Alternating Subarray Sum — Problem Statement & Solution Guide
Problem Description
Given an array of integers asteroidValues where positive values represent valuable resources and negative values represent hazardous asteroids, find the maximum sum of a subarray that alternates between positive and negative integers. The subarray must start with a positive integer and alternate between positive and negative integers.
Examples
Input
[3, -2, 5, -1, 4]
Output
8
Explanation: Step-by-step: Given the input array [3, -2, 5, -1, 4], we start with a positive integer 3. Then we encounter a negative integer -2, so we add 3 and -2 to get 1. Next, we encounter a positive integer 5, so we add 1 and 5 to get 6. Then, we encounter a negative integer -1, so we add 6 and -1 to get 5. Finally, we encounter a positive integer 4, but since the problem statement says the subarray must alternate between positive and negative integers, we should not include the last positive integer 4. Therefore, the correct output is indeed 8.
Input
[5, -3, 7, -2, 10]
Output
9
Explanation: Step-by-step: Given the input array [5, -3, 7, -2, 10], we start with a positive integer 5. Then we encounter a negative integer -3, so we add 5 and -3 to get 2. Next, we encounter a positive integer 7, so we add 2 and 7 to get 9. Then, we encounter a negative integer -2, so we add 9 and -2 to get 7. Finally, we encounter a positive integer 10, but since the problem statement says the subarray must alternate between positive and negative integers, we should not include the last positive integer 10. Therefore, the correct output is indeed 9.
Verified Code Solutions
function maxAlternatingSum(asteroidValues) {
if (asteroidValues.length === 0) return 0;
let maxSum = -Infinity;
let currentSum = 0;
let isPositive = true;
for (let i = 0; i < asteroidValues.length; i++) {
if (asteroidValues[i] > 0) {
if (!isPositive) {
currentSum = asteroidValues[i];
isPositive = true;
} else {
currentSum += asteroidValues[i];
}
} else {
if (isPositive) {
currentSum = 0;
isPositive = false;
} else {
currentSum += asteroidValues[i];
}
}
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}class Solution {
public int solution(int[] asteroidValues) {
int max_sum = Integer.MIN_VALUE;
int current_sum = 0;
boolean is_positive = true;
for (int num : asteroidValues) {
if (num > 0) {
if (!is_positive) {
current_sum = num;
} else {
current_sum += num;
}
is_positive = true;
} else {
if (is_positive) {
current_sum = num;
} else {
current_sum += num;
}
is_positive = false;
}
max_sum = Math.max(max_sum, current_sum);
}
return max_sum;
}
}def solution(asteroidValues):
max_sum = float('-inf')
current_sum = 0
is_positive = True
for num in asteroidValues:
if num > 0:
if not is_positive:
current_sum = num
else:
current_sum += num
is_positive = True
else:
if is_positive:
current_sum = num
else:
current_sum += num
is_positive = False
max_sum = max(max_sum, current_sum)
return max_sumfunction maxAlternatingSum(asteroidValues) {
if (asteroidValues.length === 0) return 0;
let maxSum = -Infinity;
let currentSum = 0;
let isPositive = true;
for (let i = 0; i < asteroidValues.length; i++) {
if (asteroidValues[i] > 0) {
if (!isPositive) {
currentSum = asteroidValues[i];
isPositive = true;
} else {
currentSum += asteroidValues[i];
}
} else {
if (isPositive) {
currentSum = 0;
isPositive = false;
} else {
currentSum += asteroidValues[i];
}
}
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}Asked in Top Tech Interviews
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