703045. Sum of alternate product
Basic
Sum of Alternate Product (Minimize sum of alternate product)
Summary
Given an array of integers, find the minimum sum of products that can be obtained by selecting every other element from the array and multiplying them together. The array is of size n. This problem involves arrays and basic algorithms, making it a fundamental exercise in computer science.
Detailed Explanation
To solve this problem, we'll follow these steps:
- Initialize two variables:
min_sumto store the minimum sum of products, andcurrent_productto keep track of the product of selected elements. - Iterate over the array from left to right, selecting every other element (i.e.,
0th,2nd,4th, ... indices). - For each selected element, multiply it with
current_product. - Update
min_sumif the current product is smaller. - Return
min_sum.
Here's a step-by-step breakdown of the algorithm:
min_sum = infinity
current_product = 1
for i from 0 to n-1 (step 2):
if i % 2 == 0:
current_product *= arr\[i]
min_sum = min(min_sum, current_product)
return min_sum
The time complexity of this algorithm is O(n), as we need to iterate over the array once. The space complexity is O(1), as we only use a constant amount of space to store min_sum and current_product.
Optimized Solutions
Java
public int minSumOfAlternateProduct(int[] arr) {
int n = arr.length;
long minSum = Long.MAX_VALUE;
long currentProduct = 1;
for (int i = 0; i < n; i += 2) {
if (i % 2 == 0) {
currentProduct *= arr[i];
minSum = Math.min(minSum, currentProduct);
}
}
return (int) minSum;
}