Calculated Target Index — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the calculated target index according to the target algorithm rules. The target index is the sum of all elements in the array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Calculated Target Index"
WHY DOES IT MATTER?
Greedy algorithms like this one are essential because they provide the simplest and fastest solution for problems where the optimal choice can be made locally. They reduce complexity, improve performance, and are easy to reason about during interviews.
OPTIMIZATION CHALLENGE
The key insight is that you only need one accumulator variable; no auxiliary data structures or repeated passes are required. This reduces both time to O(n) and space to O(1).
REAL-WORLD CONNECTION
In distributed systems, calculating the total bytes transferred across multiple servers is analogous: each server reports its byte count, and the system aggregates them in a single pass to compute the global metric, just as we sum array elements.
When explaining this to an interviewer, emphasize that the greedy choice is trivial—add each element as you read it—and that the algorithm’s elegance lies in its minimalism and clarity.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem of computing the calculated target index—defined as the sum of all elements in an array—is a classic example of a greedy algorithm. In a greedy approach, we make a locally optimal choice at each step—in this case, adding the current element to a running total—without reconsidering previous choices. Because the sum of a set of numbers is independent of the order in which they are processed, this greedy strategy is both correct and optimal.
Naive approaches that might be considered for beginners often involve nested loops or recursive calls that recompute partial sums repeatedly, leading to O(n^2) time or unnecessary stack overhead. Such methods fail on large inputs due to quadratic time complexity and potential stack overflow. The optimal paradigm is a single linear scan that accumulates the sum in a single variable, achieving O(n) time and O(1) auxiliary space.
This linear-time solution is not only efficient but also demonstrates key algorithmic principles: the greedy choice property, optimal substructure, and the importance of choosing the right data structure (a simple integer accumulator) to minimize both time and space usage. It also highlights how a problem that appears trivial can be a teaching moment for interviewers to probe a candidate’s understanding of algorithmic complexity and code clarity.
Interview Questions on This Problem
Q1What is the most efficient way to compute the sum of an array of integers in JavaScript, and what is its time complexity?
Use a single for-loop or the Array.reduce method to iterate over the array once, accumulating the sum. The time complexity is O(n), where n is the number of elements, and the space complexity is O(1).
Q2How would you modify the algorithm to handle very large integers that might cause overflow in a 32-bit environment?
Use a data type with larger capacity, such as BigInt in JavaScript or long in Java/C++. Alternatively, perform modular arithmetic if only the remainder is needed. This ensures correctness without overflow.
Q3During a coding interview at a fintech startup, the interviewer asks: "Can you explain why a recursive solution for summing an array is less desirable than an iterative one?"
A recursive solution creates a new stack frame for each element, leading to O(n) space usage and potential stack overflow for large arrays. An iterative solution uses a single loop and constant space, making it more robust and efficient for production systems.
Examples
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: Given an array [1, 2, 3, 4, 5], we calculate the sum of all elements: 1 + 2 + 3 + 4 + 5 = 15.
Input
[10, 20, 30, 40, 50]
Output
150
Explanation: Step-by-step: Given an array [10, 20, 30, 40, 50], we calculate the sum of all elements: 10 + 20 + 30 + 40 + 50 = 150.
Constraints
- 1 <= N <= 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity expected: O(N) or O(N log N)
- Space Complexity expected: O(1) or O(N)
Optimal Approach & Strategy
Traverse the array once, adding each element to a single accumulator variable. This yields O(n) time and O(1) space, the optimal solution.
Brute Force Approach
A naive approach might use nested loops or recursion to repeatedly add elements, leading to O(n^2) time or O(n) stack space. This is unnecessary for a simple sum.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
sum = 0
for num in nums:
sum += num
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.