the series stored in an array, you loop through the array and print each array element in turn. Another possibility is to print out the numbers in the series as you generate them. In that case, you may not need to store anything (depending on the series, of course).
To compute the largest value in an array, assume that the first element is the largest and store the value. Then traverse the remainder of the array. Each time a larger value is encountered, update the stored value. Once all values are traversed, return the stored value. In pseudocode, this algorithm would be implemented as follows: Algorithm: largest Input: array A of length N Output: largest value in A let largest = A[0] // store first value for index = 1 to N-1 // traverse remaining elements if A[index] > largest then largest = A[index] // update stored value if current value is larger next index return largest To determine the position of the largest value, we alter the algorithm as follows: Algorithm: largest_by_index Input: array A of length N Output: index of the largest value in A let largest = 0; // store index 0 for index = 1 to N-1 // traverse remaining elements if A[index] > A[largest] then largest = index // update stored index next index return largest We can do the same to find the position of the smallest element: Algorithm: smallest_by_index Input: array A of length N Output: index of the smallest value in A let smallest = 0; // store index 0 for index = 1 to N-1 // traverse remaining elements if A[index] < A[smallest] then smallest = index // update stored index next index return smallest To perform both algorithms simultaneously, we need to return two values. To achieve this we can use a simple data structure known as a pair: struct pair { int smallest; int largest; }; Algorithm: range_by_index Input: array A of length N Output: a pair indicating the position of the smallest and largest values in A pair result = {0, 0} // initialise the pair for index = 1 to N-1 // traverse remaining elements if A[index] < A[result.smallest] then result.smallest = index // update stored index else if A[index] > A[result.largest] then result.largest = index // update stored index next index return result
void mail ( ); { int a, b c = a+b; printf ("%d",=c); }
Arrange the any one of the order and store the memory in order vice (ie.Ascending for Descending) Then print the second data of that array it is simple way
The most efficient way to store a list is with an array.
Your best bet would probably be to iterate through the array using a for loop and compare each value to the current low and high values (which you would store in a local variable) for example: for each element in array { if current is less than lowest_value lowest_value = current else if current is greater than highest_value highest_value = current }
You data has to be stored in the array arr[] of size 10.double min(const arr[], int arrSize){double minimum = arr[0];for (int j = 0; j < arrSize; j++){if (minimum > arr[j]){minimum = arr[j];}}return minimum;}
To calculate the maximum value in an (unsorted) array, assume the first element is the largest element and store its value. If the next value is larger, store that value, otherwise continue to the next value. Repeat this process, updating the stored value each time you find a larger value. After a single pass of the array, the stored value holds the largest value. Locating the minimum is exactly the same except you're looking for values less than the stored value. The number of dimensions is immaterial. Simply treat the array as if it were one-dimensional. The following example assumes all arrays or of type int: int largest (int* a, unsigned len) { int m = a[0]; for (unsigned i=1; i<len; ++i) if (a[i]>m) m=a[i]; return m; } int smallest (int* a, unsigned len) { int m = a[0]; for (unsigned i=1; i<len; ++i) if (a[i]<m) min=a[i]; return m; } Given these functions, we can determine the largest or smallest value in any array of type int, regardless of the number of dimensions: int main () { int min, max; int a[5] = {4, 2, 5, 3, 1}; min = smallest (a, 5); // min = 1 max = largest (a, 5); // max = 5 int b[3][4] = {{1, 2, 3, 4}, {5, 4, 3, 2}, {0, 1, 2, 3}}; min = smallest ((int*) b, 3*4); // min = 0 max = largest ((int*) b, 3*4); // max = 5 return 0; } Note that the only real complication is that you must cast all multi-dimensional arrays to one-dimensional arrays. That is, for each dimension greater than 1, you add one level of indirection. E.g., a three-dimensional array of type int would be cast to int** while a four-dimensional array casts to int***.
If you have the series stored in an array, you loop through the array and print each array element in turn. Another possibility is to print out the numbers in the series as you generate them. In that case, you may not need to store anything (depending on the series, of course).
Use two variables to store the largest, and the second-largest integer. Update those in a loop, for every element in the array. Initial values might be the lowest permissible value for the int, long, double, or whatever value you use.
// 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; }
Yes, you can create array that will store class properties. But all of them have to be of the same type.
Write a console based C++ program that reads student information from a text file, build an array of objects of type class StudentInfo,