Something like (it's untested and it's quick and dirty...you can definitely make it nicer but it should give you an idea):
someMethod( int[] arr, int low, int high, int sumSoFar ) {
if( low <= high ) {
sumSoFar = arr[low] + someMethod( arr, low+1, high, sumSoFar );
}
return sumSoFar;
}
The minimum unique array sum that can be achieved is when all elements in the array are different, resulting in the sum of the array being equal to the sum of the first n natural numbers, which is n(n1)/2.
Set sum = 0, then add each of the elements of the array, one by one. Use a for loop to process each element of the array.Set sum = 0, then add each of the elements of the array, one by one. Use a for loop to process each element of the array.Set sum = 0, then add each of the elements of the array, one by one. Use a for loop to process each element of the array.Set sum = 0, then add each of the elements of the array, one by one. Use a for loop to process each element of the array.
main(){int n,a[i],s;s=0;printf("enter no of elements in array");scanf("%d",&n);printf("Enter elements in array");for(i=;i
Sum(All elements in B) - Sum(All elements in A)
It is not possible to show a flowchart in this website -- it is text only. The algorithm can be summarised as follows: int sum(std::array<int>& a) { int sum = 0; // initialise the return value for (auto i : a) // for each value in the array sum += i; // increment the sum by the value return sum; // return the sum }
#include using std::cin;using std::cout;using std::endl;int main(){int sizeOfArray = 5;int myArray[] = {0};cout myArray[i];}int sum = 0;for (int j = 0; j < sizeOfArray; j++){sum += myArray[j];}cout
int myArray[N]; //where N is a constant int mySum = 0; ... for (int arrayIndex1 = 0; arrayIndex1 < N; arrayIndex1++) { cin >> myArray[arrayIndex1]; } ... for (int int arrayIndex2 = 0; arrayIndex2 < N; arrayIndex2++) { mySum +=myArray[arrayIndex2]; } ...
int sumArray(int N, int* A) { int sum = 0; do sum += A[N-1] while (N-- > 1); return sum; }
truzi i Ghal
You add up all the array elements, then divide by the number of elements. You can use a nested for() loop in Java; inside the inner for() loop, you can both increase a counter (to count how many elements there are), and add to a "sum" variable.
By no means; you can access any random array element. If you have ever seen examples which process them in order, it is because of the following: when the order doesn't matter (for example, you want to calculate the sum of all the array elements), it is easiest to process them in order.
To find the maximum sum by selecting non-adjacent elements from a list of numbers, you can use dynamic programming. Start by creating an array to store the maximum sum up to each element. Iterate through the list of numbers and for each element, calculate the maximum sum by either including the current element or excluding it. Keep track of the maximum sum achieved so far. At the end of the iteration, the final element in the array will contain the maximum sum that can be achieved by selecting non-adjacent elements.