LintCode-Subarray Sum Closest

Given an integer array, find a subarray with sum closest to zero. Return the indexes of the first number and last number.

Example

Given [-3, 1, 1, -3, 5], return [0, 2], [1, 3], [1, 1], [2, 2] or [0, 4]

Challenge

O(nlogn) time

Analysis:

s[i] = nums[0]+....nums[i], also record the index i into s[i]. Sort array s, and the minimum difference between two consecutive element, is the the subarray.

Solution:

class Element implements Comparable<Element>{
int val;
int index;
public Element(int v, int i){
val = v;
index = i;
}

public int compareTo(Element other){
return this.val - other.val;
}

public int getIndex(){
return index;
}

public int getValue(){
return val;
}
}

public class Solution {
/**
* @param nums: A list of integers
* @return: A list of integers includes the index of the first number
*          and the index of the last number
*/
public ArrayList<Integer> subarraySumClosest(int[] nums) {
ArrayList<Integer> res = new ArrayList<Integer>();
if (nums.length==0) return res;

Element[] sums = new Element[nums.length+1];
sums[0] = new Element(0,-1);
int sum = 0;
for (int i=0;i<nums.length;i++){
sum += nums[i];
sums[i+1] = new Element(sum,i);
}

Arrays.sort(sums);
int min = Math.abs(sums[0].getValue() - sums[1].getValue());
int start =  Math.min(sums[0].getIndex(), sums[1].getIndex())+1;
int end = Math.max(sums[0].getIndex(), sums[1].getIndex());
for (int i=1;i<nums.length;i++){
int diff = Math.abs(sums[i].getValue() - sums[i+1].getValue());
if (diff<min){
min = diff;
start = Math.min(sums[i].getIndex(), sums[i+1].getIndex())+1;
end  = Math.max(sums[i].getIndex(), sums[i+1].getIndex());
}
}

res.add(start);
res.add(end);
return res;
}
}