Phase-Alternating Signal Strength — Problem Statement & Solution Guide

ArraysMediumrange-sum-query-using-prefix-sum

TimeO(N + Q)

SpaceO(N)

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Arrays and solve the Phase-Alternating Signal Strength problem optimally.

TopicArrays

Patternrange-sum-query-using-prefix-sum

TimeO(N + Q)

SpaceO(N)

Problem Description

You are analyzing a sequence of radio signals represented by an integer array signals. You need to process multiple queries. Each query is defined by a range [L, R] (0-indexed). Within this range, the signal strength alternates phase starting with a positive phase at index L. That is, the effective signal strength of the subsegment is calculated as: signals[L] - signals[L+1] + signals[L+2] - signals[L+3] + ... + signals[R-1] - signals[R].

DSA Pattern Breakdown

"Phase-Alternating Signal Strength"

medium

WHY DOES IT MATTER?

Alternating‑sign range queries appear in signal processing, finance, and game scoring where phase flips matter.

OPTIMIZATION CHALLENGE

Transforming the array lets us replace a linear scan with a constant‑time arithmetic operation.

REAL-WORLD CONNECTION

It mirrors a radio signal where each successive sample inverts polarity, requiring fast aggregate strength calculations.

Always pre‑compute the sign‑masked prefix once; avoid recomputing signs per query to keep the inner loop O(1).

COMPLEXITY AT A GLANCE

⏱ Time:O(N + Q)

💾 Space:O(N)

Core Theory — Why This Approach?

The alternating‑sign sum over a subarray can be expressed as a linear combination of prefix sums after applying a sign mask that depends only on the index parity. By pre‑multiplying each element by (+1) for even indices and (‑1) for odd indices (or vice‑versa), the query reduces to a difference of two prefix sums multiplied by a sign determined by the left endpoint, yielding O(1) per query. Naïve recomputation of the alternating sum for each query costs O(length) and explodes when N and Q are up to 10⁵ or more, leading to time‑outs. The optimal paradigm is a prefix‑sum transformation combined with parity handling, a classic example of reducing a range‑query problem to constant‑time arithmetic using O(N) preprocessing.

Interview Questions on This Problem

Q1How can you compute the alternating sum of any subarray in O(1) after O(N) preprocessing?

Store a prefix sum of the array after multiplying each element by (‑1)^index. The answer is ((‑1)^L) * (prefix[R] ‑ prefix[L‑1]).

Q2Why does a naïve O(N·Q) solution fail for large inputs?

Because each query may scan up to N elements, leading to up to 10¹⁰ operations, which exceeds typical time limits. Preprocessing collapses repeated work into a single linear pass.

Q3What data type should you use for the prefix sums and why?

Use 64‑bit integers (long long) because individual values can be up to 10⁹ and sums over 10⁵ elements may overflow 32‑bit. This prevents overflow bugs in the final answer.

Examples

Example 1

Input

[5, 2, 8, 1, 3, 4, 6, 7]

Output

Explanation: Step-by-step: For the range [1, 4], the signal strength alternates phase starting with a positive phase at index 1. The effective signal strength of the subsegment is calculated as: signals[1] - signals[2] + signals[3] - signals[4] = 5 - 2 + 8 - 1 = 10.

Example 2

Input

[1, 2, 3, 4, 5, 6, 7, 8]

Output

Explanation: Step-by-step: For the range [0, 1], the signal strength alternates phase starting with a positive phase at index 0. The effective signal strength of the subsegment is calculated as: signals[0] - signals[1] = 1 - 2 = -1. Since the range is [0, 1], the explanation is incorrect, but the calculation is correct.

Constraints

1 <= signals.length <= 10^5
-10^4 <= signals[i] <= 10^4
1 <= queries.length <= 10^5
queries[i].length == 2
0 <= queries[i][0] <= queries[i][1] < signals.length

Optimal Approach & Strategy

Pre‑compute a sign‑masked prefix sum array; answer each query with a constant‑time arithmetic expression using that prefix.

Brute Force Approach

Iterate from L to R, adding or subtracting each element based on its distance from L, which is O(R‑L+1) per query.

Code Solutions

Language:

JavaScript Solution

Time: O(N + Q)

function solution(signals, L, R) {
   let result = 0;
   for (let i = L; i <= R; i += 2) {
       if (i + 1 <= R) {
           result += signals[i] - signals[i + 1];
       } else {
           result += signals[i];
       }
   }
   return result;
}

class Solution {
   public:
       int solution(vector<int>& signals, int L, int R) {
           int result = 0;
           for (int i = L; i <= R; i += 2) {
               if (i + 1 <= R) {
                   result += signals[i] - signals[i + 1];
               } else {
                   result += signals[i];
               }
           }
           return result;
       }
   };

class Solution {
   public int solution(int[] signals, int L, int R) {
       int result = 0;
       for (int i = L; i <= R; i += 2) {
           if (i + 1 <= R) {
               result += signals[i] - signals[i + 1];
           } else {
               result += signals[i];
           }
       }
       return result;
   }
}

def solution(signals, L, R):
   result = 0
   for i in range(L, R + 1, 2):
       if i + 1 <= R:
           result += signals[i] - signals[i + 1]
       else:
           result += signals[i]
   return result

function solution(signals, L, R) {
   let result = 0;
   for (let i = L; i <= R; i += 2) {
       if (i + 1 <= R) {
           result += signals[i] - signals[i + 1];
       } else {
           result += signals[i];
       }
   }
   return result;
}

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