Cumulative Pointer Alignment — Problem Statement & Solution Guide
Problem Description
In a distributed memory system, data blocks are indexed sequentially starting from 1. To optimize cache locality, the system calculates a 'Cumulative Pointer Alignment' metric. This metric is defined as the sum of the product of each data block's value and its 1-based index position.
Given an array of integers representing the data blocks, compute the total Cumulative Pointer Alignment. The calculation follows the formula: Sum_{i=1}^{n} (nums[i-1] * i), where n is the length of the array.
Your task is to implement a function that takes the array of integers and returns the computed alignment score. Ensure your solution handles large input sizes efficiently.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Cumulative Pointer Alignment"
WHY DOES IT MATTER?
Weighted cumulative sums appear in performance metrics, financial calculations (e.g., time‑weighted returns), and scoring systems where later items carry more significance. Mastering this pattern lets engineers turn seemingly quadratic formulas into linear‑time solutions.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that each term’s index is deterministic, so you never need to recompute or store intermediate products; a single running total suffices, collapsing an O(n²) thought process into O(n).
REAL-WORLD CONNECTION
In distributed storage, each block’s access latency grows with its distance from the cache line. Summing blockSize * blockIndex mirrors the total latency, helping engineers decide optimal block placement or re‑balancing strategies.
During an interview, write the recurrence dp[i] = dp[i‑1] + a[i] * i on the whiteboard first; it shows you’re thinking in DP terms and instantly leads to the one‑pass implementation.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The Cumulative Pointer Alignment metric is essentially a weighted sum where each element’s weight equals its 1‑based position. This pattern maps directly to the classic prefix‑sum / cumulative‑sum technique, a cornerstone of dynamic programming. By maintaining a running total while iterating the array, we can compute the metric in a single pass, because each step only depends on the previously computed sum and the current element. Naïve implementations that recompute the product for every possible sub‑array or that use nested loops end up with O(n²) time, which quickly becomes infeasible for large n (10⁶+). The optimal paradigm leverages the linearity of addition and the deterministic index progression, turning the problem into a simple O(n) DP where dp[i] = dp[i‑1] + a[i] * i, with dp[0] = 0.
Interview Questions on This Problem
Q1How would you compute the sum of a[i] * (i+1) for an array of length 10⁷ without causing integer overflow in a language like Java?
Use a 64‑bit type (long) for the accumulator and each multiplication. If the problem constraints allow values up to 10⁹, the maximum possible sum fits within 64‑bit (≈10⁹ * 10⁷ * 10⁷ ≈ 10²³ < 2⁶³). In languages without built‑in big integers, you can take modulo if the problem asks for it.
Q2Explain how the prefix‑sum technique can be adapted to answer multiple queries of the form “sum of a[i] * i for i in [L, R]”.
Pre‑compute two prefix arrays: pref[i] = Σ_{k=1}^{i} a[k] and weightedPref[i] = Σ_{k=1}^{i} a[k] * k. Then the answer for [L,R] is weightedPref[R] - weightedPref[L‑1] - (L‑1) * (pref[R] - pref[L‑1]), which subtracts the extra weight contributed by the shift of indices.
Q3Why is the cumulative‑sum approach considered O(1) extra space, and when would you need O(n) extra space instead?
When you only need the final total, you keep a single accumulator and a loop index, which is O(1) auxiliary space. If you must answer arbitrary range queries later, you must store the full prefix arrays, which costs O(n) space.
Examples
Input
nums = [1, 2, 3, 4]
Output
30
Explanation: Index 1: 1 * 1 = 1 Index 2: 2 * 2 = 4 Index 3: 3 * 3 = 9 Index 4: 4 * 4 = 16 Total Sum: 1 + 4 + 9 + 16 = 30
Input
nums = [5, -2, 0, 7]
Output
23
Explanation: Index 1: 5 * 1 = 5 Index 2: -2 * 2 = -4 Index 3: 0 * 3 = 0 Index 4: 7 * 4 = 28 Total Sum: 5 + (-4) + 0 + 28 = 29
Input
nums = [100, 100, 100]
Output
600
Explanation: Index 1: 100 * 1 = 100 Index 2: 100 * 2 = 200 Index 3: 100 * 3 = 300 Total Sum: 100 + 200 + 300 = 600
Input
nums = [-1, -1, -1, -1]
Output
-10
Explanation: Index 1: -1 * 1 = -1 Index 2: -1 * 2 = -2 Index 3: -1 * 3 = -3 Index 4: -1 * 4 = -4 Total Sum: -1 + (-2) + (-3) + (-4) = -10
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The answer is guaranteed to fit in a 64-bit integer.
Optimal Approach & Strategy
Iterate once, maintaining a running total; at each step add a[i] * (i+1) to the accumulator, achieving O(n) time and O(1) extra space.
Brute Force Approach
Use two nested loops: the outer loop picks each element, the inner loop multiplies it by its index and adds to a total, resulting in O(n²) time.
Verified Code Solutions
function solution(nums) {
let result = 0;
for (let i = 0; i < nums.length; i++) {
result += nums[i] * (i + 1);
}
return result;
}class Solution {
public:
int solution(vector<int>& nums) {
int result = 0;
for (int i = 0; i < nums.size(); i++) {
result += nums[i] * (i + 1);
}
return result;
}
};class Solution {
public int solution(int[] nums) {
int result = 0;
for (int i = 0; i < nums.length; i++) {
result += nums[i] * (i + 1);
}
return result;
}
}def solution(nums):
result = 0
for i in range(len(nums)):
result += nums[i] * (i + 1)
return resultfunction solution(nums) {
let result = 0;
for (let i = 0; i < nums.length; i++) {
result += nums[i] * (i + 1);
}
return result;
}Asked in Top Tech Interviews
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