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Sum of the elements from index L to R in an array when arr[i] = i * (-1)^i in C++
In this problem, we are given two numbers L and R. We also have an array arr[] such that arr[i] = i*(-1)^i. Our task is to create a program to calculate the sum of the element from index L to R in an array when arr[i] = i*(-1)^i.
So, we need to find the sum of elements within the range [L, R] of the array.
Let’s take an example to understand the problem,
Input
L = 2 , R = 6
Output
4
Explanation
arr[] = {-1, 2, -3, 4, -5, 6}
Sum = 2+ (-3) + 4 + (-5) + 6 = 4A simple solution to the problem will be running a loop from L to R and then adding all even numbers and subtracting all odd numbers. And then finally return the sum.
Example
Program to illustrate the working of our solution,
#include <iostream>
#include <math.h>
using namespace std;
int CalcArrSumLtoR(int L, int R) {
int sum = 0;
for (int i = L; i <= R; i++){
sum += (i * pow((-1), i));
}
return sum;
}
int main() {
int L = 3, R = 15;
cout<<"Sum of elements of array from index "<<L<<" to "<<R<<" is "lt;lt;CalcArrSumLtoR(L, R);
return 0;
}Output
Sum of elements of array from index 3 to 15 is -9
This is not an effective approach and will solve the problem in O(n) time complexity.
An efficient solution would be using the formula for the sum of n odd numbers. So,
Sum of first n odd numbers = n*n
Sum of first n even numbers = n*(n+1)
Here, the final sum will be calculated as
sum = (sum of first R even number - sum of first (L-1) even number ) - (sum of first R odd number - sum of first (L-1) odd number )
* There will be N/2 even/odd number till n. i.e. there will be R/2 even number. So, we will use R/2 and L/2 to calculate the sum.
Example
Program to illustrate the working of our solution,
#include <iostream>
using namespace std;
long int findSum(int n, bool isEven) {
long int total = 0;
if(isEven == true){
total = (n) / 2;
return (total * (total+1));
}
else {
total = (n + 1) / 2;
return total * total;
}
}
int CalcArrSumLtoR(int L, int R) {
return (findSum(R, true) - findSum(L - 1, true))- (findSum(R, false) - findSum(L - 1, false));
}
int main() {
int L = 3, R = 15;
cout<<"Sum of elements of array from index "<<L<<" to "<<R<<" is "<<CalcArrSumLtoR(L, R);
return 0;
}Output
Sum of elements of array from index 3 to 15 is -9
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