Phase-Alternating Signal Strength — Problem Statement & Solution Guide
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].
Examples
Input
[5, 2, 8, 1, 3, 4, 6, 7]
Output
10
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.
Input
[1, 2, 3, 4, 5, 6, 7, 8]
Output
0
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
The optimal approach uses an alternating prefix sum array where each element signals[i] is added or subtracted depending on whether its index i is even or odd. For any query range [L, R], if L is even, the result is the difference of the prefix sums; if L is odd, the result is the negated difference. This reduces the time complexity of each query to O(1) after O(N) preprocessing.
Brute Force Approach
The brute force approach is to iterate through each query range [L, R] and compute the sum by alternating the sign of each element starting with a plus sign at L. This requires O(R - L) operations per query, resulting in a worst-case time complexity of O(Q * N).
Verified Code Solutions
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 resultfunction 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;
}Solve in Interative Editor
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