Solution.java
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Solution.java
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	import java.util.Arrays;
	import java.math.BigInteger;

	public class Solution {
	public static String solution(int[] xs) {
	//assume xs never drops below 1 element as stated in the constraint

	int negCount = 0, remove = -1;

	Arrays.sort(xs);

	for(int i = 0; i < xs.length; i++) {
	int x = xs[i];

	if(x < 0) {
	negCount++;
	remove = i; //get last negative value's index to be removed
	}
	}

	if(xs.length > 1 && remove != -1 && negCount % 2 != 0) xs[remove] = 0; //if a remove value is found and it's not the only value, and negative count is not even (multiplying will return negative)

	BigInteger result = BigInteger.ZERO; //initialize big integer since it can get pretty big according to the readme

	for(int x : xs) {
	if(x != 0)
	result = result == BigInteger.ZERO ? BigInteger.valueOf(x) : result.multiply(BigInteger.valueOf(x)); //only multiply non zero values; zero values are almost always discardable
	}

	return result.max(BigInteger.valueOf(xs[xs.length - 1])).toString(); //check if result is smaller than largest value in array (usually for catching 0s that we discarded)
	}
	}