One-dimensional arrays in C++ - Time & Space Complexity
When working with one-dimensional arrays, it's important to understand how the time to process them grows as the array gets bigger.
We want to know how the number of steps changes when the array size increases.
Analyze the time complexity of the following code snippet.
int sumArray(int arr[], int n) {
int sum = 0;
for (int i = 0; i < n; i++) {
sum += arr[i];
}
return sum;
}
This code adds up all the numbers in a one-dimensional array of size n.
- Primary operation: Adding each element of the array to the sum.
- How many times: Exactly once for each element, so n times.
As the array gets bigger, the number of additions grows in a straight line with the size.
| Input Size (n) | Approx. Operations |
|---|---|
| 10 | 10 additions |
| 100 | 100 additions |
| 1000 | 1000 additions |
Pattern observation: Doubling the array size doubles the work needed.
Time Complexity: O(n)
This means the time to sum the array grows directly with the number of elements.
[X] Wrong: "The loop runs faster because it just adds numbers, so time doesn't grow much with size."
[OK] Correct: Even simple steps add up; more elements mean more additions, so time grows with n.
Understanding how loops over arrays scale is a key skill that helps you reason about many real-world problems and algorithms.
"What if we nested another loop inside to compare each element with every other element? How would the time complexity change?"
Practice
Solution
Step 1: Understand array indexing in C++
Arrays in C++ always start indexing from 0, meaning the first element is at index 0.Step 2: Confirm the index range
Indexes go from 0 up to size-1, so the first index is always 0 regardless of array size.Final Answer:
0 -> Option CQuick Check:
Array indexing starts at 0 [OK]
- Thinking arrays start at 1
- Assuming negative indexes are valid
- Confusing with other languages
Solution
Step 1: Recall C++ array declaration syntax
In C++, a fixed-size array is declared using square brackets with size: int arr[5];Step 2: Check other options for correctness
int arr(5); uses parentheses which is invalid for arrays. array<int> arr(5); uses incorrect syntax (not standard array). int arr = new int[5]; uses dynamic allocation syntax but is incorrect without pointer.Final Answer:
int arr[5]; -> Option DQuick Check:
Fixed-size arrays use square brackets [OK]
- Using parentheses instead of brackets
- Confusing with vector syntax
- Incorrect dynamic allocation syntax
int arr[4] = {10, 20, 30, 40};
std::cout << arr[2];Solution
Step 1: Identify the array elements and indexes
The array arr has elements: arr[0]=10, arr[1]=20, arr[2]=30, arr[3]=40.Step 2: Access the element at index 2
arr[2] is 30, so std::cout will print 30.Final Answer:
30 -> Option AQuick Check:
arr[2] = 30 [OK]
- Confusing index 2 with 1 or 3
- Expecting output 20 or 40
- Thinking code causes error
int arr[3] = {1, 2, 3};
for(int i = 0; i <= 3; i++) {
std::cout << arr[i] << " ";
}Solution
Step 1: Analyze the loop range and array size
The array has size 3 with valid indexes 0,1,2. The loop runs from i=0 to i=3 inclusive, which means 4 iterations.Step 2: Identify out-of-bounds access
When i=3, arr[3] is accessed, which is outside the array bounds and causes undefined behavior.Final Answer:
Loop runs one time too many, causing out-of-bounds access -> Option BQuick Check:
Loop index must be less than array size [OK]
- Using <= instead of <
- Assuming array size is 4
- Ignoring out-of-bounds errors
arr of size n, which code snippet correctly calculates the sum of all elements?Solution
Step 1: Understand correct loop bounds for array traversal
Array indexes go from 0 to n-1, so loop must run i from 0 to less than n.Step 2: Check each option for correctness
int sum = 0; for(int i = 1; i < n; i++) sum += arr[i]; starts from 1, missing arr[0]. int sum = 0; for(int i = 0; i <= n; i++) sum += arr[i]; uses <= n, causing out-of-bounds access. int sum = 0; for(int i = 0; i < n; i++) sum += arr[i]; correctly sums all elements from 0 to n-1. int sum = arr[0]; for(int i = 1; i < n; i++) sum = arr[i]; overwrites sum instead of adding.Final Answer:
int sum = 0; for(int i = 0; i < n; i++) sum += arr[i]; -> Option AQuick Check:
Loop from 0 to n-1 sums all elements [OK]
- Starting loop at 1
- Using <= n causing errors
- Overwriting sum instead of adding
Ready to prove it?
Pass the test and this pattern moves unknown
