Resilient Target Index — Problem Statement & Solution Guide
Problem Description
The resilient target index is calculated by summing all elements in the array. If the array is empty, return 0. If the array has a single element, return that element. Otherwise, sum all elements in the array.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Resilient Target Index"
WHY DOES IT MATTER?
Summation is a foundational reduction pattern that appears in statistics, finance, and system monitoring. Mastering this pattern equips engineers to recognize and replace costly loops with O(n) linear passes, a skill that directly impacts performance at scale.
OPTIMIZATION CHALLENGE
The key insight is to avoid any form of recursion or nested iteration; a single accumulator updated during a linear scan yields the answer, collapsing both time and space overhead to their theoretical minima.
REAL-WORLD CONNECTION
Think of a distributed logging service that needs to compute total request latency across all servers each minute. Each server reports its local sum, and a central aggregator adds those partial sums—mirroring the map‑reduce paradigm used for this problem.
During an interview, write the accumulator initialization first, then loop over the array. Immediately return the accumulator after the loop—this eliminates the need for special‑case branches and demonstrates clean, defensive coding.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The Resilient Target Index problem is a classic illustration of the reduction principle in algorithm design: a seemingly complex requirement can often be expressed as a simple aggregate operation. By recognizing that the desired output is merely the sum of all elements, we can replace any multi‑step procedural logic with a single linear traversal, leveraging the associative property of addition. This transformation eliminates redundant work such as repeated scanning or conditional branching, which would otherwise inflate the time complexity.
Naïve solutions might attempt to handle each array size case separately—checking for emptiness, single‑element arrays, then iterating for larger inputs—introducing unnecessary conditional overhead and potential bugs. Moreover, a recursive formulation that splits the array into halves without memoization would incur O(2^n) calls, quickly exhausting stack space on large inputs. The optimal paradigm embraces an iterative, tail‑recursive, or divide‑and‑conquer approach that guarantees O(n) time and O(1) auxiliary space, aligning with the constraints of real‑world systems where input sizes can reach millions.
Interview Questions on This Problem
Q1How would you modify the solution if the array could contain extremely large integers that might cause overflow in a 32‑bit language?
Use a wider numeric type (e.g., long long in C++ or BigInteger in Java) for the accumulator, or perform modular arithmetic if a modulus is specified. In languages like Python, integers are arbitrary‑precision, so the built‑in sum works directly.
Q2Can you compute the Resilient Target Index in a single pass without using extra space, and how would you handle an empty array?
Yes. Initialize a variable total = 0, iterate through the array adding each element to total, and after the loop return total. If the array length is 0, the loop body never executes and total remains 0, which satisfies the empty‑array requirement.
Q3Explain how you would parallelize the sum computation for a massive distributed dataset while preserving the exact same result.
Partition the dataset across worker nodes, let each node compute a local sum, then aggregate the partial sums using a reduction (e.g., MPI_Reduce or a map‑reduce combiner). Because addition is associative and commutative, the final result is independent of the order of combination, guaranteeing correctness.
Examples
Input
[9, 2, 5, 8]
Output
24
Explanation: Step-by-step: The array [9, 2, 5, 8] has 4 elements. We sum all elements: 9 + 2 + 5 + 8 = 24.
Input
[4, 8]
Output
12
Explanation: Step-by-step: The array [4, 8] has 2 elements. We sum all elements: 4 + 8 = 12.
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
Iterate once, accumulating the sum in a variable, which yields O(n) time and O(1) auxiliary space.
Brute Force Approach
A naive method would repeatedly scan the array for each element, leading to O(n^2) time, or use recursion without memoization causing exponential calls.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
if (nums.length === 1) return nums[0];
return nums.reduce((a, b) => a + b, 0);
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 0) return 0;
if (nums.size() == 1) return nums[0];
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
if (nums.length == 1) return nums[0];
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
if len(nums) == 0:
return 0
if len(nums) == 1:
return nums[0]
return sum(nums)function solution(nums) {
if (nums.length === 0) return 0;
if (nums.length === 1) return nums[0];
return nums.reduce((a, b) => a + b, 0);
}Asked in Top Tech Interviews
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