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Solution: Big (O) of Nested Loop with Addition

Learn to evaluate the time complexity of nested loops involving addition by analyzing loop executions and applying Big O notation. This lesson breaks down the components line by line to help you understand how to simplify and determine overall algorithm efficiency.

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Solution:

C++
int main(){
int n = 10;
int sum = 0;
float pie = 3.14;
for (int i=0; i<n; i+=3){ // O(n/3)
cout << pie << endl; // O(n/3)
for (int j=0; j<n; j+=2){ // O((n/3)*(n/2))
sum += 1; // O((n/3)*(n/2))
cout << sum << endl; // O((n/3)*(n/2))
}
}
}

On line 6 in the outer loop, int i=0; runs once, i<n; gets executed n3+1\frac{n}{3}+1 ...