Optimal Target Index — Problem Statement & Solution Guide
Problem Description
Given an array or sequence of length N representing numerical values or system metrics, compute the optimal target index according to the target algorithm rules. Formally, analyze the data sequence, process edge cases, and return the exact optimal result.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Target Index"
WHY DOES IT MATTER?
Balance‑point or pivot‑index patterns appear in load‑balancing, financial reconciliation, and partitioning problems where two sides must be equal; mastering this pattern teaches you to convert global constraints into local checks using cumulative aggregates.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that the right‑hand sum can be expressed via the total sum and the already‑known left sum, eliminating the need for a second pass or nested loops.
REAL-WORLD CONNECTION
Think of a distributed cache where you want to place a coordinator node such that the total request volume on its left equals the volume on its right; the optimal target index algorithm directly models that placement decision.
During an interview, compute totalSum first, then iterate once while updating leftSum; this one‑pass pattern is a go‑to trick for any problem that asks for a split point based on sums or counts.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The Optimal Target Index problem asks for an index i in an array such that the sum of all elements strictly to the left of i equals the sum of all elements strictly to the right of i. A naive solution would recompute left and right sums for every candidate index, leading to O(N^2) time, which quickly becomes infeasible for large N (e.g., N > 10^5). The optimal paradigm leverages prefix sums: by maintaining a running total of elements seen so far (leftSum) and knowing the total sum of the array, the right sum can be derived in O(1) as totalSum - leftSum - arr[i]. This transforms the problem into a single linear scan.
The key insight is that the relationship leftSum == totalSum - leftSum - arr[i] can be rearranged to 2*leftSum + arr[i] == totalSum, allowing us to check the condition without extra storage. This approach eliminates repeated aggregation and reduces both time and auxiliary space. Moreover, handling edge cases—such as when the target index is at the boundaries (left or right sum is zero) or when multiple indices satisfy the condition—requires careful definition of the return value (first valid index, any index, or -1 if none). The linear‑time, constant‑space solution is thus the de‑facto optimal algorithm for this class of balance‑point problems.
Interview Questions on This Problem
Q1How would you modify the algorithm if the array could contain negative numbers and you needed to return all valid target indices instead of just one?
The same linear scan works because the equality condition does not depend on sign; you simply collect every index where 2*leftSum + arr[i] equals totalSum. Store results in a list while iterating, and return the list (empty if none).
Q2Can you solve the Optimal Target Index problem in a streaming context where the array elements arrive one by one and you must report the index as soon as it becomes valid?
Maintain running leftSum and totalSum (which can be updated incrementally). After reading each new element, update totalSum, compute the prospective right sum using the formula, and check the condition. If it holds, output the current index; otherwise continue. This yields O(1) per element and O(1) extra space.
Q3What is the time‑space trade‑off if you pre‑compute a prefix‑sum array instead of using a running variable?
Pre‑computing a prefix‑sum array takes O(N) time and O(N) space, after which each index can be checked in O(1) using prefix[i‑1] and total‑prefix[i]. While this simplifies code, it uses linear extra memory, whereas the running‑sum technique achieves the same O(N) time with O(1) space, which is preferable for large inputs.
Examples
Input
[9, 2, 5, 8]
Output
24
Explanation: Step-by-step: with input [9, 2, 5, 8], we first sort the array in ascending order: [2, 5, 8, 9]. Then, we iterate through the array and sum up all the elements: 2 + 5 + 8 + 9 = 24.
Input
[4, 8]
Output
12
Explanation: Step-by-step: with input [4, 8], we first sort the array in ascending order: [4, 8]. Then, we iterate through the array and sum up all the 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
First compute the total sum of the array, then iterate once while maintaining a running left sum; at each index compute right sum as total‑left‑arr[i] and check equality. This runs in O(N) time with O(1) extra space.
Brute Force Approach
For each index, recompute the sum of elements to its left and the sum to its right, then compare them. This requires O(N) work per index, leading to O(N^2) total time.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
nums.sort((a, b) => a - b);
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 0) return 0;
sort(nums.begin(), nums.end());
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
Arrays.sort(nums);
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
if not nums:
return 0
nums.sort()
return sum(nums)function solution(nums) {
if (nums.length === 0) return 0;
nums.sort((a, b) => a - b);
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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